Question

Solve the numerical .

Answer 1

Answer:

Given:

A system has been provided consisting of Normal Reaction R1 and R2 , and weight of bar is W.

To find:

Value of R1 and R2.

Concept:

In this type of questions, we shall use the concept of translational and rotational equilibrium.

For Translational equilibrium :

$\boxed{R1 + R2 - W = 0}$

$= > R1 + R2 = W \: ........(1)$

For rotational equilibrium : we shall apply equality of torques

Considering axis of rotation at R1 :

$(R2) \times l = W \times ( \dfrac{l}{2} )$

$= > R2 = \dfrac{W}{2} ..........(2)$

Putting value of R/2 in 1st Equation :

$R1 + \dfrac{W}{2} = W$

$= > R1 = \dfrac{W}{2}$

So final answer :

$\boxed{ \red{R1 = R2 = \dfrac{W}{2}}}$