problem solving on second law of motion

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Newton’s second law of motion – problems and solutions

Solved problems in Newton’s laws of motion – Newton’s second law of motion 

1. A 1 kg object accelerated at a constant 5 m/s 2 . Estimate the net force needed to accelerate the object.

Mass (m) = 1 kg

Acceleration (a) = 5 m/s 2

Wanted : net force (∑F)

We use Newton’s second law to get the net force.

∑ F = (1 kg)(5 m/s 2 ) = 5 kg m/s 2 = 5 Newton

2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….

Net force (∑F) = 2 Newton

Wanted : The magnitude and direction of the acceleration (a)

a = 2 m/s 2

The direction of the acceleration = the direction of the net force (∑F)

3. Object’s mass = 2 kg, F 1 = 5 Newton, F 2 = 3 Newton. The magnitude and direction of the acceleration is…

Mass (m) = 2 kg

F 1 = 5 Newton

F 2 = 3 Newton

net force :

∑ F = F 1 – F 2 = 5 – 3 = 2 Newton

The magnitude of the acceleration :

a = 1 m/s 2

Direction of the acceleration = direction of the net force = direction of F 1

4. Object’s mass = 2 kg, F 1 = 10 Newton, F 2 = 1 Newton. The magnitude and direction of the acceleration is…

F 2 = 1 Newton

F 1 = 10 Newton

F 1x = F 1 cos 60 o = (10)(0.5) = 5 Newton

Net force :

∑ F = F 1x – F 2 = 5 – 1 = 4 Newton

Direction of the acceleration = direction of the net force = direction of F 1x

5. F 1 = 10 Newton, F 2 = 1 Newton, m 1 = 1 kg, m 2 = 2 kg. The magnitude and direction of the acceleration is…

Mass 1 (m 1 ) = 1 kg

Mass 2 (m 2 ) = 2 kg

The net force :

∑ F = F 1 – F 2 = 10 – 1 = 9 Newton

a = ∑F / (m 1 + m 2 )

a = 9 / (1 + 2)

a = 3 m/s 2

The direction of the acceleration = the direction of the net force = direction of F 1

A 40-kg block accelerated by a force of 200 N. Acceleration of the block is 3 m/ s 2 . Determine the magnitude of friction force experienced by the block.

Mass (m) = 40 kg

Force (F) = 200 N

Acceleration (a) = 3 m/s 2

Wanted: Friction force (F g )

The equation of Newton’s second law of motion

∑ F = net force, m = mass, a = acceleration

The direction of force F rightward, the direction of friction force leftward (the direction of friction force is opposite with the direction of object’s motion).

Choose rightward as positive and leftward as negative.

F – F g = m a

200 – F g = (40)(3)

200 – F g = 120

F g = 200 – 120

F g = 80 Newton

The correct answer is D.

7. Block A with a mass of 100-gram place above block B with a mass of 300 gram, and then block b pushed with a force of 5 N vertically upward. Determine the normal force exerted by block B on block A.

Force (F) = 5 Newton

Mass of block A (m A ) = 100 gram = 0.1 kg

Mass of block B (m B ) = 300 gram = 0.3 kg

Acceleration of gravity (g) = 10 m/s 2

Weight of block A (w A ) = (0.1 kg)(10 m/s 2 ) = 1 kg m/s 2 = 1 Newton

Weight of block B (w B ) = (0.3 kg)(10 m/s 2 ) = 3 kg m/s 2 = 3 Newton

Wanted : Normal force exerted by block B to block A

F = push force (act on block B)

w A = weight of block A (act on block A)

w B = weight of block B (act on block B)

N A = normal force exerted by block B on block A (Act on block A)

N A ’ = normal force exerted by block A on block B (Act on block B)

Apply Newton’s second law of motion on both blocks :

F – w A – w B + N A – N A ’ = (m A + m B ) a

N A and N A ’ are action-reaction forces that have the same magnitude but opposite in direction so eliminated from the equation.

F – w A – w B = (m A + m B ) a

5 – 1 – 3 = (0.1 + 0.3) a

5 – 4 = (0.4) a

1 = (0.4) a

a = 1 / 0.4

a = 2.5 m/s 2

Apply Newton’s second law of motion on block A :

N A – w A = m A a

N A – 1 = (0.1)(2.5)

N A – 1 = 0.25

N A = 1 + 0.25

N A = 1.25 Newton

The correct answer is B.

8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.

B. 4 N downward

C. 9 N upward

D. 9 N downward

Weight of X (w X ) = 4 Newton

Pull force (F x ) = 2 Newton

Tension force (F T ) = 9 Newton

Wanted: Net force acts on object X

Vertically upward forces that act on object X :

The tension force has the same magnitude in all part of the cord. So the tension force is 9 N.

Vertically downward forces that act on object X :

There are two forces that act on object X and both forces are vertically downward, the horizontal component of weight w x and the horizontal component of force F x .

Net force act on the object X :

F T – w X – F x = 9 – 4 – 2 = 9 – 6 = 3

The net force act on the object X is 3 Newton, vertically upward.

The correct answer is A.

9. An object initially at rest on a smooth horizontal surface. A force of 16 N acts on the object so the object accelerated at 2 m/s 2 . If the same object at rest on a rough horizontal surface so the friction force acts on the object is 2 N, then determine the acceleration of the object if the same force of 16 N acts on the object.

A. 1.75 m/s 2

B. 1.50 m/s 2

C. 1.00 m/s 2

D. 0.88 m/s 2

Force (F) = 16 Newton = 16 kg m/s 2

Acceleration (a) = 2 m/s 2

Friction force (F fric ) = 2 Newton = 2 kg m/s 2

Wanted : Object’s acceleration ?

Smooth horizontal surface (no friction force) :

Mass of object is 8 kilogram.

Rough horizontal surface (there is a friction force) :

F – F fric = m a

16 – 2 = 8 a

a = 1.75 m/s 2

Object’s acceleration is 1.75 m/s 2 .

10. Tom and Andrew push an object on the smooth floor. Tom push the object with a force of 5.70 N. If the mass of the object is 2.00 kg and acceleration experienced by the object is 2.00 ms -2 , then determine the magnitude and direction of force act by Tom.

A. 1.70 N and its direction is opposite with force acted by Andre.w

B. 1.70 N and its direction same as force acted by Andrew

C. 2.30 N and its direction is opposite with force acted by Andrew.

D. 2.30 N and its direction same as force acted by Andrew.

Push force acted by Andrew (F 1 ) = 5.70 Newton

Mass of object (m) = 2.00 kg

Acceleration (a) = 2.00 m/s 2

Wanted : Magnitude and direction of force acted by Tom (F 2 ) ?

Apply Newton’s second law of motion :

F 1 + F 2 = m a

5.70 + F 2 = (2)(2)

5.70 + F 2 = 4

F 2 = 4 – 5.70

F 2 = – 1.7 Newton

Minus sign indicated that (F 2 ) is opposite with push force act by Andrew (F 1 ).

11. If the mass of the block is the same, which figure shows the smallest acceleration?

Net force A :

ΣF = 4 N + 2 N – 3 N = 6 N – 3 N = 3 Newton, leftward

Net force B :

ΣF = 2 N + 3 N – 4 N = 5 N – 4 N = 1 Newton, rightward

Net force C :

ΣF = 4 N + 3 N – 2 N = 7 N – 2 N = 5 Newton, rightward

Net force D :

ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward

The equation of Newton’s second law :

a = acceleration, ΣF = net force, m = mass

Based on the above formula, the acceleration (a) is directly proportional to the net force (ΣF) and inversely proportional to mass (m). If the mass of an object is the same, the greater the resultant force, the greater the acceleration or the smaller the resultant force, the smaller the acceleration. Based on the above calculation, the smallest net force is 1 Newton so the acceleration is also smallest.

12. Some forces act on an object with a mass of 20 kg, as shown in the figure below.

Determine the object’s acceleration.

Mass of object (m) = 20 kg

Net force (ΣF) = 25 N + 30 N – 15 N = 40 N

Wanted: Acceleration of an object

Object’s acceleration calculated using the equation of Newton’s second law :

a = ΣF / m = 40 N / 20 kg = 2 N/kg = 2 m/s 2

13. Which statements below describes Newton’s third law?

(1) Passengers pushed forward when the bus braked suddenly

(2) B ooks on paper are not falling when the paper is pulled quickly

(3) When playing skateboard when the foot pushes the ground back then the skateboard will slide forward

(4) O ars pushed backward, boats moving forward

(1) Newton’s first law

(2) Newton’s first law

(3) Newton’s third law

(4) Newton’s third law

[wpdm_package id=’470′]

• Mass and weight
• Normal force
• Newton’s second law of motion
• Friction force
• Motion on the horizontal surface without friction force
• The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
• Motion on the inclined plane without friction force
• Motion on the rough inclined plane with the friction force
• Motion in an elevator
• The motion of bodies connected by cord and pulley
• Two bodies with the same magnitude of accelerations
• Rounding a flat curve – dynamics of circular motion
• Rounding a banked curve – dynamics of circular motion
• Uniform motion in a horizontal circle
• Centripetal force in uniform circular motion

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6 Applications of Newton’s Laws

6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of the section, you will be able to:

• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy: Applying Newton’s Laws of Motion

• Identify the physical principles involved by listing the givens and the quantities to be calculated.
• Sketch the situation, using arrows to represent all forces.
• Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
• Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
• Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure (a). Then, as in Figure (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set [latex]{a}_{y}=0.[/latex]) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Different Tensions at Different Angles

Consider the traffic light (mass of 15.0 kg) suspended from two wires as shown in Figure . Find the tension in each wire, neglecting the masses of the wires.

The system of interest is the traffic light, and its free-body diagram is shown in Figure (c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in Figure (d). There are two unknowns in this problem ([latex]{T}_{1}[/latex] and [latex]{T}_{2}[/latex]), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero.

First consider the horizontal or x -axis:

Thus, as you might expect,

This gives us the following relationship:

Note that [latex]{T}_{1}[/latex] and [latex]{T}_{2}[/latex] are not equal in this case because the angles on either side are not equal. It is reasonable that [latex]{T}_{2}[/latex] ends up being greater than [latex]{T}_{1}[/latex] because it is exerted more vertically than [latex]{T}_{1}.[/latex]

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for [latex]{T}_{2}[/latex] in terms of [latex]{T}_{1}[/latex] reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of [latex]{T}_{1}[/latex] to be

Finally, we find the magnitude of [latex]{T}_{2}[/latex] by using the relationship between them, [latex]{T}_{2}=1.225{T}_{1}[/latex], found above. Thus we obtain

Significance

Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker in Newton’s Laws of Motion .

