A uniform meter scale of mass 0.24 \text{ kg} is made of steel. It is kept on two wedges, W_{1} and W_{2}, in a horizontal position. W_{1} is at a distance of 0.2 \text{ m} from one of its ends, while W_{2} is at a distance of 0.4 \text{ m} from the other end. If the force on the scale is N_{1} due to W_{1} and N_{2} due to W_{2}, then : (take g=10.0 \text{ m s}^{-2})

  1. A. N_{1}=1.6 \text{ N} and N_{2}=0.8 \text{ N}
  2. B. N_{1}=0.8 \text{ N} and N_{2}=1.6 \text{ N}
  3. C. N_{1}=0.6 \text{ N} and N_{2}=1.8 \text{ N} ✓
  4. D. N_{1}=1.8 \text{ N} and N_{2}=0.6 \text{ N}

Correct Answer: C. N_{1}=0.6 \text{ N} and N_{2}=1.8 \text{ N}

Explanation

Balancing torques about the center of mass (at 0.5 m) gives 0.3 N_1 = 0.1 N_2. Balancing vertical forces gives N_1 + N_2 = mg = 2.4 \text{ N}. Solving these equations yields N_1 = 0.6 \text{ N} and N_2 = 1.8 \text{ N}.

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