A uniform metre rule pivoted at 60 cm carries a 5 N weight at 20 cm. What single force applied at the 100 cm end will balance the rule? Ignore the rule's own weight.
A5 N upward
B5 N downward
C2.5 N downward
D2.5 N upward
Answer & Solution
Correct answer: B. 5 N downward
1. Pivot at 60 cm. The 5 N weight is at 20 cm, so its arm is 60 − 20 = 40 cm on the left.
2. The 5 N weight pulls down on the left of the pivot, creating an anticlockwise moment of 5 × 40 = 200 N·cm.
3. The 100 cm end is 100 − 60 = 40 cm to the right of the pivot.
4. A downward force F at the 100 cm end creates a clockwise moment of F × 40.
5. Balance: F × 40 = 200, so F = 5 N.
6. Direction must be downward to oppose the anticlockwise turn.
7. Hence the answer is 5 N downward.
_Source: Selina Concise Physics ICSE Class 10, Ch 1 'Force' (icsehelp.com chapter extract)_
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