Question

A uniform rod of length $L$ and mass $2M$ rests over two supports $A$ and $B$. A particle having mass, $M$ is placed at the right end. Determine the reactions at supports $A$ and $B$.

Answer 1

$R_A+R_B=2Mg+Mg=3mg$

Taking moment about A , $M_A=0$

$\Rightarrow Mg-R_B\left(\frac{2L}{3}\right)+2Mg\left(\frac{L}{2}\right)=0$

$\Rightarrow \frac{2R_B}{3}=2Mg$

$\Rightarrow R_B=3Mg$ (N)

$\Rightarrow R_A=0$ (N)

Hence, solved.