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Question
A uniform metre rule of mass 100 g is balanced on the fulcrum at mark 40 cm by suspending an unknown mass m at the mark 20 cm.
- Find the value of m.
- To which side the rule will tilt if the mass m is moved to the mark 10 cm?
- What is the resultant moment now?
- How can it be balanced by another mass of 50 g?
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Solution
i. The 100 g weight of the rule produces a clockwise motion around the knife edge. Using the principle of moments, O. Wt. mg generates an anticlockwise moment.
Anticlockwise moment = Clockwise moment
(40 − 20) × M g = 100 g (50 − 40)
20 M g = 100 × 10
M = `(100 xx 10)/20`
= 50 g
ii. Suppose mass M is shifted to the left, or to the 10 cm mark. The rule will tilt to the left, or M's side.
iii. When M (50 g) is moved to left from 20 cm mark to E 10 cm mark by 10 cm.
Resultant moment now = M g × 10 cm
= (50 × 10) gf × cm
= 500 gf × cm anticlockwise
iv.
Let a mass of 50 g be suspended at a distance of x cm from O to the right side to balance the mass of 50 g at the 10 cm mark.
∴ Clockwise moments = Anticlockwise moment
100 × (50 − 40) + (50 × x) = 50 (40 − 10)
100 × 10 + 50x = 50 × 30
1000 + 50x = 1500
50x = 1500 − 1000
50x = 500
x = `500/50 = 10` cm to the right from the 40 cm mark, i.e., at 50 cm mark.
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How can it be balanced by another mass 50 g ?
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