(4)

Given that: Mass of rod, $m=20\text{kg,}$

Length of rod, $l=5\text{m}$

[IMAGE 57]

Weight of the uniform rod = mg

$=20\times 10=200\text{N}$

For vertical direction, $W=N_2$

For horizontal direction, $f=N_1$

(where $f$ is the frictional force)

Since rod is stationary, so net torque is zero (about B)

$\Rightarrow$  Distance from wall to the centre of mass of rod (L/2) is

  $D=\frac{L}{2}\cos 30°=\frac{1}{2}\left(\frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}}\right)=\frac{\sqrt{3}}{4}$

$\text{⇒Now, distance }OA=L\sin 30°=\frac{L}{2}$

Taking torque about point B to be zero

$\Rightarrow mg\times \frac{L}{2}\cos 30°=N_1\times L\sin 30°$

$\Rightarrow mg\times \frac{1}{2}\times \frac{\sqrt{3}}{2}=f\times \frac{1}{2} \text{[As, }{N}_1=f\text{]}$

$\Rightarrow 200\times \frac{\sqrt{3}}{4}=\frac{f}{2} \text{or} f=100\sqrt{3}\text{N}$

(1)

Here, $AD=20\text{cm}$

$AE=160\text{cm}$

$g=10{\text{m/s}}^2$

Take moments about $C$

[IMAGE 59]

Clockwise moment = Anticlockwise moment

       $2\times g\times DC=0.50\times g\times C{C}_1+m\times g\times CE$

or    $2\times (40-20)=0.500\times (100-40)+m\times (60+60)$

or    $40-30=120m$

or    $m=\frac{1}{12}\text{kg}$

(2)

Given situation is shown in figure.

[IMAGE 60]

$N_1$ = Normal reaction on A

$N_2$ = Normal reaction on B

W = Weight of the rod

In vertical equilibrium,

            $N_1+N_2=W$                             ...(i)

Torque balance about centre of mass of the rod,

           $N_{1}x=N_2(d-x)$

Putting value of $N_2$ from equation (i)

$N_{1}x=(W-N_1)(d-x)\Rightarrow N_{1}x=Wd-Wx-N_{1}d+N_{1}x$

$\Rightarrow N_{1}d=W(d-x);$     $\therefore$    $N_1=\frac{W(d-x)}{d}$