Question

The forces $F_1\text{and}{F}_2$ in a brake and the direction of rotation of the drum are as shown in the figure. The coefficient of friction is $\mathrm{0.25.}$ The angle of wrap is $\frac{3\pi }{2}$ radians. It is given that $R=1$m and $F_2=1N.$ The torque (in Nm) exerted on the drum is

• A. 2.2482

Answer 1

The correct option is A 2.2482

i.e $F_1$ is tight side tension.

$F_1>F_2$

$\frac{F_1}{F_2}=e^{\mu \theta }=e^{0.25\times \frac{3\pi }{2}}$

$F_1=F_2.e^{0.25\times \frac{3\pi }{2}}$

$=1\times e^{0.25\times \frac{3\times \pi }{2}}=3.2482N$

Torque $=(F_1-F_2)\times r$

$=(3.2482-1)\times 1=2.2482Nm$