Newton's Second Law for a System of Particles — Question 7

Question 7

Support reading for an internally moving load

A $60\,\mathrm{kg}$ person stands on a platform of mass $20\,\mathrm{kg}$ that rests on a scale. At one instant the platform acceleration is zero and the person's center accelerates upward at $0.50\,\mathrm{m/s^2}$. Use $g=9.8\,\mathrm{m/s^2}$.

1. Find the center acceleration of person plus platform and the scale reading.

2. Would internal forces cancel if the system were the person alone?

Question 7 - Solution

Support reading for an internally moving load

(a) Derivation and calculation

The system center acceleration is $$\boxed{a_C=\frac{60(0.50)+20(0)}{80}=0.375\ \mathrm{m/s^2}}.$$ Upward force balance for person plus platform gives $$N-80g=80a_C\Rightarrow\boxed{N=80(9.8)+30=814\ \mathrm N}.$$

(b) Result and interpretation

For the person alone, the platform contact is external: $$N_p-60g=60(0.50)\Rightarrow\boxed{N_p=618\ \mathrm N}.$$ The stationary platform satisfies $N-N_p-20g=0$, recovering $N=618+196=814$ N. Internal versus external depends on the selected system.

Physical check

The scale reads total weight plus the rate of change of the system vertical momentum.