A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a h

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6.System of Particles and Rotational Motion

medium

A rod of weight $W$ is supported by two parallel knife edges $A$ and $B$ and is in equilibrium in a horizontal position. The knives are at a distance $d$ from each other. The centre of mass of the rod is at distance $x$ from $A.$ The normal reaction on $A$ is

$\frac{{Wx}}{d}$

$\;\frac{{Wd}}{x}$

$\;\frac{{W\left( {d - x} \right)}}{x}$

$\;\frac{{W\left( {d - x} \right)}}{d}$

(AIPMT-2015)

Solution

Given situation is shown in figure.

${N_1}$ = Normal reaction on $A$

${N_2}$ = Normal reaction on $B$

$W$ $=$ Weight of the rod

In vertical equilibrium,

${N_1} + {N_2} = W$

Torque balance about center of mass of the rod,

${N_1}x = {N_2}\left( {d – x} \right)$

Putting value of ${N_2}$ from equation $(i)$

${N_1}x = \left( {W – {N_1}} \right)(d – x)$

$\Rightarrow {N_1}x = Wd – Wx – {N_1}d + {N_1}x$
$\Rightarrow {N_1}d = W\left( {d – x} \right)$
$\therefore {N_1} = \frac{{W\left( {d – x} \right)}}{d}$

Standard 11

Physics

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hard

A uniform disc with mass $M=4\,kg$ and radius $R=$ $10\,cm$ is mounted on a fixed horizontal axle as shown in figure. A block with mass $m =2\,kg$ hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in the cord is_______ $N$

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hard

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medium

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