The rod shown in figure has a mass of `2kg` and length `4m`. In equilibrium, find the hinge force (or its two component) acting on the rod and tension in the string. Tage `g=10 m//s^(2) , sin37^(@) = (3)/(5)` and `cos 37^(@)=(4)/(5)`

Answer

A, B, C

Explanation

In the figure , only those forces which are acting on the rod has been shown. Here `H` and `V` are horizontal and vertical components of the hinge force.

`sumF_(x) =0 rArr H-0.8T=0` ...(i)
` sumF_(Y) =0 rArr V+0.6T-20=0` ...(ii)
`sum pi_(0)=0`
`rArr` Clockwire torque of `20N=`anticlock-wire torque of `0.6T`.
All other forces`(H,V and 0.8T` pass through `O`, hence their torque are zero).
`20xx 2=0.6Txx4` ...(iii)
Solving these three equations, we get
`T=16.67N`,
`H=13.33N`
and `V=10N`
Hing force `(F)`
`F = sqrt(13.33)^(2)+(10)^(2) = 16.67 N`
`tan theta = (10)/(13.33)`
`theta = tan^(-1)((10)/(13.33)) = 37^(@)`