Equilibrium And Torque Balance

Question

A uniform plank of negligible mass rests on a central pivot acting as a seesaw. A 30 kg child sits 1.5 m from the pivot on one side; how far from the pivot must a 45 kg child sit on the other side to keep the plank balanced?

RankGuru · Accepted Answer

Rotational equilibrium requires that the net torque about the pivot be zero, so the clockwise and anticlockwise torques must balance: $m_1 g d_1 = m_2 g d_2$. The gravitational field $g$ appears on both sides and cancels, leaving the simple condition $m_1 d_1 = m_2 d_2$. Substituting the data for the first child, $30 	imes 1.5 = 45 	imes d_2$, so $45 = 45 d_2$ and $d_2 = 1.0	ext{ m}$. The value 2.25 m reverses the proportionality, wrongly placing the heavier child farther out. The value 1.5 m ignores the mass difference and just copies the first child's distance. The value 0.5 m underestimates the distance by applying an incorrect ratio. This is the NCERT principle of moments applied to a balanced rigid body in static equilibrium. As a plausibility check, the heavier 45 kg child must sit closer to the pivot than the lighter 30 kg child in order to balance the torques, and indeed $1.0	ext{ m} < 1.5	ext{ m}$, confirming the result. Because the plank itself is massless and symmetric about the pivot it contributes no net torque, so only the two children's weights enter the balance condition, which is the very lever principle exploited in beam balances, crowbars, and steelyards to trade force for distance.

Equilibrium And Torque Balance

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