Newton’s third law and what it means

Different textbooks state Newton’s third law in different ways:

To every action there is always opposed an equal reaction; or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.
— Newton, quoted in Fundamentals of Physics, Halliday and Resnick, 1981 edition

If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction.
— Classical Dynamics of Particles and Systems, J. B. Marion, 1970 edition

If body A exerts a force F on body B, then body B exerts a force F on body A of the same size and along the same line, but in the opposite direction.
— Essential Principles of Physics, Whelan and Hodgson, 1979 edition

If one body exerts a force on a second, the second exerts an equal and opposite force on the first.
— Development of Concepts of Physics, A. Arons, 1965 edition

The terms “action” and “reaction” — often quoted, because they stem from Newton himself — may be misleading to students in that they might imply cause and effect, but in fact, either force can be the “action” or the “reaction” and the forces act simultaneously.

On the “action and reaction” phrase, physics educator Arnold Arons had this to say in A Guide to Introductory Physics Teaching, 1990 edition:

“The old, conventional jargon ‘for every action there is an equal and opposite reaction’ has always been gibberish to the majority of students, and, fortunately, many authors are abandoning it. It is best to say “if one object exerts a force on a second, the second exerts an equal and opposite force on the first” — or some other, equally simple and straightforward, form. Even this simple a statement is not initially understood. Students, even when repeating the words correctly, do not do so with the clear realization that one is talking about two different forces, each acting on a different body . . .”

We will use the following form:

Newton’s third law:
If body A exerts a force F on body B, then body B exerts an equal but oppositely directed force F on body A; the forces lie along a line joining the two bodies.

When does Newton’s third law apply?

Newton’s third law applies to gravitational and electrostatic forces, and to elastic forces between two objects connected by a straight spring (as long as we ignore rotation and vibration of the objects). It does not apply to forces that depend on the velocities of the interacting objects (e.g., the force between two moving electric charges).

Newton’s third law generally applies in problems of the kind encountered in CSET Science I.

Examples of Newton’s third law

Now let’s look at some examples of Newton’s third law.

Example 1:

To get a feel for Newton’s third law, sit in a chair with wheels, and hold a soccer ball or other ball. Hold your feet up so they don’t drag on the floor. Throw the ball away from you. (You might want to try this outside, if you can find a smooth surface for the chair to roll on.)

When the ball moves forward away from you, you move back. The harder you throw the ball, the stronger the force that pushes you backward.

What has exerted the force on the ball? You (that is, your muscles).

What has exerted the force on you? The ball itself.

Example 2:

An ice skater stands facing a wall and pushes against it. The ice skater accelerates backward.

The ice skater exerted a force on the wall by pushing on it. The wall exerted a force of equal magnitude on the skater, but in the opposite direction.

Note that if a person wearing sneakers and standing on rough ground pushes against a wall, the person will not accelerate backward, even though the wall does exert the “equal and opposite” force. That’s because of friction between the shoes and the ground.

Example 3:

An astronaut doing repairs on a space vehicle accidentally becomes un-tethered, and floats in space near the vehicle. To move toward the vehicle, she must throw something (maybe a tool from her tool belt!) in the direction away from the vehicle.

The astronaut, in throwing the tool, exerts a force on it that causes it to accelerate away from the vehicle. The tool, however, exerts an equal and opposite force on the astronaut, directed toward the vehicle.

Example 4:

A cannon is fired. As the cannonball flies off, the cannon itself recoils backward.

The cannon, in firing, exerted a force on the cannonball. But the cannonball, in flying out, pushed the cannon back with an equal and opposite force.

Example 5:

Probably the most-quoted example of Newton’s third law is how a rocket moves in space. The engine generates thrust by burning fuel. Hot gases accelerate out the back (or bottom) of the rocket, and they exert a force accelerating the rocket forward (or up).

The horse and wagon problem

The examples given in the previous section seem rather straightforward. However, Newton’s third law is notorious for causing confusion. The “horse and wagon problem” is a good example of this confusion.

