Question

A uniform rod of mass m and length L is held at rest by a force F applied at it's end as shown in the vertical plane. The ground is sufficiently rough.
Find
(a) Force F
(b) Normal reaction exerted by the ground
(c) Frictional force exerted by the ground (magnitude and dirtection)

Answer 1

N is the reactional force on rod by the ground C is the centre of mass of the rod .

f is the friction force $=\mu N$

Length of Rod $=AB=L$

The equation for translation of rod are

$Fcos\alpha =mg-N---------\left(1\right)$

$Fsin\alpha =\mu N-----------\left(2\right)$

Now for rotational equilibrium torque about point Q should be balance

$Fcos\alpha .BD+Fsin\alpha .AD=mg\times BE$

So,$Fcos\alpha .Lcos\alpha +Fsin\alpha .sin\alpha =mg\left(\frac{L}{2}cos\alpha \right)$

$⟹FL(co{s}^2\alpha +si{n}^2\alpha )=\frac{mgL}{2}cos\alpha$

$F=\frac{mg}{2}cos\alpha$

Putting this value in one we get

$\frac{mg}{2}co{s}^2\alpha =mg-N$

$⟹N=\frac{mg}{2}(2-co{s}^2\alpha )$

$⟹N=\frac{mg}{2}(1+si{n}^2\alpha )$

Frictional force

$f=\mu N$

$f=\frac{\mu mg}{2}(1+si{n}^2\alpha )$

The direction of this force towards right to stop the sliding.