Particle Acceleration

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Drag Force on a Barge

Two tugboats push on a barge at different angles ( Figure ). The first tugboat exerts a force of [latex]2.7\times {10}^{5}\,\text{N}[/latex] in the x -direction, and the second tugboat exerts a force of [latex]3.6\times {10}^{5}\,\text{N}[/latex] in the y -direction. The mass of the barge is [latex]5.0\times {10}^{6}\,\text{kg}[/latex] and its acceleration is observed to be [latex]7.5\times {10}^{-2}\,{\text{m/s}}^{2}[/latex] in the direction shown. What is the drag force of the water on the barge resisting the motion? ( Note: Drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object. Since the barge is flat bottomed, we can assume that the drag force is in the direction opposite of motion of the barge.)

The directions and magnitudes of acceleration and the applied forces are given in Figure (a). We define the total force of the tugboats on the barge as [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] so that

The drag of the water [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] is in the direction opposite to the direction of motion of the boat; this force thus works against [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] as shown in the free-body diagram in Figure (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x – and y -axes are in the same direction as [latex]{\mathbf{\overset{\to }{F}}}_{1}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{2}.[/latex] The problem quickly becomes a one-dimensional problem along the direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex], since friction is in the direction opposite to [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}.[/latex] Our strategy is to find the magnitude and direction of the net applied force [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] and then apply Newton’s second law to solve for the drag force [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}.[/latex]

Since [latex]{F}_{x}[/latex] and [latex]{F}_{y}[/latex] are perpendicular, we can find the magnitude and direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}[/latex] directly. First, the resultant magnitude is given by the Pythagorean theorem:

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] is in the opposite direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] but its magnitude is slightly less than [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}}.[/latex] The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water [latex]{F}_{\text{D}}[/latex] in terms of known quantities:

Substituting known values gives

The direction of [latex]{\mathbf{\overset{\to }{F}}}_{\text{D}}[/latex] has already been determined to be in the direction opposite to [latex]{\mathbf{\overset{\to }{F}}}_{\text{app}},[/latex] or at an angle of [latex]53^\circ[/latex] south of west.

The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where [latex]{F}_{\text{D}}[/latex] is less than 1/600th of the weight of the ship.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

What Does the Bathroom Scale Read in an Elevator?

Figure shows a 75.0-kg man (weight of about 165 lb.) standing on a bathroom scale in an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of [latex]1.20\,{\text{m/s}}^{2},[/latex] and (b) if the elevator moves upward at a constant speed of 1 m/s.

If the scale at rest is accurate, its reading equals [latex]{\mathbf{\overset{\to }{F}}}_{\text{p}}[/latex], the magnitude of the force the person exerts downward on it. Figure (a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn, as in Figure (b). Analysis of the free-body diagram using Newton’s laws can produce answers to both Figure (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight [latex]\mathbf{\overset{\to }{w}}[/latex] and the upward force of the scale [latex]{\mathbf{\overset{\to }{F}}}_{\text{s}}.[/latex] According to Newton’s third law, [latex]{\mathbf{\overset{\to }{F}}}_{\text{p}}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{\text{s}}[/latex] are equal in magnitude and opposite in direction, so that we need to find [latex]{F}_{\text{s}}[/latex] in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law,

From the free-body diagram, we see that [latex]{\mathbf{\overset{\to }{F}}}_{\text{net}}={\mathbf{\overset{\to }{F}}}_{s}-\mathbf{\overset{\to }{w}},[/latex] so we have

Solving for [latex]{F}_{s}[/latex] gives us an equation with only one unknown:

or, because [latex]w=mg,[/latex] simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes [latex]{F}_{s}-w=\text{−}ma.[/latex])

which gives

The scale reading in Figure (a) is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight:

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding

Now calculate the scale reading when the elevator accelerates downward at a rate of [latex]1.20\,{\text{m/s}}^{2}.[/latex]

[latex]{F}_{\text{s}}=645\,\text{N}[/latex]

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Two Attached Blocks

Figure shows a block of mass [latex]{m}_{1}[/latex] on a frictionless, horizontal surface. It is pulled by a light string that passes over a frictionless and massless pulley. The other end of the string is connected to a block of mass [latex]{m}_{2}.[/latex] Find the acceleration of the blocks and the tension in the string in terms of [latex]{m}_{1},{m}_{2},\,\text{and}\,g.[/latex]

We draw a free-body diagram for each mass separately, as shown in Figure . Then we analyze each one to find the required unknowns. The forces on block 1 are the gravitational force, the contact force of the surface, and the tension in the string. Block 2 is subjected to the gravitational force and the string tension. Newton’s second law applies to each, so we write two vector equations:

For block 1: [latex]\mathbf{\overset{\to }{T}}+{\mathbf{\overset{\to }{w}}}_{1}+\mathbf{\overset{\to }{N}}={m}_{1}{\mathbf{\overset{\to }{a}}}_{1}[/latex]

For block 2: [latex]\mathbf{\overset{\to }{T}}+{\mathbf{\overset{\to }{w}}}_{2}={m}_{2}{\mathbf{\overset{\to }{a}}}_{2}.[/latex]

Notice that [latex]\mathbf{\overset{\to }{T}}[/latex] is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

The component equations follow from the vector equations above. We see that block 1 has the vertical forces balanced, so we ignore them and write an equation relating the x -components. There are no horizontal forces on block 2, so only the y -equation is written. We obtain these results:

When block 1 moves to the right, block 2 travels an equal distance downward; thus, [latex]{a}_{1x}=\text{−}{a}_{2y}.[/latex] Writing the common acceleration of the blocks as [latex]a={a}_{1x}=\text{−}{a}_{2y},[/latex] we now have

From these two equations, we can express a and T in terms of the masses [latex]{m}_{1}\,\text{and}\,{m}_{2},\,\text{and}\,g:[/latex]

Notice that the tension in the string is less than the weight of the block hanging from the end of it. A common error in problems like this is to set [latex]T={m}_{2}g.[/latex] You can see from the free-body diagram of block 2 that cannot be correct if the block is accelerating.

Calculate the acceleration of the system, and the tension in the string, when the masses are [latex]{m}_{1}=5.00\,\text{kg}[/latex] and [latex]{m}_{2}=3.00\,\text{kg}.[/latex]

[latex]a=3.68\,{\text{m/s}}^{2},[/latex] [latex]T=18.4\,\text{N}[/latex]

Atwood Machine

A classic problem in physics, similar to the one we just solved, is that of the Atwood machine , which consists of a rope running over a pulley, with two objects of different mass attached. It is particularly useful in understanding the connection between force and motion. In Figure , [latex]{m}_{1}=2.00\,\text{kg}[/latex] and [latex]{m}_{2}=4.00\,\text{kg}\text{.}[/latex] Consider the pulley to be frictionless. (a) If [latex]{m}_{2}[/latex] is released, what will its acceleration be? (b) What is the tension in the string?

We draw a free-body diagram for each mass separately, as shown in the figure. Then we analyze each diagram to find the required unknowns. This may involve the solution of simultaneous equations. It is also important to note the similarity with the previous example. As block 2 accelerates with acceleration [latex]{a}_{2}[/latex] in the downward direction, block 1 accelerates upward with acceleration [latex]{a}_{1}[/latex]. Thus, [latex]a={a}_{1}=\text{−}{a}_{2}.[/latex]

(The negative sign in front of [latex]{m}_{2}a[/latex] indicates that [latex]{m}_{2}[/latex] accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is

Solving for a :

• Observing the first block, we see that [latex]\begin{array}{c}T-{m}_{1}g={m}_{1}a\hfill \\ T={m}_{1}(g+a)=(2\,\text{kg})(9.8\,{\text{m/s}}^{2}+3.27\,{\text{m/s}}^{2})=26.1\,\text{N}\text{.}\hfill \end{array}[/latex]

The result for the acceleration given in the solution can be interpreted as the ratio of the unbalanced force on the system, [latex]({m}_{2}-{m}_{1})g[/latex], to the total mass of the system, [latex]{m}_{1}+{m}_{2}[/latex]. We can also use the Atwood machine to measure local gravitational field strength.

Determine a general formula in terms of [latex]{m}_{1},{m}_{2}[/latex] and g for calculating the tension in the string for the Atwood machine shown above.

[latex]T=\frac{2{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}g[/latex] (This is found by substituting the equation for acceleration in Figure (a), into the equation for tension in Figure (b).)

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

Substituting the known values yields

Substituting the known values of m and a gives

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

What Force Acts on a Model Helicopter?

A 1.50-kg model helicopter has a velocity of [latex]5.00\mathbf{\hat{j}}\,\text{m/s}[/latex] at [latex]t=0.[/latex] It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of [latex](6.00\mathbf{\hat{i}}+12.00\mathbf{\hat{j}})\text{m/s}\text{.}[/latex] What is the magnitude of the resultant force acting on the helicopter during this time interval?

We can easily set up a coordinate system in which the x -axis [latex](\mathbf{\hat{i}}[/latex] direction) is horizontal, and the y -axis [latex](\mathbf{\hat{j}}[/latex] direction) is vertical. We know that [latex]\Delta t=2.00s[/latex] and [latex](6.00\mathbf{\hat{i}}+12.00\mathbf{\hat{j}}\,\text{m/s})-(5.00\mathbf{\hat{j}}\,\text{m/s}).[/latex] From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

The magnitude of the force is now easily found:

The original problem was stated in terms of [latex]\mathbf{\hat{i}}-\mathbf{\hat{j}}[/latex] vector components, so we used vector methods. Compare this example with the previous example.

Find the direction of the resultant for the 1.50-kg model helicopter.

49.4 degrees

Baggage Tractor

Figure (a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by [latex]F=(820.0t)\,\text{N,}[/latex] find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force [latex]\mathbf{\overset{\to }{T}},[/latex] which is our objective.

