Question

A ladder of length 5 m is placed against a smooth wall as shown in figure. The coefficient of friction is $\mu$ between ladder and ground. What is minimum value of $\mu$ if the ladder is not to slip?

• A. $\mu =\frac{1}{2}$
• B. $\mu =\frac{1}{4}$
• C. $\mu =\frac{3}{8}$
• D. $\mu =\frac{5}{8}$

Answer 1

Answer: (C)

$\mu =\frac{3}{8}$

Vertical force on ladder are gravitational force, $mg$ at center of ladder and Normal force from ground $N_1$ at point $B$. Balancing forces $N_1=mg$

Horizontal forces are, frictional force $f=\mu N_1=\mu mg$ at B, and normal force from the wall $N_2$ at $A$.

Balancing forces $N_2=f=\mu mg$.

To prevent slipping, the torque about point $B$ must be zero.

Torque of gravitational force is $mg\times \frac{3}{2}$.

Torque of Normal force from ground $N_1$ is $mg\times 0=0$.

Torque of Normal force from wall $N_2$ is $\mu mg\times 4$.

Torque of frictional force is $\mu mg\times 0=0$.

Balancing torques, $\mu =3/8$