Question

Find the force $\overrightarrow{F}$ exerted by the cylinder on the ceiling (through the threads).

• A. $\overrightarrow{F}=2\frac{1}{3}m(g-\overrightarrow{w_0})$
• B. $\overrightarrow{F}=3\frac{1}{3}m(g-\overrightarrow{w_0})$
• C. $\overrightarrow{F}=4\frac{1}{3}m(g-\overrightarrow{w_0})$
• D. $\overrightarrow{F}=\frac{1}{3}m(g-\overrightarrow{w_0})$

Answer 1

The correct option is D $\overrightarrow{F}=\frac{1}{3}m(g-\overrightarrow{w_0})$

If the observer is in the elevator the gravitational force as
observed by the observer will be = $m(g-w_o)$
where g= acceleration due to gravity $w_o$= acceleration of the elevator
From Newton's Laws :
$m(g-w_o)-2T=ma$ .......(1) a = acceleration of the cylinder relative to the car
(because the observer is in the car he will see relative motion)
As slipping is not there:
$I\alpha =2Tr\frac{1}{2}m{r}^2\times \frac{a}{r}=2TrT=\frac{ma}{4}$
Substitutung the value of T in equation (1)
$m(g-w_o)-\frac{ma}{2}=ma$
on solving we get
$\frac{3ma}{2}=m(g-w_o)$
or a=$\frac{2}{3}(g-w_o)$
Force exerted by the cylinder on the ceiling of the elevator =2T
using the value of T from above which is also equal to $\frac{ma}{4}$
Substituting the value of "a" from above and solving
$2T=\frac{ma}{2}a=\frac{2}{3}(g-{\omega }_o)2T=\frac{m\frac{2}{3}(g-{\omega }_o)}{2}2T=\frac{1}{3}m(g-{\omega }_o)orF=\frac{1}{3}m(g-{\omega }_o)$