Since acceleration is a function of time, we can determine the velocity of the tractor by using [latex]a=\frac{dv}{dt}[/latex] with the initial condition that [latex]{v}_{0}=0[/latex] at [latex]t=0.[/latex] We integrate from [latex]t=0[/latex] to [latex]t=3\text{:}[/latex]

• Refer to the free-body diagram in Figure (b). [latex]\begin{array}{ccc}\hfill \sum {F}_{x}& =\hfill & {m}_{\text{tractor}}{a}_{x}\hfill \\ \hfill 820.0t-T& =\hfill & {m}_{\text{tractor}}(0.7805)t\hfill \\ \hfill (820.0)(3.00)-T& =\hfill & (650.0)(0.7805)(3.00)\hfill \\ \hfill T& =\hfill & 938\,\text{N}.\hfill \end{array}[/latex]

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving Figure (a), whereas only the mass of the truck (since it supplied the force) was of use in Figure (b).

Recall that [latex]v=\frac{ds}{dt}[/latex] and [latex]a=\frac{dv}{dt}[/latex]. If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have [latex]dt=\frac{ds}{v}[/latex] and [latex]dt=\frac{dv}{a}.[/latex] Now, equating these expressions, we have [latex]\frac{ds}{v}=\frac{dv}{a}.[/latex] We can rearrange this to obtain [latex]{a}^{}ds={v}^{}dv.[/latex]

Motion of a Projectile Fired Vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure ). Determine the maximum height it will travel if atmospheric resistance is measured as [latex]{F}_{\text{D}}=(0.0100{v}^{2})\,\text{N,}[/latex] where v is the speed at any instant.

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Initially, [latex]{y}_{0}=0[/latex] and [latex]{v}_{0}=50.0\,\text{m/s}\text{.}[/latex] At the maximum height [latex]y=h,v=0.[/latex] The free-body diagram shows [latex]{F}_{\text{D}}[/latex] to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

The acceleration depends on v and is therefore variable. Since [latex]a=f(v)\text{,}[/latex] we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, [latex]h=114\,\text{m}\text{.}[/latex]

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t ), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

• Newton’s laws of motion can be applied in numerous situations to solve motion problems.
• Some problems contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether [latex]{F}_{\text{net}}=ma[/latex] or [latex]{F}_{\text{net}}=0.[/latex]
• The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating vertically, the normal force is less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force is always less than the full weight of the object.
• Some problems contain several physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics to solve these problems.

Conceptual Questions

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at g . Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

The scale is in free fall along with the astronauts, so the reading on the scale would be 0. There is no difference in the apparent weightlessness; in the aircraft and in orbit, free fall is occurring.

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force [latex]\mathbf{\overset{\to }{F}}[/latex] so that the swing ropes are [latex]30.0^\circ[/latex] with respect to the vertical. (a) Calculate the tension in each of the two ropes supporting the swing under these conditions. (b) Calculate the magnitude of [latex]\mathbf{\overset{\to }{F}}.[/latex]

a. 170 N; b. 170 N

Find the tension in each of the three cables supporting the traffic light if it weighs 2.00 × 10 2 N.

Three forces act on an object, considered to be a particle, which moves with constant velocity [latex]v=(3\mathbf{\hat{i}}-2\mathbf{\hat{j}})\,\text{m/s}\text{.}[/latex] Two of the forces are [latex]{\mathbf{\overset{\to }{F}}}_{1}=(3\mathbf{\hat{i}}+5\mathbf{\hat{j}}-6\mathbf{\hat{k}})\,\text{N}[/latex] and [latex]{\mathbf{\overset{\to }{F}}}_{2}=(4\mathbf{\hat{i}}-7\mathbf{\hat{j}}+2\mathbf{\hat{k}})\,\text{N}\text{.}[/latex] Find the third force.

[latex]{\mathbf{\overset{\to }{F}}}_{3}=(-7\mathbf{\hat{i}}+2\mathbf{\hat{j}}+4\mathbf{\hat{k}})\,\text{N}[/latex]

A flea jumps by exerting a force of [latex]1.20\times {10}^{-5}\,\text{N}[/latex] straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of [latex]0.500\times {10}^{-6}\,\text{N}[/latex] on the flea while the flea is still in contact with the ground. Find the direction and magnitude of the acceleration of the flea if its mass is [latex]6.00\times {10}^{-7}\,\text{kg}[/latex]. Do not neglect the gravitational force.

Two muscles in the back of the leg pull upward on the Achilles tendon, as shown below. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force?

After a mishap, a 76.0-kg circus performer clings to a trapeze, which is being pulled to the side by another circus artist, as shown here. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution.

A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow the first dolphin if it was moving horizontally? (The gravitational force is balanced by the buoyant force of the water.)

When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel?

A large rocket has a mass of [latex]2.00\times {10}^{6}\,\text{kg}[/latex] at takeoff, and its engines produce a thrust of [latex]3.50\times {10}^{7}\,\text{N}.[/latex] (a) Find its initial acceleration if it takes off vertically. (b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust?

a. [latex]7.70\,{\text{m/s}}^{2}[/latex]; b. 4.33 s

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110.0 kg.

A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110.0 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell.

a. 46.4 m/s; b. [latex]2.40\times {10}^{3}\,{\text{m/s}}^{2}\text{;}[/latex] c. 5.99 × 10 3 N; ratio of 245

A 0.500-kg potato is fired at an angle of [latex]80.0^\circ[/latex] above the horizontal from a PVC pipe used as a “potato gun” and reaches a height of 110.0 m. (a) Neglecting air resistance, calculate the potato’s velocity when it leaves the gun. (b) The gun itself is a tube 0.450 m long. Calculate the average acceleration of the potato in the tube as it goes from zero to the velocity found in (a). (c) What is the average force on the potato in the gun? Express your answer in newtons and as a ratio to the weight of the potato.

An elevator filled with passengers has a mass of [latex]1.70\times {10}^{3}\,\text{kg}[/latex]. (a) The elevator accelerates upward from rest at a rate of [latex]1.20\,{\text{m/s}}^{2}[/latex] for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of [latex]0.600\,{\text{m/s}}^{2}[/latex] for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

a. [latex]1.87\times {10}^{4}\,\text{N;}[/latex] b. [latex]1.67\times {10}^{4}\,\text{N;}[/latex] c. [latex]1.56\times {10}^{4}\,\text{N;}[/latex] d. 19.4 m, 0 m/s

A 20.0-g ball hangs from the roof of a freight car by a string. When the freight car begins to move, the string makes an angle of [latex]35.0^\circ[/latex] with the vertical. (a) What is the acceleration of the freight car? (b) What is the tension in the string?

A student’s backpack, full of textbooks, is hung from a spring scale attached to the ceiling of an elevator. When the elevator is accelerating downward at [latex]3.8\,{\text{m/s}}^{2}[/latex], the scale reads 60 N. (a) What is the mass of the backpack? (b) What does the scale read if the elevator moves upward while slowing down at a rate [latex]3.8\,{\text{m/s}}^{2}[/latex]? (c) What does the scale read if the elevator moves upward at constant velocity? (d) If the elevator had no brakes and the cable supporting it were to break loose so that the elevator could fall freely, what would the spring scale read?

a. 10 kg; b. 90 N; c. 98 N; d. 0

A service elevator takes a load of garbage, mass 10.0 kg, from a floor of a skyscraper under construction, down to ground level, accelerating downward at a rate of [latex]1.2\,{\text{m/s}}^{2}[/latex]. Find the magnitude of the force the garbage exerts on the floor of the service elevator?

A roller coaster car starts from rest at the top of a track 30.0 m long and inclined at [latex]20.0^\circ[/latex] to the horizontal. Assume that friction can be ignored. (a) What is the acceleration of the car? (b) How much time elapses before it reaches the bottom of the track?

a. [latex]3.35\,{\text{m/s}}^{2}[/latex]; b. 4.2 s

The device shown below is the Atwood’s machine considered in Figure . Assuming that the masses of the string and the frictionless pulley are negligible, (a) find an equation for the acceleration of the two blocks; (b) find an equation for the tension in the string; and (c) find both the acceleration and tension when block 1 has mass 2.00 kg and block 2 has mass 4.00 kg.

Two blocks are connected by a massless rope as shown below. The mass of the block on the table is 4.0 kg and the hanging mass is 1.0 kg. The table and the pulley are frictionless. (a) Find the acceleration of the system. (b) Find the tension in the rope. (c) Find the speed with which the hanging mass hits the floor if it starts from rest and is initially located 1.0 m from the floor.

Shown below are two carts connected by a cord that passes over a small frictionless pulley. Each cart rolls freely with negligible friction. Calculate the acceleration of the carts and the tension in the cord.

A 2.00 kg block (mass 1) and a 4.00 kg block (mass 2) are connected by a light string as shown; the inclination of the ramp is [latex]40.0^\circ[/latex]. Friction is negligible. What is (a) the acceleration of each block and (b) the tension in the string?

6.1 Solving Problems with Newton’s Laws Copyright © 2016 by OpenStax. All Rights Reserved.

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Mechanics: Newton's Laws of Motion

Calculator pad, version 2, newton's laws of motion: problem set.

An African elephant can reach heights of 13 feet and possess a mass of as much as 6000 kg. Determine the weight of an African elephant in Newtons and in pounds. (Given: 1.00 N = .225 pounds)

• Audio Guided Solution
• Show Answer

About twenty percent of the National Football League weighs more than 300 pounds. At this weight, their Body Mass Index (BMI) places them at Grade 2 obesity, which is one step below morbid obesity. Determine the mass of a 300 pound (1330 N) football player.

With fuel prices for combustible engine automobiles increasing, researchers and manufacturers have given more attention to the concept of an ultralight car. Using carbon composites, lighter steels and plastics, a fuel-efficient car can be manufactured at 540 kg. How much less does an ultralight car weigh compared to a 1450-kg Honda Accord (2007)?

According to the National Center for Health Statistics, the average mass of an adult American male is 86 kg. Determine the mass and the weight of an 86-kg man on the moon where the gravitational field is one-sixth that of the Earth.

