Question

A uniform ladder of length $10.0m$ and mass $16.0kg$ is resting against a vertical wall making an angle of ${37}^o$ with it. The vertical wall is frictionless but the ground is rough. An electrician weighing $60.0kg$ climbs up the ladder. If he stays on the ladder at a point $8.00m$ from the lower end, what will be the normal force and the force of friction on the ladder by the ground? What should be the minimum of friction for the electrician to work safely?

Answer 1

Let the forces $F_1$ and $F_2$ as shown in the above figure.

Since, the ladder should not slip or rotate.

The torques expression is given as,

$mg\left(8\sin {37}^{\circ }\right)+Mg\left(5\sin {37}^{\circ }\right)-F_2\left(10\cos {37}^{\circ }\right)=0$

$60\times 9.8\times \left(8\sin {37}^{\circ }\right)+16\times 9.8\times \left(5\sin {37}^{\circ }\right)-F_2\left(10\cos {37}^{\circ }\right)=0$

$F_2=412N$

Since, from the equilibrium of the ladder the force $F_2$ will be equal to the friction force.

The friction force is given as,

$f=412N$

The normal force is given as,

$F_1=\left(m+M\right)g$

$=\left(60+16\right)\times 9.8$

$=744.8N$

The minimum coefficient of friction is given as,

$\mu =\frac{f}{F_1}$

$=\frac{412}{744.8}$

$=0.553$

Thus, the minimum coefficient of friction is $0.553$.