Applying Newton's Laws — Question 6

Question 6

An elevator’s acceleration changes tensions through a hanging chain

Take $g=10$ m/s$^2$. Two masses, 2 kg above and 3 kg below, hang in series from an elevator ceiling by light cords. The elevator accelerates upward at 1 m/s$^2$; both masses stay fixed relative to it.

1. Find the lower cord tension.

2. Find the upper cord tension.

3. Find both tensions if the elevator instead descends at constant velocity.

Question 6 - Solution

An elevator’s acceleration changes tensions through a hanging chain

(a) Isolate the lower mass.

$$T_l-3g=3a\ \Rightarrow\ \boxed{T_l=3(10+1)=33\,\mathrm N}$$ $T_l-30=3(1)$ gives $T_l=33$ N.

(b) Support and accelerate both masses.

$$T_u-(2+3)g=(2+3)a\ \Rightarrow\ T_u=5(11)=55\,\mathrm N$$ $T_u-50=5(1)$ gives $T_u=55$ N. Equivalently, the upper mass obeys $T_u-T_l-20=2(1)$.

(c) Use acceleration rather than travel direction.

$$a=0:\quad T_l=3g=30\,\mathrm N,\quad T_u=5g=50\,\mathrm N$$ Constant velocity means zero acceleration, so $T_l=30$ N and $T_u=50$ N. Downward motion alone does not reduce the load.