The rising concern among athletic trainers and health advocates (and parents) regarding concussions and multiple concussions among high school football players has prompted numerous studies of the effectiveness of protective head gear and the forces and accelerations experienced by players. One study suggested that there is a 50% chance of concussions for impacts rated at 75 g's of acceleration (i.e., 75 multiplied by 9.8 m/s/s). (The average head impact results in 22 to 24 g's of acceleration.) If a player's head mass (with helmet) is 6.0 kg and considered to be a free body , then what net force would be required to produce an acceleration of 75 g's?

Captain John Stapp of the U.S. Air Force tested the human limits of acceleration by riding on a rocket sled of his own design, known as the Gee Whiz. What net force would be required to accelerate the 82-kg Stapp at 450 m/s/s (the highest acceleration tested by Stapp)?

Sophia, whose mass is 52 kg, experienced a net force of 1800 N at the bottom of a roller coaster loop during her school's physics field trip to the local amusement park. Determine Sophia's acceleration at this location.

The Top Thrill Dragster stratacoaster at Cedar Point Amusement Park in Ohio uses a hydraulic launching system to accelerate riders from 0 to 54 m/s (120 mi/hr) in 3.8 seconds before climbing a completely vertical 420-foot hill . Determine the net force required to accelerate an 86-kg man.

a. Determine the net force required to accelerate a 540-kg ultralight car from 0 to 27 m/s (60 mph) in 10.0 seconds. b. Determine the net force required to accelerate a 2160-kg Ford Expedition from 0 to 27 m/s (60 mph) in 10.0 seconds.

Problem 10:

Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They determine that a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s. What is the acceleration value of a cart with …

a. a mass of M when acted upon by a net force of 2F ? b. a mass of 2M when acted upon by a net force of F ? c. a mass of 2M when acted upon by a net force of 2F ? d. a mass of 4M when acted upon by a net force of 2F ? e. a mass of 2M when acted upon by a net force of 4F ?

Problem 11:

F grav = F norm = 60.5 N F app = 40.2 N F frict = 5.7 N.

Problem 12:

F grav = F norm = 207 N F tens = 182 N F frict = 166 N.

Problem 13:

F tens = 2340 N F grav = 2120 N F norm1 = F norm2 = 276 N.

Problem 14:

It's Friday night and Skyler has been assigned the noble task of baby-sitting Casey, his 2-year old brother. He puts a crash helmet on Casey, places him in the red wagon and takes him on a stroll through the neighborhood. As Skyler starts across the street, he exerts a 52 N forward force on the wagon. There is a 24 N resistance force and the wagon and Casey have a combined weight of 304 N. Construct a free body diagram depicting the types of forces acting upon the wagon. Then determine the net force, mass and acceleration of the wagon.

Problem 15:

After a lead-off single in the 8 th inning, Earl makes an effort to steal second base. As he hits the dirt on his head first dive, his 73.2 kg body encounters 249 N of friction force. Construct a free body diagram depicting the types of forces acting upon Earl. Then determine the net force and acceleration.

Problem 16:

Mira and Tariq are lab partners for the Pulley and Bricks Lab. They have determined that the 2.15-kg brick is experiencing a forward tension force of 9.54 N and a friction force of 8.69 N as it is accelerated across the table top. Construct a free body diagram depicting the types of forces acting upon the brick. Then determine the net force and acceleration of the brick.

Problem 17:

Moments after making the dreaded decision to jump out the door of the airplane, Darin's 82.5-kg body experiences 118 N of air resistance. Determine Darin's acceleration at this instant in time. HINT: begin by drawing a free body diagram and determine the net force.

Problem 18:

Kelli and Jarvis are members of the stage crew for the Variety Show. Between acts, they must quickly move a Baby Grand Piano onto stage. After the curtain closes, they exert a sudden forward force of 524 N to budge the piano from rest and get it up to speed. The 158-kg piano experiences 418 N of friction.

a. What is the piano's acceleration during this phase of its motion? b. If Kelli and Jarvis maintain this forward force for 1.44 seconds, then what speed will the piano have?

Problem 19:

Skydiving tunnels have become popular attractions, appealing in part to those who would like a taste of the skydiving experience but are too overwhelmed by the fear of jumping out of a plane at several thousand feet. Skydiving tunnels are vertical wind tunnels through which air is blown at high speeds, allowing visitors to experience bodyflight . On Natalya's first adventure inside the tunnel, she changes her orientation and for an instant, her 46.8-kg body momentarily experiences an upward force of air resistance of 521 N. Determine Natalya's acceleration during this moment in time.

Problem 20:

A rope is used to pull a 2.89-kg bucket of water out of a deep well.

a. What is the acceleration of the bucket when the tension in the rope is 30.2 N? b. If starting from rest, what speed will the bucket have after experiencing this force for 2.16 seconds?

Problem 21:

A 0.104-kg model rocket accelerates at 45.9 m/s/s on takeoff. Determine the upward thrust experienced by the rocket.

Problem 22:

Brandon is the catcher for the Varsity baseball team. He exerts a forward force on the 0.145-kg baseball to bring it to rest from a speed of 38.2 m/s. During the process, his hand recoils a distance of 0.135 m. Determine the acceleration of the ball and the force which is applied to it by Brandon.

Problem 23:

Alejandra is attempting to drag her 32.6-kg Golden Retriever across the wooden floor by applying a horizontal force. What force must she apply to move the dog with a constant speed of 0.95 m/s? The coefficient of friction between the dog and the floor is 0.72.

Problem 24:

The coefficient of friction between the wheels of Dawson's 1985 Ford Coupe and the dry pavement is 0.85. Determine the acceleration which the 1300-kg Coupe experiences while skidding to a stop.

Problem 25:

Nicholas, Brianna, Dylan and Chloe are practicing their hockey on frozen Bluebird Lake. As Dylan and Chloe chase after the 0.162 kg puck, it decelerates from 10.5 m/s to 8.8 m/s in 14 seconds.

a. Determine the acceleration of the puck. b. Determine the force of friction experienced by the puck. c. Determine the coefficient of friction between the ice and the puck.

Problem 26:

Unbeknownst to most students, every time the school floors are waxed, the physics teachers get together to have a barrel of phun doing friction experiments in their socks (uhm - they do have clothes on; its just that they don't have any shoes on their feet). On one occasion, Mr. London applies a horizontal force to accelerate Mr. Schneider (mass of 84 kg) rightward at a rate of 1.2 m/s/s. If the coefficient of friction between Mr. Schneider 's socks and the freshly waxed floors is 0.35, then with what force (in Newtons) must Mr. London be pulling?

Problem 27:

Dexter Eius is running through the cafeteria when he slips on some mashed potatoes and falls to the floor. (Let that be a lesson for Dexter.) Dexter lands in a puddle of milk and skids to a stop with an acceleration of -4.8 m/s/s. Dexter weighs 780 Newtons. Determine the coefficient of friction between Dexter and the milky floor.

Problem 28:

The Harrier Jump Jet is a fixed wing military jet designed for vertical takeoff and landing (VTOL). It is capable of rotating its jets from a horizontal to a vertical orientation in order to takeoff, land and conduct horizontal maneuvers. Determine the vertical thrust required to accelerate an 8600-kg Harrier upward at 0.40 m/s/s.

Problem 29:

While skydiving, Dee Selerate opens her parachute and her 53.4-kg body immediately accelerates upward for an instant at 8.66 m/s/s. Determine the upward force experienced by Dee during this instant.

Problem 30:

A 1370-kg car is skidding to a stop along a horizontal surface. The car decelerates from 27.6 m/s to a rest position in 3.15 seconds. Assuming negligible air resistance, determine the coefficient of friction between the car tires and the road surface.

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Newton Second Law of Motion Example Problems with Answers

Newton's 2nd law of motion involves force, mass and acceleration of an object. It is the acceleration of an object produced by an action or force which is directly proportional to the magnitude of the net force in the same direction and inversely proportional to the object mass. Calculate net force, mass and acceleration of an object by referring the below Newton second law of motion example problems with answers.

Newton 2nd Law of Motion Problems with Solutions

Let us consider the problem: A 15 kg object moving to the west with an acceleration of 10m/s 2 . What is the net force acting on an object?

We can calculate Force, Mass and Acceleration using the given formula.

Newton's Second Law of Motion Formulae:

Substituting the values in the above given formula,

Net Force (F net )= 15 x 10 = 150 N Therefore, the value of Net force = 150 N

Refer the newton 2nd law of motion problems with solutions: A softball has a mass of 1.5 kg and hits the catcher's glove with a force of 30 N? What is the acceleration of the softball?

Substituting the values in the above given formula, Acceleration = 30 / 1.5 = 20 m/s 2 Therefore, the value of Acceleration is 20 m/s 2

Refer the problem with solution: What is the mass of a truck if it produces a force of 15000 N while accelerating at a rate of 6 m/s 2 ?

Substituting the values in the above given formula, Mass = 15000 / 6 = 2500 kg Therefore, the value of Mass is 2500 kg

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Newton's Second Law Of Motion

Newton’s second law of motion, unlike the first law of motion, pertains to the behaviour of objects for which all existing forces are unbalanced. The second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. This article discusses Newton’s second law in detail.

What is Newton’s Second Law of Motion?

Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration.

Sir Issac newton

Defining Newton’s Second Law of Motion

Newton’s second law states that the acceleration of an object depends upon two variables – the net force acting on the object and the mass of the object.  The acceleration of the body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body. This means that as the force acting upon an object is increased, the acceleration of the object is increased. Likewise, as the mass of an object is increased, the acceleration of the object is decreased.

Newton’s second law can be formally stated as,

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

This statement is expressed in equation form as,

Deriving Newton’s Second Law

For Changing Mass

Let us assume that we have a car at a point (0) defined by location X 0 and time t 0 . The car has a mass m 0 and travels with a velocity v 0 . After being subjected to a force F, the car moves to point 1 which is defined by location X 1 and time t 1 . The mass and velocity of the car change during the travel to values m 1 and v 1 . Newton’s second law helps us determine the new values of m 1 and v 1 if we know the value of the acting force.

Taking the difference between point 1 and point 0, we get an equation for the force acting on the car as follows:

Let us assume the mass to be constant. This assumption is good for a car because the only change in mass would be the fuel burned between point “1” and point “0”. The weight of the fuel is probably small relative to the rest of the car, especially if we only look at small changes in time. Meanwhile, if we were discussing the flight of a bottle rocket, then the mass does not remain constant, and we can only look at changes in momentum.

