Q.A 2 m long wooden plank of mass 20 kg is pivoted (supported from below) at 0.5 m from either end. A person of mass 40 kg starts walking from one of these pivots towards the farther end. How far can the person walk before the plank topples?
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Start your 14-day free trial to unlock the full solution →The plank is 2 m long, pivoted at 0.5 m from EACH end (so the two pivots are 1 m apart, symmetric about the plank's centre, which is also the plank's own centre of mass location). The plank's own weight (20 kg, acting at the exact centre, i.e. exactly midway between the two pivots) contributes ZERO net torque about either pivot on its own additional tipping tendency from the pivot symmetric position -- but as the person walks from the near pivot toward the far (unsupported) end, at some point the plank tips about the FAR pivot (the one nearer the person), lifting off the NEAR pivot. Taking torques about the far pivot (the one closer to the person's direction of travel) at the moment of just-about-to-topple (normal reaction at the near pivot = 0): let x be the distance the person has walked FROM the far pivot (i.e., toward the overhanging end beyond that pivot), so the person's weight (40 kg) acts at distance x beyond the far pivot, tending to topple the plank about it, while the plank's own weight (20 kg, acting at the plank's centre, which is m on the OTHER side of this same far pivot, since the pivot is 0.5 m from each end and the plank's centre is 1 m from that end) provides a restoring torque. Balancing at the tipping point: $40(x)=20(0.5)\Rightarrow x=\frac{10}{40 …
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