Question:
A uniform meter scale of mass 0.24 kg is made
of steel. It is kept on two wedges, W1 and W2, in
a horizontal position. W1 is at a distance of 0.2
m from one of its ends, while W2 is at distance
of 0.4 m from the other end. If the force on the
scale is N1 due to W1 and N2 due to W2, then :
(take g = 10.0 m s–2)
A uniform meter scale of mass 0.24 kg is made
of steel. It is kept on two wedges, W1 and W2, in
a horizontal position. W1 is at a distance of 0.2
m from one of its ends, while W2 is at distance
of 0.4 m from the other end. If the force on the
scale is N1 due to W1 and N2 due to W2, then :
(take g = 10.0 m s–2)
Options:
- N1 = 1.6 N and N2 = 0.8 N
- N1 = 0.8 N and N2 = 1.6 N
- N1 = 0.6 N and N2 = 1.8 N
- N1 = 1.8 N and N2 = 0.6 N
- N1 = 1.6 N and N2 = 0.8 N
- N1 = 0.8 N and N2 = 1.6 N
- N1 = 0.6 N and N2 = 1.8 N
- N1 = 1.8 N and N2 = 0.6 N
Correct Answer: C) N1 = 0.6 N and N2 = 1.8 N
Explanation:
Net torque about W1: N2*(0.6) – W*(0.3) = 0 => N2 = 0.5 * W = 0.5 * (0.24 * 10) = 1.2 N.
Wait, let’s verify torque:
Meter scale center of gravity is at 0.5m.
W1 is at 0.2m. W2 is at 0.4m from the other end (which means at 0.6m).
CG (0.5m) is between W1 (0.2m) and W2 (0.6m).
Distance from W1 to CG is 0.3m. Distance from W1 to W2 is 0.4m.
Torque about W1: N2 * 0.4 = W * 0.3 => N2 = 0.75 * W = 0.75 * 2.4 = 1.8 N.
N1 + N2 = 2.4 N => N1 = 0.6 N. So option (c) is correct.
Net torque about W1: N2*(0.6) – W*(0.3) = 0 => N2 = 0.5 * W = 0.5 * (0.24 * 10) = 1.2 N.
Wait, let’s verify torque:
Meter scale center of gravity is at 0.5m.
W1 is at 0.2m. W2 is at 0.4m from the other end (which means at 0.6m).
CG (0.5m) is between W1 (0.2m) and W2 (0.6m).
Distance from W1 to CG is 0.3m. Distance from W1 to W2 is 0.4m.
Torque about W1: N2 * 0.4 = W * 0.3 => N2 = 0.75 * W = 0.75 * 2.4 = 1.8 N.
N1 + N2 = 2.4 N => N1 = 0.6 N. So option (c) is correct.