Question

1. Turning force
a) A uniform metre rod is balanced on a sharp knife edge at its centre. A mass of 0.02 kg is suspended a
0.30 m mark. Where another weight of 0.05 kg should be suspended so that the rod remains
horizontal. Ans: 0.12 m)
02
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Answer 1

A mass of 0.02kg is suspended a 0.3m mark. Let another weight of 0.05kg should be suspended x m mark so that rod remains horizontal.

rod remains horizontal only when net torque acting on rod equals zero.

we know, torque = vector product of Force and displacement.

so, 0.02kg × 10m/s² × 0.3m (→) + 0.05kg × 10m/s² × x m (→) = 0

or, 0.02 × 0.3 = -0.05 × x

or, x = -0.12 m

negative sign indicates that another mass is suspended opposite side of first mass on rod.

so, answer is 0.12m

Answer 2

various forces acting on the metre rod is shown in fig

P= 0.02 * 9.8N & Q= 0.05 * 9.8N

Take moments about F.

Moment of P about F = P * AF = 0.02 * 9.8 * 0.3 Nm (anti-clockwise)

Moment of Q about F = Q * BF = 0.05 * 9.8 * x Nm (clockwise)

Moment of W about F = W * 0= 0

according to the principle of moments,

clockwise moments = anticlockwise moments

0.05 * 9.8 * X = 0.02 * 9.8 * 0.3

X = 0.02 * 0.3 / 0.05 = 0.12m

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