For Constant Mass

For a constant mass, Newton’s second law can be equated as follows:

Application of Second Law

Newton’s second law is applied to identify the amount of force needed to make an object move or make it stop. Following are a few examples that we have listed to help you understand this point:

Kicking a ball

When we kick a ball, we exert force in a specific direction. The stronger the ball is kicked, the stronger the force we put on it and the further away it will travel.

Pushing a cart

It is easier to push an empty cart in a supermarket than a loaded one, and more mass requires more acceleration.

Two people walking

Among the two people walking, if one is heavier than the other, the one weighing heavier will walk slower because the acceleration of the person weighing lighter is greater.

Get a glimpse of Newton’s second law of motion being taught in BYJU’S classes.

Newton’s Second Law Solved Examples

If there is a block of mass 2kg, and a force of 20 N is acting on it in the positive x-direction, and a force of 30 N in the negative x-direction, then what would be its acceleration?

We first have to calculate the net force acting on it to calculate its acceleration.

The negative acceleration indicates that the block is slowing and its acceleration vector is moving in an opposite direction directed opposite to the direction of motion.

How much horizontal net force is required to accelerate a 1000 kg car at 4 m/s 2 ? Solution: Newton’s 2nd Law relates an object’s mass, the net force on it, and its acceleration: Therefore, we can find the force as follows: F net = ma Substituting the values, we get 1000 kg × 4 m/s 2 = 4000 N Therefore, the horizontal net force is required to accelerate a 1000 kg car at 4 m/s -2 is 4000 N.

Newton’s second law is applied in daily life to a great extent. For instance, in Formula One racing, the engineers try to keep the mass of cars as low as possible. Low mass will imply more acceleration, and the more the acceleration, the chances to win the race are higher.

Top 10 Most Important and Expected Questions on Laws of Motion.

Frequently Asked Questions – FAQs

How does newton’s second law of motion apply to rockets.

According to Newton’s second law of motion, we know that force is a product of mass and acceleration. When a force is applied to the rocket, the force is termed as thrust. The greater the thrust, the greater will be the acceleration. Acceleration is also dependent on the rocket’s mass, and the lighter the rocket faster is the acceleration.

How does Newton’s second law apply to a car crash?

According to the definition of Newton’s second law of motion, force is the dot product of mass and acceleration. The force in a car crash is dependent either on the mass or the acceleration of the car. As the acceleration or mass of the car increases, the force with which a car crash takes place will also increase.

What is the other name for Newton’s second law?

What are some daily life examples of newton’s second law of motion.

• Acceleration of the rocket is due to the force applied, known as thrust, and is an example of Newton’s second law of motion.
• Another example of Newton’s second law is when an object falls from a certain height, the acceleration increases because of the gravitational force.

Write the formula for Newton’s second law of motion?

State newton’s second law of motion, for a constant mass, how is newton’s second law equated, define net force., state true or false:net force is the vector sum of all forces acting on a body., watch the video and learn more about newton’s laws of motion.

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Newton's Second Law of Motion - Basic Problem

Please do provide a detailed step-by-step solution and a final answer.

• A truck with a mass of 3,500 kg, including the passengers, has an engine that produces a net horizontal force of 525 N. Assuming that there is no other forces involve in the motion, find the following:

a.) acceleration of the truck

b.) starting from rest, how long will it take for the truck to reach the velocity of 11.5 m/s?

2. An object with a mass m1 accelerates at 8.0 m/s2 when a force F is applied. A second object with a mass m2 accelerates at 4.0 m/s2 when the same force F is applied to it.

a.) Find the value of the ratio m2/m1.

b.) Find the acceleration of the combined mass under the action of force F.

1 Expert Answer

John B. answered • 10/20/21

Licensed Physics Instructor with Industry Experience

a) Since the engine is the only force acting, the net force is 525 N.

a=0.15 m/s 2

b) Using v=v 0 +at

11.5=0+(0.15)t

t=11.5/0.15

t=76.7 seconds

For the first mass F=m1(8)

For the second mass F=m2(4)

a) Therefore m2/m1 =(F/4)/(F/8) = 8/4 = 2

b) The combined mass is now m2 + m1 = F/8 + F/4 = 3F/8

Using F = (Combined Mass) x a

a= F/(3F/8)

a= 8/3 m/s 2 or 2.67 m/s 2

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Newton's First Law of Motion

Claimed by Raisa-Claire Aghomo (Fall '23)

This page describes Newton's first law of motion, the first of his three famous laws of motion published in his work Principia Mathematica .

• 1.1 A Mathematical Model
• 2.3 Difficult
• 3.1 Scenario: Tablecloth Party Trick
• 3.2 Scenario: Objects in Space
• 3.3 Scenario: Turning car
• 5.1 External links
• 6 References

The Main Idea

Newton's first law states that an object at rest will stay at rest and an object in motion will stay in motion with the same speed and direction of travel unless the object is acted upon by an unbalanced external force.

Newton's first law states that it is the natural tendency for objects to remain on their current course. The tendency of matter to obey this law is called Inertia , so it is also sometimes called the Law of Inertia.

A Mathematical Model

The first law states that if the Net Force acting on the object is zero, then its velocity [math]\displaystyle{ \vec{v} }[/math] will not change over time. Velocity is a vector, which has both direction and magnitude, therefore if the Net Force is zero, neither the direction or magnitude can be changing. In other words, if the net force acting on an object is zero, it will not accelerate.

This idea can be expressed in the following manner:

Newton's first law applies to:

1) Objects at rest ( [math]\displaystyle{ |\vec{v}| = 0 }[/math] ), which will stay at rest unless a nonzero force acts upon it.

2) Objects in motion ( [math]\displaystyle{ |\vec{v}| \neq 0 }[/math] ), which will continue to be in motion with the same velocity, proceeding in the same straight line, unless a nonzero force acts upon it.

Because of the qualitative nature of Newton's first law, some of these example problems are conceptual questions rather than mathematical calculations.

Question 1: Suppose you want to push a box across a table in a straight line at a constant speed. What force, if any, would you have to exert on the box? (Describe it qualitatively- not enough information is supplied for a numerical answer.)

Answer: The moving box would experience 3 forces (besides any you exert on it): gravity, normal force from the table, and friction with the table and air. Gravity points downwards, normal force points upwards, and friction opposes the direction of motion. Because of the nature of normal force, the normal force takes on whatever magnitude necessary to cancel gravity. The only unbalanced force is therefore friction. Since your objective is to keep the box moving in a straight line at a constant speed (that is, at a constant velocity), the net force acting on the box must be 0 according to Newton's first law. The force you exert should therefore balance the friction force by being equal in magnitude and opposite in direction.

Question 2: Is a change in position an indicator of interaction?

Answer: On its own, a change in position is not enough to indicate an interaction because an object can have a nonzero velocity (that is, have a changing position) even with no forces acting on it, so long as that velocity is constant. With some additional information, however, a change in position can indicate an interaction. For example, if an object is initially at rest and is later found at another position, its velocity must have changed and it must have been acted on by a nonzero unbalanced external force.

Question: A 16kg traffic light is suspended by two cables, each 22 [math]\displaystyle{ ^\circ }[/math] from horizontal, as shown below:

What is the tension in each of the cables?

Solution: Because the traffic light is at rest and not accelerating, by Newton's first law, any forces acting on it must be balanced (that is, the net force acting on it must be 0). The forces acting on the traffic light are gravity and tension in the 2 cables. The horizontal components of the cables' tension forces must be equal in magnitude, or the traffic light would be accelerating to the left or the right. Combining that information with the fact that the two cables are inclined by the same amount leads to the conclusion that the tension in the two cables must be the same. Finally, we know that the combined vertical components of the two cables' tension forces must equal the traffic light's weight, or the light would be accelerating vertically. This allows us to create the following equation:

[math]\displaystyle{ 2T\sin(22) = 16*9.8 }[/math]

[math]\displaystyle{ T = \frac{16*9.8}{2\sin(22)} = 209.3 }[/math] N.

(Requires knowledge of Static Friction .)

Question: Suppose there exists a car of mass 9000 kg that is moving at a constant speed of 90 m/s in an easterly direction. The car is being buffeted by a strong wind, which exerts a 1000N force on it in the northerly direction. From this information, you know that the coefficient of static friction between the road and teh car's tires [math]\displaystyle{ \mu_s }[/math] must be at least what value?

Solution: Because the car is travelling at a constant speed without changing direction, by Newton's first law, any forces acting on it must be balanced (that is, the net force acting on it must be 0). The forces acting on the car are gravity, the normal force from the road, the wind, and static friction with the road. Gravity acts in the downward direction, the normal force acts in the upward direction, the wind acts in a northerly direction, and static friction acts in a southerly direction. We know that the magnitude of the normal force must be equal to the magnitude of the gravitational force, or the car would be accelerating vertically. That is, the magnitude of the normal force [math]\displaystyle{ N }[/math] must be 9,000kg * 9.8m/s = 88200N. We also know that the magnitude of the static friction force must be equal to the magnitude of the wind force, or the car would be accelerating along the north-south axis. That is, the magnitude of the friction force [math]\displaystyle{ f_s }[/math] must be 1000N.

[math]\displaystyle{ f_s \leq \mu_s * N }[/math]

[math]\displaystyle{ \mu_s \geq \frac{f_s}{N} }[/math]

[math]\displaystyle{ \mu_s \geq \frac{1000}{88200} = .0113 }[/math]

Connectedness

Newton's first law is applicable to any situation where the net force on an object is 0 and its velocity remains constant. There are nearly limitless examples of such situations, as well as nearly limitless applications. A few can be found below.

Scenario: Tablecloth Party Trick

A classic demonstration of Newton's first law is a party trick in which a tablecloth is yanked out from underneath an assortment of dinnerware, which barely moves and remains on the table. The tablecloth accelerates because a strong external force- a person's arm- acts on it, but the only force acting on the dinnerware is kinetic friction with the sliding tablecloth. This force is significantly weaker, and if the tablecloth is pulled quickly enough, does not have enough time to impart a significant impulse on the dinnerware. This trick demonstrates Newton's first law because the dinnerware begins at rest and remains so because no significant forces act on it.

