A uniform rod of length $L$ and mass $1.8\ kg$ is made to rest on two measuring scale at its two ends. A uniform block of mass $2.7\ kg$ is placed on the rod at a distance $L/4$ from the left end. The force experienced by the measuring scale on the right end is

Question

Dr. Abubacker · Accepted Answer

(a) Mass of a rod, $m = 1.8kg$
$	herefore$ Weight of a rod, $W = mg = 1.8kg 	imes 10ms^{- 2} = 18\ N$
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As the rod is uniform, therefore weight of the rod is acting at its midpoint
Taking moments about $A$,
$27 	imes \frac{L}{4} + 18 	imes \frac{L}{2} = F 	imes L$
$\Rightarrow FL = \frac{L}{4}\lbrack 27 + 36brack = \frac{63L}{4} \Rightarrow F = \frac{63}{4} = 16N$

Q. A uniform rod of length $L$ and mass $1.8\ kg$ is made to rest on two measuring scale at its two ends. A uniform block of mass $2.7\ kg$ is placed on the rod at a distance $L/4$ from the left end. The force experienced by the measuring scale on the right end is

Explanation

Reviewed by Dr. Abubacker

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