Calculate force transmitted by the man on the floor of lift, Mechanical Engineering

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Calculate force transmitted by the man on the floor of lift:

A vertical lift having total mass 500Kg acquires an upward velocity of 2m/sec over a distance of 3m of motion with the constant acceleration, starting from rest. Determine the tension in the cable supporting lift. If lift while stopping moves with the constant deceleration and comes to rest in 2sec, calculate force transmitted by the man having mass 75kg on the floor of lift during the interval.

Sol: Given,

Mass of lift ML  = 500Kg

Final Velocity v = 2m/sec

Distance covered s = 3m

Initial velocity u = 0

Cable tension T = ?

Apply relation v2  = u2  + 2as

22=0+ 2a X 3a = 2/3 m/sec2                                                      ...(i)

As lift moves up, T > ML X g Net accelerating force = T - MLg, and is equal to,T - MLg = maT- 500 X 9.81 = 500 X 2/3

T = 5238.5N                                                                     .......ANS

Let force transmitted by the man having mass of 75Kg, is FF - mg = ma

For finding acceleration, by sing relation v = u + at0 = 2 + a X 2

219_Calculate force transmitted by the man on the floor of lift.png

 a = -1 m/sec2                                                                                                               ..(ii)

By putting value in equation, F - mg = ma

F - 75 X 9.81 = 75(-1)

F = 660.75N                                                             .......ANS

 


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