Question

A uniform ladder AB rest against a smooth wall at B and among a rough ground at A. a boy whose weight is twise of the ladder climbs. prove that the force of the friction at the top of the ladder B is 5 as great as when he is at the bottom.

Answer 1

answer

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explanation Let m and M be mass of man and ladder.

From FBD normal reaction at A is N

1

=(m+M)g which remains constant as the man ascends.

Thus C is correct.

Net torque of all the forces is zero about B. As shown in the figure if x decreases(as the man climbs up) then torque of mg about B will decrease. To balance this out the opposite torque due to the frictional force at A should increase. As the distance between the forces(reaction force at B and friction force at A) is constant, the magnitude of frictional force f must increase.

Thus A and D are correct.