Answer :
Final answer:
The force exerted on the plank by the second pivot is option (d) 131 N, derived from the principle of static equilibrium in Physics, where the sum of all forces and torques equals zero.
Explanation:
This question is a classic example of static equilibrium in Physics. The entire system here is in static equilibrium. That means the sum of all forces and torques about any point equals zero. The plank is in equilibrium, so the sum of the forces equals 0: that is, the total downward force due to the weight of the plank balances out with the total force exerted by the pivots. You have a pivot at the left end which we can take as pivot A and another one 1 m from the right end which we'll call pivot B. The weight of the plank, being centrally distributed, acts at the center of the plank, which is 2 m from the left end. Let's denote the force from pivot B as 'F' and 'g' the acceleration due to gravity (g ~9.8 m/s^2). The weight of plank is equal to its mass times 'g': 20 kg * 9.8 m/s^2, equals 196 N. The horizontal plank is in a rotational equilibrium - so the sum of torques should be zero.
Therefore, Torque due to force B + Torque due to weight of plank = 0. Put in the distances and forces, we get F * (4m -1m) = 196N * 2m => F = (196N * 2m)/3m = 131 N.
Learn more about static equilibrium here:
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