An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally

Question

2 years ago

8

moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42 m.

1 answer:

amid [387] · Answer 1 · 2023-07-02T20:56:32+03:00

The tension in the supporting cable when the cab originally moves downward is 18422.4 N

What is tension?

Tension is described as the pulling force by the means of a three-dimensional object.

Tension might also be described as the action-reaction pair of forces acting at each end of said elements.

Here,

m =combined mass = 1600 kg

s = Displacement of the elevator = 42 m

g = Acceleration due to gravity = 9.81 m/s²

u = Initial velocity = 12 m/s

v = Final velocity = 0

According to the equation of motion:

$v^{2} - u^{2} = 2as$

0 - 12^2 = 2*a*42

a = - 144 / 84

a = - 1.714 m/s^2

Now let's write the equation of the forces acting on the elevator. Taking upward as positive direction:

T-mg = ma

T = m(g-a)

T = 1600 ( 9.8-(-1.74))

T=18422.4 N

Hence,

The tension in the supporting cable when the cab, originally moving downward is 18422.4 N

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