A 274-N plank, of length L = 6.23 m, rests on the ground and on a frictionless roller at the top of a wall of height h = 2.87 m (see Fig). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ

Question

Aditi Chauhan · Answer

The plank is in rotational equilibrium; therefore the net torque about the point say O must be equal to zero, that is,

The plank is in horizontal equilibrium, therefore the frictional force must be equal to the horizontal component of normal force N , that is

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A 274-N plank, of length L = 6.23 m, rests on the ground and on a frictionless roller at the top of a wall of height h = 2.87 m (see Fig). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ

A 274-N plank, of length L = 6.23 m, rests on the ground and on a frictionless roller at the top of a wall of height h = 2.87 m (see Fig). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ

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A 274-N plank, of length L = 6.23 m, rests on the ground and on a frictionless roller at the top of a wall of height h = 2.87 m (see Fig). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ

A 274-N plank, of length L = 6.23 m, rests on the ground and on a frictionless roller at the top of a wall of height h = 2.87 m (see Fig). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ

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