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A weight of 5 kg is suspended 4 m left of a balance point. In order for the system to be in equilibrium, how heavy must another weight suspensed 8 m right of the same balance point be?
  • A
    10 kg
  • B
    1.66 kg
  • C
    20 kg
  • D
    2.5 kg

To obtain equilibrium on both sides of balance point:
The (mass x arm) on both sides should be equal.

5 x 4 = 8 x (unknown mass)
Unknown mass = 20 / 8 = 2.5 kg

Correct Answer: 2.5 kg

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