Newton's Second Law for a System of Particles — Question 2

Question 2

Unequal accelerations can share one center

Two particles of $2.0$ and $4.0\,\mathrm{kg}$ have instantaneous accelerations $(3,-1)$ and $(-1,2)\,\mathrm{m/s^2}$ in an inertial frame.

1. Find the center acceleration and total external force.

2. Can the net force on each particle be inferred? Can the separate external force on each be inferred without interaction information?

Question 2 - Solution

Unequal accelerations can share one center

(a) Derivation and calculation

Form the mass-weighted acceleration: $$M\vec a_C=2(3,-1)+4(-1,2)=(2,6)\ \mathrm N.$$ $$\boxed{\vec a_C=(1/3,1)\ \mathrm{m/s^2}},\qquad \boxed{\vec F_{\mathrm{ext}}=(2,6)\ \mathrm N}.$$

(b) Result and interpretation

Each measured acceleration fixes the corresponding total force: $$\boxed{\vec F_{1,\mathrm{total}}=(6,-2)\ \mathrm N},\qquad \boxed{\vec F_{2,\mathrm{total}}=(-4,8)\ \mathrm N}.$$ But $\vec F_{i,\mathrm{total}}=\vec F_{i,\mathrm{ext}}+\vec F_{i,\mathrm{int}}$. Without the interaction force, the separate external contributions are not identifiable.

Physical check

The total particle forces add to the net external force because internal action-reaction pairs cancel.