The “horse and wagon” problem boils down to this question: How can a horse pull a wagon if the wagon pulls back on the horse with an equal and opposite force?

Here’s a hint: in the examples discussed above, we ignored the force of friction. However, when a horse pulls on a wagon, friction is essential. There are not two objects involved in this problem, but three: the horse, the wagon, and the ground. There are two sets of “action-reaction” pairs of forces:

the force of the horse on the wagon, and the force of the wagon on the horse
the force of the horse on the ground, and the force of the ground on the horse

The solution is explained here:

Basically, the horse must push on the ground such that the “reaction” force of the ground on the horse (which acts to push the horse forward) is greater than the force of the wagon pulling back on the horse.

Read a humorous take on this problem and a very clear explanation.

Other difficulties with the third law

When two billiard balls collide, or when a skater pushes against a wall, we have contact between two objects that are exerting forces on one another. In the case of gravity acting on, say, a book resting on a table or a ball falling toward the Earth, we have a non-contact force.

Some students have difficulty with the concept of a non-contact force. It may be helpful to demonstrate non-contact interactions such as the acceleration of small nails toward a magnet (or the magnet toward the nails), or the acceleration of small bits of paper toward a balloon that has been charged by friction (or the acceleration of the balloon toward the paper).

With gravity we have another potential source of confusion: it is easy to see the acceleration of a book toward the Earth, when the book is not supported by a table; but we can’t detect the acceleration of the Earth toward the book. The force of the book on the Earth is the same as the force of the Earth on the book, but because of the very large difference in masses, the acceleration of the Earth toward the book is not detectable.

Electrostatic repulsion: A non-contact force

Sample problem: Verifying Newton's third law

A block of mass M1 = 3 kg and a smaller block of mass M2 = 1.5 kg are in contact on a table as shown here in (A). A force F of 3 N pushes on M1 from the left.

Assuming the table surface is frictionless, find the force of M2 acting on M1 (which, as we know from Newton's third law, should be the same as the force of M1 acting on M2). These forces are shown in (B).

First think about what is happening overall in this problem.

The blocks have an unbalanced force acting in the horizontal direction and will accelerate in the horizontal direction. We will not need to consider gravity or the normal force of the surface on a block because the blocks are not accelerating in the vertical direction and so have no net force in the vertical direction.
The two blocks will move to the right together. Their acceleration to the right is given by the force F of 3 N and the combined mass of both blocks.

Now think about block M1 separately.

M1 has two forces acting on it, as shown by the red and blue arrows in part B) of the diagram: the force F of 3 N pushing it from the left (blue arrow), and the force of M2 pushing it or resisting its movement from the right (red arrow). Call this force from M2 (the red arrow) "F_fromM2."
The net force on block M1 is the vector sum of the F and F_M2: the net force F_netM1 on M1 = F - F_M2

In other words, we have F_netM1 = F - F_fromM2

We want to solve for FM2:

F_fromM2 = F – F_netM1

We know F (it is 3 N) and we can find the net force on M1 by considering what it must be judging from its mass and acceleration.

F_netM1 = (mass of M1) (acceleration of M1)

As noted above, the two blocks move together because they are in contact, so they have the same acceleration. This acceleration is given by the force that is applied to them, 3 N.

The mass of the two blocks together is M1 + M2 = 3 kg + 1.5 kg = 4.5 kg. The acceleration is the force per unit mass, or
a = F/m = 3 N/(4.5 kg) = 0.67 ms^-2.

Now we use this value of the acceleration to find the net force on M1:

F_netM1 = (mass of M1) (acceleration of M1) = (3 kg) (0.67 ms^-2) = 2 N

Think about what this means: the system of two blocks has a force of 3 N applied to it. But considering just block M1, we find it has the force of 3 N pushing it to the right and a force from M2 pushing it back to the left, so that a net force of 2 N pushes M1 to the right.

The situation depicted above is equivalent to this one:

You can see without even solving the equation that F_fromM2 must be 1 N.

Formally:

F_fromM2 = F - F_netM1 = 3 N - 2 N = 1 N

In other words, block M2 exerts a force of 1 N on block M1.

