Question

A uniform bar of length l m is suspended at its ends and loaded by a weight W kgf at its middle.In equilibrium ,find the reactions of R1 and R2 at its both ends

Answer 1

For mechanical equilibrium, R1 + R2 = W
For rotational equilibrium, R1(l/2) + W(0) -R2(l/2) = 0 => R1 = R2

Therefore, R1 = R2 = W/2

Answer 2

The reactions of $\bold{R_{1} and R_{2}}$ is $\bold{\frac{W}{2}}$

Solution:

As given the uniform bar has a length l which has mass suspended at its both the ends and is loaded by a weight as W. the system given is in equilibrium, thus the reaction at both the end is given by,

Given: Length = l   Mass = m Load = W

Reaction at one end = $R_{1}$

Reaction at other end = $R_{2}$

For the mechanical equilibrium to occur,

$R_{1} \text { and } R_{2}$ should be balanced by Weight.

Therefore, $R_{1}+R_{2}=W$  

And the weight is suspended in between at the position of $\frac{l}{2}$  

We can say that the $R_{1}$  is balanced by $\frac{W}{2}$  and $R_{2}$ is balanced by $\frac{W}{2}$

$\begin{array}{l}{R_{1}=\frac{W}{2} \text { and } R_{2}=\frac{W}{2}} \\ {R_{1}=R_{2}=\frac{W}{2}}\end{array}$