Question

A uniform rod of mass $m = 15kg$ leans against a smooth vertical wall making an angle $\theta = {37^ \circ }$ with horizontal. The other end rests on a rough horizontal floor. Calculate the normal force and the friction force that the floor exerts on the rod.

Answer 1

Hint: In this problem we are going to write first all the forces which are acting on the rod. These include gravitational force, normal force and frictional force. And then by applying force balance we are going to get normal and friction force that the floor exerts on the rod.

Complete answer:
The forces acting on the rod are
(a) Its weight W,
(b) normal force N, by the vertical wall,
(c) normal force $N_2$ by the floor and
(d) frictional force f by the floor.

Taking horizontal and vertical components,
${N_1} = f$------(1)
And ${N_2} = mg$------(2)

Taking torque about ground we get,
${N_1}x\cos \theta = mg\dfrac{x}{2}\sin \theta $
$ \Rightarrow {N_1}\dfrac{3}{5} = mg\dfrac{1}{2} \times \dfrac{4}{5}$

$ \Rightarrow {N_1} = mg \times \dfrac{2}{3}$
So, ${N_1} = 100N$ also the$f = 100N$

The normal force by floor ${N_2} = mg = 150N$

Note: While attempting this type of question, always remember to apply force balance. The balance forces plays an important role to solve the above question. It is important note that, in this case we considered the object to be at static motion thus we haven’t used any value related to motion and velocity.