A uniform rod of length `1m` having mass `1 kg` rests against a smooth wall at an angle of `30^(@)` with the ground. Calculate the force exerted by the ground on the rod. Take `g = 10 ms^(-2)`.

Explanation

In Fig `AB` is a uniform rod of length `1 m` with centre at `C, /_BAE = 30^(@), AC = 0.5m`

The various force involved are :
(i) Weight `mg` of the rod acting vertically downwards at `C`.
(ii) Horizontal force `F` due to the wall at `B`.
(iii) Reaction of floor `f` acting along `AB`. This is the force extered by the ground on the rod. It is the resultant of normal reaction `f_(y)` of floor and force of friction `f_(x)` along `AE` (that prevents slioding away of the rod from the wall).
In eqilibrium,
(i) net force on the rod along the ground is zero.
`f_(x) - F = 0 or f_(x) = F` ...(1)
(ii) net force on the rod perpendicular to the ground is zero.
`f_(y) - mg = 0 or f_(y) = mg = 1 xx 10 = 10N`
(iii) Algebraic sum of moments of force about `A = 0`
`F xx BE = mg xx AD`
`F xx (AB sin 30^(@)) = mg xx AC cos 30^(@)`
`F xx 1 xx (1)/(2) = 1 xx 10 xx 0.5 xx (sqrt(3))/(2)`
`F = 5sqrt(3) = 8.66N`
From (1), `f_(x) = F = 8.66N`
Force exerted by the ground on the rod
`f = sqrt(f_(x)^(2) + f_(y)^(2)) = sqrt(8.66^(2) + 10^(2)) = 13.22 N`