Question

In the system shown, a force of $100$ N is applied at the end shown. What is the magnitude $\tau$ (in N-m) of the torque which was produced on the drum for starting motion? (Given that the coefficient of static friction is $0.1$)

• A. $10$
• B. $5$
• C. $20$
• D. $25$

Answer 1

Answer: (C)

$20$

For the motion to start, the 'moment-couple' on the drum must be equal to the product of frictional force of the rod against the drum and the drum radius. Since the rod is in equilibrium, we can take torque about the fulcrum to find the normal force of the drum on the rod (Normal force N acts at centre of rod)

$\sum {\tau }_{fulcrum}=0$
or
$-100(2+2)+N\left(2\right)=0$
or
$N=200N$
Friction force$f=\mu N$ $=\left(0.1\right)\left(200\right)=20N$

$\tau =$ Torque produced by couple$=$$\left(20\right)1$
or $\tau =20Nm$