Determine tension in the rope - motion of lift, Mechanical Engineering

Determine tension in the rope:

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

Sol: 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 man Fm  =?

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

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

Cage moves down only when WL + Wm  >T

Net accelerating force = (WL  + Wm)- T

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

T = 7135.98N                                                   .......ANS

Calculation for the force exerted by the man. Consider the weight of man only, Fm - Wm = maFm  - 600 = (600/9.81) X 1.67Fm  = 714.37N

 

2048_Determine tension in the rope.png

As Newton's third law states that the force of action and reaction between interacting bodies are equal in magnitude but opposite in direction and have the same line of action.

That is, Force exerted by the man = F = 714.37N                                        ........ANS

Posted Date: 10/19/2012 7:48:46 AM | Location : United States







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