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#### A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?

8 viewed last edited 1 year ago An0nym0us
0  Krishna
0

Given that

Mass of the block M = 25kg

Mass of the man m = 50 kg

Acceleration due to gravity g = 9.8 m/s^2

Normal force required to floor yielding N_y = 700 N

Case 1: A man's action on the floor

When a man lifts a block, he is directly applying force in the upward direction. Causing the weight of the mass to grow. Normal force and force applied to the block act vertically upward against the man's and block's weight.

Normal force applied on the floor = weight of the man + weight of the block

N = mg + Mg

N = (25)* 9.8 m/s^2 + 50 * 9.8m/s^2

N = 245 + 490

N = 735 N

Case 1: A man's action on the floor In this case weight of the man is balanced by the weight of the block since we are applying force on the block in downward direction using the pulley.

Normal force applied on the floor = weight of the block -  weight of the man

N = Mg - mg

N=50\ m\ *9.8m/s^2\ -\ (25\ m\ )*9.8m/s^2

N = 490 - 245

N = 245 N

Choosing the mode to lift the block

Normal force applied on the floor in first case N = 735 N

Normal force required to floor yielding N_y = 700 N

N > N_y (735 > 700),  which will lead to floor yielding,

Normal force applied on the floor in first case N = 245 N

N < N_y (245< 700), which will not causing floor yielding, So Case 2 should adopt to lift the block.