A rod of mass m is placed on two supports as shown- Calculate the normal reaction of each support- N-1 - 13mg-N-2-23mgN-1 - 23mg-N-2-13mgN-1 - mg2-N-2-mg2N-1 - mg4-N-2-3mg4

The correct option is A N_1 = 13mg, N_2=23mgAs th rod is in equilibrium, ∵ τ_A = 0 ∴ (N_1 × 0) + ( N_2 ×3L4) nbsp; ( mg ×L2) = 0 N_2 = 23mg As, N_1+N_2=mg ...

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Standard XII

Physics

The Complete Equilibrium

A rod of mass...

Question

A rod of mass $m$ is placed on two supports as shown. Calculate the normal reaction of each support.

N_{1} = \frac{1}{3} m g, N_{2} = \frac{2}{3} m g

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N_{1} = \frac{2}{3} m g, N_{2} = \frac{1}{3} m g

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N_{1} = \frac{m g}{2}, N_{2} = \frac{m g}{2}

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N_{1} = \frac{m g}{4}, N_{2} = \frac{3 m g}{4}

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Solution

The correct option is A $N_{1} = \frac{1}{3} m g, N_{2} = \frac{2}{3} m g$
As th rod is in equilibrium,
$∵ τ_{A} = 0$
$∴ (N_{1} \times 0) + (N_{2} \times \frac{3 L}{4}) - (m g \times \frac{L}{2}) = 0$
$N_{2} = \frac{2}{3} m g$
As, $N_{1} + N_{2} = m g$
and $N_{1} = \frac{1}{3} m g$

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A rod of mass m is placed on two supports as shown. Calculate the normal reaction of each support.

The correct option is A N1=13mg, N2=23mgAs th rod is in equilibrium, ∵ τA=0 ∴(N1×0)+(N2×3L4)−(mg×L2)=0 N2=23mg As, N1+N2=mg and N1=13mg

A rod of mass $m$ is placed on two supports as shown. Calculate the normal reaction of each support.

The correct option is A $N_{1} = \frac{1}{3} m g, N_{2} = \frac{2}{3} m g$
As th rod is in equilibrium,
$∵ τ_{A} = 0$
$∴ (N_{1} \times 0) + (N_{2} \times \frac{3 L}{4}) - (m g \times \frac{L}{2}) = 0$
$N_{2} = \frac{2}{3} m g$
As, $N_{1} + N_{2} = m g$
and $N_{1} = \frac{1}{3} m g$