Scenario: Objects in Space

In space, some intergalactic objects exist so far away from other bodies of matter that gravitational forces acting on them are negligible in magnitude. These objects continue to move in a straight line at a constant speed for very long periods due to Newton's first law.

Scenario: Turning car

You have probably experienced a situation in which you were driving or riding in a car when the driver takes a sharp turn. As a result, you were pressed against the side of the car to the outside of the turn. This is the result of Newton's first law; when the car changed direction, your body's natural tendency to continue moving in a straight line caused it to collide with the side of the car (or your seatbelt), which then applied enough normal force to cause your body to turn along with the car.

The nature of the tendencies of matter regarding motion has been the subject of much thought throughout human history. Aristotle (384–322 BCE) famously believed that all objects have a "natural place" towards which they tend: heavy objects belong on the earth and therefore tend to move downwards, while lighter substances such as smoke belong in the sky and therefore tend to move upwards. Once an object reached its natural place, Aristotle believed, it would remain there at rest. Aristotle believed objects could not continue to move forever without being acted on by a force to keep it in motion, which is consistent with any observations he could have made on the surface of the earth, although today, we know this to be the result of Friction .

Galileo Galilei (1564-1642), who studied the motion of celestial bodies, was the first to propose that perpetual motion was actually the natural state of objects, and that forces such as friction were necessary to bring them to rest or otherwise change their velocities. Galileo performed an experiment with two ramps and a bronze ball. The two ramps were set up at the same angle of incline, facing each other. Galileo observed that if a ball was released on one of the ramps from a certain height, it would roll down that ramp and up the other and reach that same height. He then experimented with altering the angle of the second ramp. He observed that even when the second ramp was less steep than the first, the ball would reach the same height it was dropped from. (Today, this is known to be the result of conservation of energy.) Galileo reasoned that if the second ramp were removed entirely, and the ball rolled down the first ramp and onto a flat surface, it would never be able to reach the height it was dropped from, and would therefore never stop moving if conditions were ideal. Galileo was essentially the first person to propose the idea that we now know as Newton's first law.

Isaac Newton (1643-1727) confirmed Galileo's idea with his own experiments and published it along with his 2 other laws in his 1687 work Principia Mathematica . Although he gave credit to Galileo, today the law is known by Newton's name.

Newton's first law appears in Principia (which was written in Latin) as follows:

Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter in directum, nisi quatenus a viribus impressis cogitur statum illum mutare.

Translated to English, this reads:

"Law I: Every body persists in its state of being at rest or of moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed."

• Acceleration
• Newton's Second Law of Motion
• Newton's Third Law of Motion
• Galileo Galilei

External links

Science of NFL Football: https://www.youtube.com/watch?v=08BFCZJDn9w

Real world application of Newton's First Law: https://www.youtube.com/watch?v=8zsE3mpZ6Hw

Everything you want to know about Newton's First Law of Motion: http://swift.sonoma.edu/education/newton/newton_1/html/newton1.html

NASA page on Newton's first law: https://www.grc.nasa.gov/www/k-12/airplane/newton1g.html

http://teachertech.rice.edu/Participants/louviere/Newton/law1.html

http://education.seattlepi.com/galileos-experiments-theory-rolling-balls-down-inclined-planes-4831.html

https://science.howstuffworks.com/innovation/scientific-experiments/newton-law-of-motion2.htm

Matter and Interactions: Modern Mechanics. Volume One. 4th Edition.

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4.6: Problem-Solving Strategies

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Learning Objectives

By the end of this section, you will be able to:

• Understand and apply a problem-solving procedure to solve problems using Newton’s laws of motion.

Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy for Newton’s Laws of Motion

Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation . Such a sketch is shown in Figure (a). Then, as in Figure (b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists).

Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest . This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well.

A diagram showing the system of interest and all of the external forces is called a free-body diagram . Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem . This is done in Figure (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known.

APPLYING NEWTON'S SECOND LAW

Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then the net force is described by the equation: \(F_{net} = ma\)

For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the following conclusions: \[ F_{net \, x} = ma \]

\[ F_{net \, y} = 0 \]

You will need this information in order to determine unknown forces acting in a system.

Step 4. As always, check the solution to see whether it is reasonable . In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake.

• To solve problems involving Newton’s laws of motion, follow the procedure described:
• Draw a sketch of the problem.
• Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram.
• Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the \(x\)-direction) then \(F_{net \, x} = 0 \). If the object does accelerate in that direction, \(F_{net \, x} = ma \).
• Check your answer. Is the answer reasonable? Are the units correct?

10.7 Newton’s Second Law for Rotation

Learning objectives.

By the end of this section, you will be able to:

• Calculate the torques on rotating systems about a fixed axis to find the angular acceleration
• Explain how changes in the moment of inertia of a rotating system affect angular acceleration with a fixed applied torque

In this section, we put together all the pieces learned so far in this chapter to analyze the dynamics of rotating rigid bodies. We have analyzed motion with kinematics and rotational kinetic energy but have not yet connected these ideas with force and/or torque. In this section, we introduce the rotational equivalent to Newton’s second law of motion and apply it to rigid bodies with fixed-axis rotation.

Newton’s Second Law for Rotation

We have thus far found many counterparts to the translational terms used throughout this text, most recently, torque, the rotational analog to force. This raises the question: Is there an analogous equation to Newton’s second law, Σ F → = m a → , Σ F → = m a → , which involves torque and rotational motion? To investigate this, we start with Newton’s second law for a single particle rotating around an axis and executing circular motion. Let’s exert a force F → F → on a point mass m that is at a distance r from a pivot point ( Figure 10.37 ). The particle is constrained to move in a circular path with fixed radius and the force is tangent to the circle. We apply Newton’s second law to determine the magnitude of the acceleration a = F / m a = F / m in the direction of F → F → . Recall that the magnitude of the tangential acceleration is proportional to the magnitude of the angular acceleration by a = r α a = r α . Substituting this expression into Newton’s second law, we obtain

Multiply both sides of this equation by r ,

Note that the left side of this equation is the torque about the axis of rotation, where r is the lever arm and F is the force, perpendicular to r . Recall that the moment of inertia for a point particle is I = m r 2 I = m r 2 . The torque applied perpendicularly to the point mass in Figure 10.37 is therefore

The torque on the particle is equal to the moment of inertia about the rotation axis times the angular acceleration . We can generalize this equation to a rigid body rotating about a fixed axis.

If more than one torque acts on a rigid body about a fixed axis, then the sum of the torques equals the moment of inertia times the angular acceleration:

The term I α I α is a scalar quantity and can be positive or negative (counterclockwise or clockwise) depending upon the sign of the net torque. Remember the convention that counterclockwise angular acceleration is positive. Thus, if a rigid body is rotating clockwise and experiences a positive torque (counterclockwise), the angular acceleration is positive.

Equation 10.25 is Newton’s second law for rotation and tells us how to relate torque, moment of inertia, and rotational kinematics. This is called the equation for rotational dynamics . With this equation, we can solve a whole class of problems involving force and rotation. It makes sense that the relationship for how much force it takes to rotate a body would include the moment of inertia, since that is the quantity that tells us how easy or hard it is to change the rotational motion of an object.

Deriving Newton’s Second Law for Rotation in Vector Form

As before, when we found the angular acceleration, we may also find the torque vector. The second law Σ F → = m a → Σ F → = m a → tells us the relationship between net force and how to change the translational motion of an object. We have a vector rotational equivalent of this equation, which can be found by using Equation 10.7 and Figure 10.8 . Equation 10.7 relates the angular acceleration to the position and tangential acceleration vectors:

We form the cross product of this equation with r → r → and use a cross product identity (note that r → · α → = 0 r → · α → = 0 ):

We now form the cross product of Newton’s second law with the position vector r → , r → ,

Identifying the first term on the left as the sum of the torques, and m r 2 m r 2 as the moment of inertia, we arrive at Newton’s second law of rotation in vector form:

This equation is exactly Equation 10.25 but with the torque and angular acceleration as vectors. An important point is that the torque vector is in the same direction as the angular acceleration.

Applying the Rotational Dynamics Equation

Before we apply the rotational dynamics equation to some everyday situations, let’s review a general problem-solving strategy for use with this category of problems.

Problem-Solving Strategy

Rotational dynamics.

• Examine the situation to determine that torque and mass are involved in the rotation. Draw a careful sketch of the situation.
• Determine the system of interest.
• Draw a free-body diagram. That is, draw and label all external forces acting on the system of interest.
• Identify the pivot point. If the object is in equilibrium, it must be in equilibrium for all possible pivot points––chose the one that simplifies your work the most.
• Apply ∑ i τ i = I α ∑ i τ i = I α , the rotational equivalent of Newton’s second law, to solve the problem. Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation.
• As always, check the solution to see if it is reasonable.

Example 10.16

Calculating the effect of mass distribution on a merry-go-round.

• The moment of inertia of a solid disk about this axis is given in Figure 10.20 to be 1 2 M R 2 . 1 2 M R 2 . We have M = 50.0 kg M = 50.0 kg and R = 1.50 m R = 1.50 m , so I = ( 0.500 ) ( 50.0 kg ) ( 1.50 m ) 2 = 56.25 kg-m 2 . I = ( 0.500 ) ( 50.0 kg ) ( 1.50 m ) 2 = 56.25 kg-m 2 . To find the net torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that τ = r F sin θ = ( 1.50 m ) ( 250.0 N ) = 375.0 N-m . τ = r F sin θ = ( 1.50 m ) ( 250.0 N ) = 375.0 N-m . Now, after we substitute the known values, we find the angular acceleration to be α = τ I = 375.0 N-m 56.25 kg-m 2 = 6.67 rad s 2 . α = τ I = 375.0 N-m 56.25 kg-m 2 = 6.67 rad s 2 .
• We expect the angular acceleration for the system to be less in this part because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I , we first find the child’s moment of inertia I c I c by approximating the child as a point mass at a distance of 1.25 m from the axis. Then I c = m R 2 = ( 18.0 kg ) ( 1.25 m ) 2 = 28.13 kg-m 2 . I c = m R 2 = ( 18.0 kg ) ( 1.25 m ) 2 = 28.13 kg-m 2 . The total moment of inertia is the sum of the moments of inertia of the merry-go-round and the child (about the same axis): I = 28.13 kg-m 2 + 56.25 kg-m 2 = 84.38 kg-m 2 . I = 28.13 kg-m 2 + 56.25 kg-m 2 = 84.38 kg-m 2 . Substituting known values into the equation for α gives α = τ I = 375.0 N-m 84.38 kg-m 2 = 4 .44 rad s 2 . α = τ I = 375.0 N-m 84.38 kg-m 2 = 4 .44 rad s 2 .

