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An elevator cage of a mineshaft, weighing 8KN when, is lifted or lowered by means of a wire rope. Once a man weighing 600N, entered it and lowered with uniform acceleration such that when a distance of 187.5m was covered, the velocity of cage was 25m/sec. Determine the tension in the rope and the force exerted by the man on the floor of the cage.

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Given data;

Weight of empty lift WL = 8KN = 8000N

Weight of man Wm = 600N

Distance covered by lift s = 187.5m

Velocity of lift after 187.5m v = 25m/sec

Tension in rope T = ?

Force exerted on the man Fm =?

Apply the relation v2 = u2 + 2as, for finding acceleration.

(25)2 = 0 + 2a(187.5)a = 1.67m/sec2

Cage moves down only when WL + Wm >T

Net accelerating force = (WL + Wm)- T

Using the relation F = ma, we get (WL + Wm)- T = ma = [(WL + Wm)/g]a(8000 + 600) – T = [(8000 + 600)/9.81] X 1.67

T = 7135.98N

Calculation for force exerted by the man Consider only the weight of the man,

Fm – Wm = maFm – 600 = (600/9.81) X 1.67Fm = 714.37N

Since Newton’s third law i.e The force of action and reaction between interacting bodies are equal in magnitude, opposite in direction and have the same line of action. 

i.e., Force exerted by the man = F = 714.37N

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