Now, just to verify our work, consider the force of M1 on M2. (We expect this to be the same, based on Newton's third law.) We can see from the diagram that M2 has only one horizontal force on it: the force from M1.

The net force on M1 must be the force from M2 because there are no other forces.

F_fromM2 = F_netM2

The acceleration of M2 is, as we worked out above, 0.67 ms^-2.
The mass of M2 is 1.5 kg.
Putting these two facts together, we have

F_netM2 = force from M1 = (mass of M2) (acceleration of M2) = (1.5 kg) (0.67 ms^-2) = 1 N

Indeed, the force from M1 on M2 is the same as the force from M2 on M1.

Putting it all together: More applications of Newton's laws

Let's see how all of Newton's laws can be applied to common problems.

Sample problem 1:

How can a 100 lb object be lowered from a roof using a rope with a breaking strength of 60 lb, without breaking the rope?

We are working this problem in imperial units. Weight is measured in pounds, mass in slugs, and the acceleration due to gravity (9.8 ms^-2) is 32 ft s^-2.

The essential fact is that the force of the object on the rope (which is the same as the force of the rope of the object) cannot exceed 60 lb, or the rope will break.

The force of the rope on the object, or of the object on the rope, depends on the weight of the object (the force of gravity acting on it) and also on how fast the object is accelerating downward.

Consider what would happen if the object were simply dropped from the roof, without holding on to the rope: in that case, the rope would be slack, and the rope and object would exert no force on each other. The acceleration of the object would be 32 ft/s².

On the other hand, if the object were held still by the rope, the force of the rope on the object would have to balance the weight of the object, and the force of the rope on the object (or object on rope) would have to be 100 lb. The acceleration would be 0.

In between these two extremes (letting the object fall freely, or holding it still with the rope) there must be an acceleration of the object such that the force of the rope on the object is between 0 and 100 lb. We want to find the acceleration when the force is 60 lb, the critical breaking point.

If the object is held with the rope but the object accelerates downward, the force of the rope on the object is reduced compared to what it would be if the object were held still.

Draw a free-body diagram showing the forces acting on the object:

(Note that the object weighs 100 lb, so its mass is (100lb)/32 fts^-2) = 3.13 slugs.)

The net force on the object (which is always equal to the object’s mass times its acceleration) is the result of vectorially adding the two forces acting on it:

F_net = F_grav-F_rope

We have a constraint on F_rope. Solving for F_rope, we have

F_rope = F_grav-F_net

The force of the rope on the object (or of object on rope) cannot be more than 60 lb (i.e. F_rope <= 60 lb)

F_rope = F_grav-F_net <= 60 lb

We want to know the acceleration of the object that will fit this constraint. As mentioned before, the acceleration of the object times the object’s mass is the net force on the object. So we solve for the net force:

F_net >= F_grav-60 lb

Therefore (mass of object) (acceleration of object) >= F_grav-60 lb

So

acceleration of the object (so as not to break rope) >= (F_grav-60 lb)/(3.13 slugs)

= (100 lb-60 lb)/3.13 slugs = 12.8 ft s^-2

We have solved the problem. As long as the acceleration of the object is at least 12.8 ft s^-2 (which is not as high as the acceleration would be in free fall) the rope will not break.

Sample problem 2:

A block of mass 8.16 kg is pushed to the right across a rough surface with a force of 50 N. The frictional force has a magnitude of 10 N. Draw the free body diagram showing all forces including the normal force. What is the net force on the block? What is the acceleration of the block?
We take the positive-x direction to be to the right and the positive-y direction to be up.

The net force in the vertical direction is zero because the block is not moving in this direction. The force of gravity (pointing down) balances the normal force (pointing up). The force of gravity and the normal force thus both have the magnitude Fgravity = (mass) g = (8.16 kg) (9.8 ms^-2) = 80 N.

The net force in the horizontal direction is 50 N-10N = 40 N, pointing to the right.

The acceleration to the right is F/m = (40 N)/(8.16 kg) = 4.9 ms^-2.