A uniform rod of length 1 m is pivoted at its midpoint. If the weight of 40 gf is suspended at one of its ends. Where should the weight of 120 gf be suspended to keep the rod horizontal?
A uniform rod of length 1 m is pivoted at its midpoint. If the weight of 40 gf is suspended at one of its ends. Where should the weight of 120 gf be suspended to keep the rod horizontal? Correct Answer 16 cm from the midpoint
The correct answer is option 1) i.e. 16 cm from the midpoint
CONCEPT:
- Equilibrium is a condition when the body is not undergoing a change in rotational or translational motion.
- The conditions for equilibrium of a body are as follows:
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The algebraic sum of the forces must be zero.
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The algebraic sum of the moments of the forces about any point in their plane must be zero.
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CALCULATION:
- Let us assume 100 gf is located at distance x from the pivoted point.
- If the rod has to stay horizontal, the system must be in equilibrium.
- Therefore, the algebraic sum of the moments of the forces about the pivoted point must be zero.
ΣM = 0
⇒ 40(0.5) - 120(x) = 0
⇒ 20 = 120x
⇒ x = 0.16 cm
- Therefore, the weight of 120 gf must be suspended 16 cm from the midpoint.
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Feb 20, 2025
