Question

A uniform rod of mass m= 15 kg leans against a smooth vertical wall making an angle $\theta ={37}^{\circ }$ with horizontal. The other ends rests on a rough horizontal floor. Calculate the normal force and the friction force that the floor exerts on the rod. [Take g= 10 $m/s^2$)

Answer 1

The forces acting em the rod are shown in figure. They 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,
$N_1=f$ .......... (i)
and $N_2=mg$ .......... (ii)
Taking torque about B,
$N_1\left(AO\right)=mg\left(CB\right)$
or, $N_1\left(AB\right)cos\theta =mg\frac{AB}{2}sin\theta N_1\frac{3}{5}=\frac{W}{2}\frac{4}{5}$
or, $N_1=\frac{2}{3}W$ ........... (iii)
The normal force force by the floor is $N_2=W=\left(15kg\right)\left(10m/s^2\right)=150N$
The frictional force is $f=N_1=\frac{2}{3}W=100N$