Significance

Check your understanding 10.7.

The fan blades on a jet engine have a moment of inertia 30.0 kg-m 2 30.0 kg-m 2 . In 10 s, they rotate counterclockwise from rest up to a rotation rate of 20 rev/s. (a) What torque must be applied to the blades to achieve this angular acceleration? (b) What is the torque required to bring the fan blades rotating at 20 rev/s to a rest in 20 s?

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The Possible Collapse of the U.S. Home Insurance System

A times investigation found climate change may now be a concern for every homeowner in the country..

This transcript was created using speech recognition software. While it has been reviewed by human transcribers, it may contain errors. Please review the episode audio before quoting from this transcript and email [email protected] with any questions.

From “The New York Times,” I’m Sabrina Tavernise. And this is “The Daily.”

[MUSIC PLAYING]

Today, my colleague, Christopher Flavelle, on a “Times” investigation into one of the least known and most consequential effects of climate change — insurance — and why it may now be a concern for every homeowner in the country.

It’s Wednesday, May 15.

So, Chris, you and I talked a while ago about how climate change was really wreaking havoc in the insurance market in Florida. You’ve just done an investigation that takes a look into the insurance markets more broadly and more deeply. Tell us about it.

Yeah, so I cover climate change, in particular the way climate shocks affect different parts of American life. And insurance has become a really big part of that coverage. And Florida is a great example. As hurricanes have gotten worse and more frequent, insurers are paying out more and more money to rebuild people’s homes. And that’s driving up insurance costs and ultimately driving up the cost of owning a home in Florida.

So we’re already seeing that climate impact on the housing market in Florida. My colleagues and I started to think, well, could it be that that kind of disruption is also happening in other states, not just in the obvious coastal states but maybe even through the middle of the US? So we set out to find out just how much it is happening, how much that Florida turmoil has, in fact, become really a contagion that is spreading across the country.

So how did you go about reporting this? I mean, where did you start?

All we knew at the start of this was that there was reason to think this might be a problem. If you just look at how the federal government tracks disasters around the country, there’s been a big increase almost every year in the number and severity of all kinds of disasters around the country. So we thought, OK, it’s worth trying to find out, what does that mean for insurers?

The problem is getting data on the insurance industry is actually really hard. There’s no federal regulation. There’s no government agency you can go to that holds this data. If you talk to the insurers directly, they tend to be a little reluctant to share information about what they’re going through. So we weren’t sure where to go until, finally, we realized the best people to ask are the people whose job it is to gauge the financial health of insurance companies.

Those are rating agencies. In particular, there’s one rating company called AM Best, whose whole purpose is to tell investors how healthy an insurance company is.

Whoa. So this is way down in the nuts and bolts of the US insurance industry.

Right. This is a part of the broader economy that most people would never experience. But we asked them to do something special for us. We said, hey, can you help us find the one number that would tell us reporters just how healthy or unhealthy this insurance market is state by state over time? And it turns out, there is just such a number. It’s called a combined ratio.

OK, plain English?

Plain English, it is the ratio of revenue to costs, how much money these guys take in for homeowner’s insurance and how much they pay out in costs and losses. You want your revenue to be higher than your costs. If not, you’re in trouble.

So what did you find out?

Well, we got that number for every state, going back more than a decade. And what it showed us was our suspicions were right. This market turmoil that we were seeing in Florida and California has indeed been spreading across the country. And in fact, it turns out that in 18 states, last year, the homeowner’s insurance market lost money. And that’s a big jump from 5 or 10 years ago and spells real trouble for insurance and for homeowners and for almost every part of the economy.

So the contagion was real.

Right. This is our first window showing us just how far that contagion had spread. And one of the really striking things about this data was it showed the contagion had spread to places that I wouldn’t have thought of as especially prone to climate shocks — for example, a lot of the Midwest, a lot of the Southeast. In fact, if you think of a map of the country, there was no state between Pennsylvania and the Dakotas that didn’t lose money on homeowner’s insurance last year.

So just huge parts of the middle of the US have become unprofitable for homeowner’s insurance. This market is starting to buckle under the cost of climate change.

And this is all happening really fast. When we did the Florida episode two years ago, it was a completely new phenomenon and really only in Florida. And now it’s everywhere.

Yeah. And that’s exactly what’s so striking here. The rate at which this is becoming, again, a contagion and spreading across the country is just demolishing the expectations of anyone I’ve spoken to. No one thought that this problem would affect so much of the US so quickly.

So in these states, these new places that the contagion has spread to, what exactly is happening that’s causing the insurance companies to fold up shop?

Yeah. Something really particular is happening in a lot of these states. And it’s worth noting how it’s surprised everyone. And what that is, is formally unimportant weather events, like hailstorms or windstorms, those didn’t used to be the kind of thing that would scare insurance companies. Obviously, a big problem if it destroys your home or damages your home. But for insurers, it wasn’t going to wipe them out financially.

Right. It wasn’t just a complete and utter wipeout that the company would then have to pony up a lot of money for.

Exactly. And insurers call them secondary perils, sort of a belittling term, something other than a big deal, like a hurricane.

These minor league weather events.

Right. But those are becoming so frequent and so much more intense that they can cause existential threats for insurance companies. And insurers are now fleeing states not because of hurricanes but because those former things that were small are now big. Hailstorms, wildfires in some places, previous annoyances are becoming real threats to insurers.

Chris, what’s the big picture on what insurers are actually facing? What’s happening out there numbers-wise?

This is a huge threat. In terms of the number of states where this industry is losing money, it’s more than doubled from 10 years ago to basically a third of the country. The amount they’re losing is enormous. In some states, insurers are paying out $1.25 or even $1.50 for every dollar they bring in, in revenue, which is totally unsustainable.

And the result is insurers are making changes. They are pulling back from these markets. They’re hiking premiums. And often, they’re just dropping customers. And that’s where this becomes real, not just for people who surf balance sheets and trade in the stock market. This is becoming real for homeowners around the country, who all of a sudden increasingly can’t get insurance.

So, Chris, what’s the actual implication? I mean, what happens when people in a state can’t get insurance for their homes?

Getting insurance for a home is crucial if you want to sell or buy a home. Most people can’t buy a home without a mortgage. And banks won’t issue a mortgage without home insurance. So if you’ve got a home that insurance company doesn’t want to cover, you got a real problem. You need to find insurance, or that home becomes very close to unsellable.

And as you get fewer buyers, the price goes down. So this doesn’t just hurt people who are paying for these insurance premiums. It hurts people who want to sell their homes. It even could hurt, at some point, whole local economies. If home values fall, governments take in less tax revenue. That means less money for schools and police. It also means people who get hit by disasters and have to rebuild their homes all of a sudden can’t, because their insurance isn’t available anymore. It’s hard to overstate just how big a deal this is.

And is that actually happening, Chris? I mean, are housing markets being dragged down because of this problem with the insurance markets right now?

Anecdotally, we’ve got reports that in places like Florida and Louisiana and maybe in parts of California, the difficulty of getting insurance, the crazy high cost of insurance is starting to depress demand because not everyone can afford to pay these really high costs, even if they have insurance. But what we wanted to focus on with this story was also, OK, we know where this goes eventually. But where is it beginning? What are the places that are just starting to feel these shocks from the insurance market?

And so I called around and asked insurance agents, who are the front lines of this. They’re the ones who are struggling to find insurance for homeowners. And I said, hey, is there one place that I should go if I want to understand what it looks like to homeowners when all of a sudden insurance becomes really expensive or you can’t even find it? And those insurance agents told me, if you want to see what this looks like in real life, go to a little town called Marshalltown in the middle of Iowa.

We’ll be right back.

So, Chris, you went to Marshalltown, Iowa. What did you find?

Even before I got to Marshalltown, I had some idea I was in the right spot. When I landed in Des Moines and went to rent a car, the nice woman at the desk who rented me a car, she said, what are you doing here? I said, I’m here to write a story about people in Iowa who can’t get insurance because of storms. She said, oh, yeah, I know all about that. That’s a big problem here.

Even the rental car lady.

Even the rental car lady knew something was going on. And so I got into my rental car and drove about an hour northeast of Des Moines, through some rolling hills, to this lovely little town of Marshalltown. Marshalltown is a really cute, little Midwestern town with old homes and a beautiful courthouse in the town square. And when I drove through, I couldn’t help noticing all the roofs looked new.

What does that tell you?

Turns out Marshalltown, despite being a pastoral image of Midwestern easy living, was hit by two really bad disasters in recent years — first, a devastating tornado in 2018 and then, in 2020, what’s called a derecho, a straight-line wind event that’s also just enormously damaging. And the result was lots of homes in this small town got severely damaged in a short period of time. And so when you drive down, you see all these new roofs that give you the sense that something’s going on.

So climate had come to Marshalltown?

Exactly. A place that had previously seemed maybe safe from climate change, if there is such a thing, all of a sudden was not. So I found an insurance agent in Marshalltown —

We talked to other agents but haven’t talked to many homeowners.

— named Bobby Shomo. And he invited me to his office early one morning and said, come meet some people. And so I parked on a quiet street outside of his office, across the street from the courthouse, which also had a new roof, and went into his conference room and met a procession of clients who all had versions of the same horror story.

It was more — well more of double.

A huge reduction in coverage with a huge price increase.

Some people had faced big premium hikes.

I’m just a little, small business owner. So every little bit I do feel.

They had so much trouble with their insurance company.

I was with IMT Insurance forever. And then when I moved in 2020, Bobby said they won’t insure a pool.

Some people had gotten dropped.

Where we used to see carriers canceling someone for frequency of three or four or five claims, it’s one or two now.

Some people couldn’t get the coverage they needed. But it was versions of the same tale, which is all of a sudden, having homeowner’s insurance in Marshalltown was really difficult. But I wanted to see if it was bigger than just Marshalltown. So the next day, I got back in my car and drove east to Cedar Rapids, where I met another person having a version of the same problem, a guy named Dave Langston.

