Question

The diagram in Fig. 1.34 shows a uniform metre rule
weighing 100 gf, pivoted at its centre O. Two weights
150 gf and 250 gf hang from the points A and B
respectively of the metre rule such that OA = 40 cm
and OB = 20 cm. Calculate : (i) the total anticlockwise
moment about O, (ii) the total clockwise moment about
O, (iii) the difference of anticlockwise and clockwise
moments, and (iv) the distance from O where a 100 gf
weight should be placed to balance the metre rule.
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Answer 1

Hey there!

i) Total aniticlockwise moment = Force * perpendicular distance

                                                    = 150 * 40 = 6000 gf cm

ii) Total clockwise moment = Force * perpendicular distance

                                             = 250 * 20 = 5000 gf cm

iii) Difference between anticlockwise and clockwise moment :

= Anticlockwise - clockwise

= 6000 - 5000

= 1000 gf cm

iv) For a body in equilibrium,

Anticlockwise moment = clockwise moment

6000 = 5000 + 100 * d

6000 - 5000 = 100 * d

1000 / 100 = d

d = 10 cm

Hence, 100 gf should be placed on 10 cm right side of O to balance the meter scale.

Answer 2

Answer:

Here's the Answer

Explanation:

Hope it helps you!