Newton’s third law is easy to state but harder to grasp. For example, consider what happens when you release a ball. Not surprisingly, it falls down. But if the ball and the earth exert equal and opposite forces on each other, as Newton’s third law alleges, why doesn’t the earth “fall up” to meet the ball?

The key to understanding this and many similar puzzles is that the forces are equal but the accelerations are not. Equal causes can produce very unequal effects. FIGURE 7.11 shows equal-magnitude forces on the ball and the earth. The force on ball B is simply the gravitational force of Chapter 6:

where $m_{\text{B}}$ is the mass of the ball. According to Newton’s second law, this force gives the ball an acceleration

This is just the familiar free-fall acceleration.

FIGURE 7.11 The action/reaction forces of a ball and the earth are equal in magnitude.

According to Newton’s third law, the ball pulls up on the earth with force $\vec{F}{}_{\text{ball on earth}}$. Because $\vec{F}{}_{\text{earth on ball}}$ and $\vec{F}{}_{\text{ball on earth}}$ are an action/reaction pair, $\vec{F}{}_{\text{ball on earth}}$ must be equal in magnitude and opposite in direction to $\vec{F}{}_{\text{earth on ball}}$. That is,

Using this result in Newton’s second law, we find the upward acceleration of the earth as a whole is

The upward acceleration of the earth is less than the downward acceleration of the ball by the factor $m_{\text{B}}\text{/}{m}_{\text{E}}$. If we assume a 1 kg ball, we can estimate the magnitude of $\vec{a}{}_{\text{E}}\text{:}$

With this incredibly small acceleration, it would take the earth $8\times {10}^{15}$ years, approximately 500,000 times the age of the universe, to reach a speed of 1 mph! So we certainly would not expect to see or feel the earth “fall up” after we drop a ball.

EXAMPLE 7.3 The forces on accelerating boxes

The hand shown in FIGURE 7.12 pushes boxes A and B to the right across a frictionless table. The mass of B is larger than the mass of A.

1. Draw free-body diagrams of A, B, and the hand H, showing only the horizontal forces. Connect action/reaction pairs with dashed lines.

2. Rank in order, from largest to smallest, the horizontal forces shown on your free-body diagrams.

FIGURE 7.12 Hand H pushes boxes A and B.

FIGURE 7.12 Full Alternative Text

VISUALIZE a. The hand H pushes on box A, and A pushes back on H. Thus $\vec{F}{}_{\text{H on A}}$ and $\vec{F}{}_{\text{A on H}}$ are an action/reaction pair. Similarly, A pushes on B and B pushes back on A. The hand H does not touch box B, so there is no interaction between them. There is no friction. FIGURE 7.13 shows five horizontal forces and identifies two action/reaction pairs. Notice that each force is shown on the free-body diagram of the object that it acts on.

FIGURE 7.13 The free-body diagrams, showing only the horizontal forces.

FIGURE 7.13 Full Alternative Text

b. According to Newton’s third law, $F_{\text{A on H}}=F_{\text{H on A}}$ and $F_{\text{A on B}}=F_{\text{B on A}}$. But the third law is not our only tool. The boxes are accelerating to the right, because there’s no friction, so Newton’s second law tells us that box A must have a net force to the right. Consequently, $F_{\text{H on A}}>F_{\text{B on A}}$. Similarly, $F_{\text{arm on H}}>F_{\text{A on H}}$ is needed to accelerate the hand. Thus

$$F_{\text{arm on H}}>F_{\text{A on H}}=F_{\text{H on A}}>F_{\text{B on A}}=F_{\text{A on B}}$$

REVIEW You might have expected $F_{\text{A on B}}$ to be larger than $F_{\text{H on A}}$ because $m_{\text{B}}>m_{\text{A}}$. It’s true that the net force on B is larger than the net force on A, but we have to reason more closely to judge the individual forces. Notice how we used both the second and the third laws to answer this question.