Question

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.

Answer 1

Given data;

Weight of empty lift W_L = 8KN = 8000N

Weight of man W_m = 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 F_m =?

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

(25)² = 0 + 2a(187.5)a = 1.67m/sec²

Cage moves down only when W_L + W_m >T

Net accelerating force = (W_L + W_m)- T

Using the relation F = ma, we get (W_L + W_m)- T = ma = [(W_L + W_m)/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,

F_m – W_m = maF_m – 600 = (600/9.81) X 1.67F_m = 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