Question

A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 gram mass be attached so that the meterstick will be balanced in rotational equilibrium

Answer 1

The 40g mass will be attached at 69 cm

Step-by-step explanation:

First, make a sketch of the meterstick with the masses placed on it;

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↓ Δ ↓

20 g.................50 cm.................40g

38 cm y cm

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

12cm 50 cm 69cm

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↓ Δ ↓

20 g.................50 cm.................40g

38 cm 19 cm