Newton's Third Law — Question 9

Question 9

A third-law pair cancels even when neither body is in equilibrium

In common axes A exerts $(3,-4,2)$ N on B. A also has external force $(1,2,0)$ N; B also has external force $(-2,1,5)$ N.

1. Find B’s force on A.

2. Find the individual net forces.

3. Find the combined-system external resultant and verify the cancellation.

Question 9 - Solution

A third-law pair cancels even when neither body is in equilibrium

(a) Reverse every component.

$$\boxed{\vec F_{B\to A}=-\vec F_{A\to B}=(-3,4,-2)\,\mathrm N}$$ $\vec F_{B\to A}=(-3,4,-2)$ N.

(b) Add forces for each recipient.

$$\vec R_A=(1,2,0)+(-3,4,-2)=(-2,6,-2),\quad\vec R_B=(1,-3,7)\quad\mathrm N$$ A’s resultant is $(1,2,0)+(-3,4,-2)=(-2,6,-2)$ N. B’s is $(-2,1,5)+(3,-4,2)=(1,-3,7)$ N.

(c) Sum the body equations.

$$\vec R_A+\vec R_B=(-1,3,5)=(1,2,0)+(-2,1,5)\quad\mathrm N$$ The resultants sum to $(-1,3,5)$ N, equal to the sum of the two external forces. The mutual pair cancels regardless of either body’s acceleration or mass.