A uniform beam $A O B$ , $O$ being the mid point of AB, mass $M$ and length $L$ rests on three identical vertical springs with stiffness constant $k_{1}$ , $k_{2}$ and $k_{3}$ at A, $O$ and $B$ respectively. The bases of the springs are fixed to a horizontal platform. Assume that springs remain vertical.

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the compression in each spring is $\frac{M g}{4 k}$

the torque of spring $3$ about $A$ is $\frac{M g l}{2}$ in anticlockwise direction

the torque of spring $2$ about $B$ is $\frac{M g l}{4}$ in anticlockwise direction

the force exerted by spring $2$ is twice that exerted by spring $1$

answer is A, B.

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Detailed Solution

x-same

x= $\Rightarrow F_{1} = k x$ $F_{2} = 2 k x$

$k x + k x + 2 k x = M g$

$x = \frac{M g}{4 k}$

$τ_{A} = k x \times L = \frac{M g}{4} \times L$

$τ_{B} = \frac{M g}{2 k} \times \frac{L}{2} (c w)$

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