Tell me about Dave.

Dave lives in a handsome, modest, little townhouse on a quiet cul-de-sac on a hill at the edge of Cedar Rapids. He’s the president of his homeowners association. There’s 17 homes on this little street. And this is just as far as you could get from a danger zone. It looks as safe as could be. But in January, they got a letter from the company that insures him and his neighbors, saying his policy was being canceled, even though it wasn’t as though they’d just been hit by some giant storm.

So then what was the reason they gave?

They didn’t give a reason. And I think people might not realize, insurers don’t have to give a reason. Insurance policies are year to year. And if your insurance company decides that you’re too much of a risk or your neighborhood is too much of a risk or your state is too much of a risk, they can just leave. They can send you a letter saying, forget it. We’re canceling your insurance. There’s almost no protection people have.

And in this case, the reason was that this insurance company was losing too much money in Iowa and didn’t want to keep on writing homeowner’s insurance in the state. That was the situation that Dave shared with tens of thousands of people across the state that were all getting similar letters.

What made Dave’s situation a little more challenging was that he couldn’t get new insurance. He tried for months through agent after agent after agent. And every company told him the same thing. We won’t cover you. Even though these homes are perfectly safe in a safe part of the state, nobody would say yes. And it took them until basically two days before their insurance policy was going to run out until they finally found new coverage that was far more expensive and far more bare-bones than what they’d had.

But at least it was something.

It was something. But the problem was it wasn’t that good. Under this new policy, if Dave’s street got hit by another big windstorm, the damage from that storm and fixing that damage would wipe out all the savings set aside by these homeowners. The deductible would be crushingly high — $120,000 — to replace those roofs if the worst happened because the insurance money just wouldn’t cover anywhere close to the cost of rebuilding.

He said to me, we didn’t do anything wrong. This is just what insurance looks like today. And today, it’s us in Cedar Rapids. Everyone, though, is going to face a situation like this eventually. And Dave is right. I talked to insurance agents around the country. And they confirmed for me that this kind of a shift towards a new type of insurance, insurance that’s more expensive and doesn’t cover as much and makes it harder to rebuild after a big disaster, it’s becoming more and more common around the country.

So, Chris, if Dave and the people you spoke to in Iowa were really evidence that your hunch was right, that the problem is spreading and rapidly, what are the possible fixes here?

The fix that people seem most hopeful about is this idea that, what if you could reduce the risk and cause there to be less damage in the first place? So what some states are doing is they’re trying to encourage homeowners to spend more money on hardening their home or adding a new roof or, if it’s a wildfire zone, cut back the vegetation, things that can reduce your risk of having really serious losses. And to help pay for that, they’re telling insurers, you’ve got to offer a discount to people who do that.

And everyone who works in this field says, in theory, that’s the right approach. The problem is, number one, hardening a home costs a fantastic amount of money. So doing this at scale is hugely expensive. Number two, it takes a long time to actually get enough homes hardened in this way that you can make a real dent for insurance companies. We’re talking about years or probably decades before that has a real effect, if it ever works.

OK. So that sounds not particularly realistic, given the urgency and the timeline we’re on here. So what else are people looking at?

Option number two is the government gets involved. And instead of most Americans buying home insurance from a private company, they start buying it from government programs that are designed to make sure that people, even in risky places, can still buy insurance. That would be just a gargantuan undertaking. The idea of the government providing homeowner’s insurance because private companies can’t or won’t would lead to one of the biggest government programs that exists, if we could even do it.

So huge change, like the federal government actually trying to write these markets by itself by providing homeowner’s insurance. But is that really feasible?

Well, in some areas, we’re actually already doing it. The government already provides flood insurance because for decades, most private insurers have not wanted to cover flood. It’s too risky. It’s too expensive. But that change, with governments taking over that role, creates a new problem of its own because the government providing flood insurance that you otherwise couldn’t get means people have been building and building in flood-prone areas because they know they can get that guaranteed flood insurance.

Interesting. So that’s a huge new downside. The government would be incentivizing people to move to places that they shouldn’t be.

That’s right. But there’s even one more problem with that approach of using the government to try to solve this problem, which is these costs keep growing. The number of billion-dollar disasters the US experiences every year keeps going up. And at some point, even if the government pays the cost through some sort of subsidized insurance, what happens when that cost is so great that we can no longer afford to pay it? That’s the really hard question that no official can answer.

So that’s pretty doomsday, Chris. Are we looking at the end of insurance?

I think it’s fair to say that we’re looking at the end of insurance as we know it, the end of insurance that means most Americans can rest assured that if they get hit by a disaster, their insurance company will provide enough money they can rebuild. That idea might be going away. And what it shows is maybe the threat of climate change isn’t quite what we thought.

Maybe instead of climate change wrecking communities in the form of a big storm or a wildfire or a flood, maybe even before those things happen, climate change can wreck communities by something as seemingly mundane and even boring as insurance. Maybe the harbinger of doom is not a giant storm but an anodyne letter from your insurance company, saying, we’re sorry to inform you we can no longer cover your home.

Maybe the future of climate change is best seen not by poring over weather data from NOAA but by poring over spreadsheets from rating firms, showing the profitability from insurance companies, and how bit by bit, that money that they’re losing around the country tells its own story. And the story is these shocks are actually already here.

Chris, as always, terrifying to talk to you.

Always a pleasure, Sabrina.

Here’s what else you should know today. On Tuesday, the United Nations has reclassified the number of women and children killed in Gaza, saying that it does not have enough identifying information to know exactly how many of the total dead are women and children. The UN now estimates that about 5,000 women and about 8,000 children have been killed, figures that are about half of what it was previously citing. The UN says the numbers dropped because it is using a more conservative estimate while waiting for information on about 10,000 other dead Gazans who have not yet been identified.

And Mike Johnson, the Speaker of the House, gave a press conference outside the court in Lower Manhattan, where Michael Cohen, the former fixer for Donald Trump, was testifying for a second day, answering questions from Trump’s lawyers. Trump is bound by a gag order. So Johnson joined other stand-ins for the former president to discredit the proceedings. Johnson, one of the most important Republicans in the country, attacked Cohen but also the trial itself, calling it a sham and political theater.

Today’s episode was produced by Nina Feldman, Shannon Lin, and Jessica Cheung. It was edited by MJ Davis Lin, with help from Michael Benoist, contains original music by Dan Powell, Marion Lozano, and Rowan Niemisto, and was engineered by Alyssa Moxley. Our theme music is by Jim Brunberg and Ben Landsverk of Wonderly.

That’s it for “The Daily.” I’m Sabrina Tavernise. See you tomorrow.

• May 16, 2024   •   30:47 The Make-or-Break Testimony of Michael Cohen
• May 15, 2024   •   27:03 The Possible Collapse of the U.S. Home Insurance System
• May 14, 2024   •   35:20 Voters Want Change. In Our Poll, They See It in Trump.
• May 13, 2024   •   27:46 How Biden Adopted Trump’s Trade War With China
• May 10, 2024   •   27:42 Stormy Daniels Takes the Stand
• May 9, 2024   •   34:42 One Strongman, One Billion Voters, and the Future of India
• May 8, 2024   •   28:28 A Plan to Remake the Middle East
• May 7, 2024   •   27:43 How Changing Ocean Temperatures Could Upend Life on Earth
• May 6, 2024   •   29:23 R.F.K. Jr.’s Battle to Get on the Ballot
• May 3, 2024   •   25:33 The Protesters and the President
• May 2, 2024   •   29:13 Biden Loosens Up on Weed
• May 1, 2024   •   35:16 The New Abortion Fight Before the Supreme Court

Hosted by Sabrina Tavernise

Featuring Christopher Flavelle

Produced by Nina Feldman ,  Shannon M. Lin and Jessica Cheung

Edited by MJ Davis Lin

With Michael Benoist

Original music by Dan Powell ,  Marion Lozano and Rowan Niemisto

Engineered by Alyssa Moxley

Listen and follow The Daily Apple Podcasts | Spotify | Amazon Music | YouTube

Across the United States, more frequent extreme weather is starting to cause the home insurance market to buckle, even for those who have paid their premiums dutifully year after year.

Christopher Flavelle, a climate reporter, discusses a Times investigation into one of the most consequential effects of the changes.

On today’s episode

Christopher Flavelle , a climate change reporter for The New York Times.

Background reading

As American insurers bleed cash from climate shocks , homeowners lose.

See how the home insurance crunch affects the market in each state .

Here are four takeaways from The Times’s investigation.

There are a lot of ways to listen to The Daily. Here’s how.

We aim to make transcripts available the next workday after an episode’s publication. You can find them at the top of the page.

Christopher Flavelle contributed reporting.

The Daily is made by Rachel Quester, Lynsea Garrison, Clare Toeniskoetter, Paige Cowett, Michael Simon Johnson, Brad Fisher, Chris Wood, Jessica Cheung, Stella Tan, Alexandra Leigh Young, Lisa Chow, Eric Krupke, Marc Georges, Luke Vander Ploeg, M.J. Davis Lin, Dan Powell, Sydney Harper, Mike Benoist, Liz O. Baylen, Asthaa Chaturvedi, Rachelle Bonja, Diana Nguyen, Marion Lozano, Corey Schreppel, Rob Szypko, Elisheba Ittoop, Mooj Zadie, Patricia Willens, Rowan Niemisto, Jody Becker, Rikki Novetsky, John Ketchum, Nina Feldman, Will Reid, Carlos Prieto, Ben Calhoun, Susan Lee, Lexie Diao, Mary Wilson, Alex Stern, Dan Farrell, Sophia Lanman, Shannon Lin, Diane Wong, Devon Taylor, Alyssa Moxley, Summer Thomad, Olivia Natt, Daniel Ramirez and Brendan Klinkenberg.

Our theme music is by Jim Brunberg and Ben Landsverk of Wonderly. Special thanks to Sam Dolnick, Paula Szuchman, Lisa Tobin, Larissa Anderson, Julia Simon, Sofia Milan, Mahima Chablani, Elizabeth Davis-Moorer, Jeffrey Miranda, Renan Borelli, Maddy Masiello, Isabella Anderson and Nina Lassam.

Christopher Flavelle is a Times reporter who writes about how the United States is trying to adapt to the effects of climate change. More about Christopher Flavelle

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