A piece of metal was put on a elevator to travel from the 5th floor to the groun

Question

A piece of metal was put on a elevator to travel from the 5th floor to the ground. Figure 1 shows the vertical velocity v of the elevator during the journey. The metal Kept on the same position in the elevator in the time shown. The total mass of the lift and the metal is 1500kg. Take g = 10ms-2.

a) The elevator is suspended from above by several cables, Sketch a graph to show how the total tension T in the cables changes  with time.

b)If unfortunately the elevator is out of control such that its downward speed reaches 12 ms-1, the lift will be braked by a device in a distance of 7.5m. Find the average net force on the elevator when the elevator is braking. Take the downward direction to be positive.

A piece of metal was put on a elevator to travel from the 5th floor to the ground. Figure 1 shows the vertical velocity v of the elevator during the journey. The metal Kept on the same position in the elevator in the time shown. The total mass of the lift and the metal is 1500kg. Take g = 10ms-2. The elevator is suspended from above by several cables, Sketch a graph to show how the total tension T in the cables changes with time. If unfortunately the elevator is out of control such that its downward speed reaches 12 ms-1, the lift will be braked by a device in a distance of 7.5m. Find the average net force on the elevator when the elevator is braking. Take the downward direction to be positive.

Explanation / Answer

Since acceleration is defined as the rate of change of velovity, this the acceleration of elevstor from t = 0 to 3 is zero.

from t = 3 to 5 the acceleration will be given by the slope of the curve i.e.(4-0)/(5-3) = 2 m/ss

From t = 5 to 8 again there is no change in velocity, thus acceleration is zero

for t = 8 to 9.5 it is (0-4)/(9.5 -8) = -2.67 m/ss (negative sign denotes that the elevator is decelerating)

for t = 9.5 onwards, there is no change in velocity, hence no acceleration.

The acceleration calculated above is done agains the acceleration of gravity, hence a force due to gravity always acts in the downward direction as shown in the diagram:

Here T shows the tention, a is the acceeleration of the elevator and Mg is the net force due to gravity on the elevato+box system.

Thus the equation of motion of the system will be:

Mg-T = Ma

(with downward as positive direction)

HEnce for t = 0 to 3, a = 0

or

T = 1500*10 - 0 = 15000 N

for t = 3 to 5, a = 2 m/ss

or

T = 1500(10-2) = 12000 N

for t = 5 to 8, a = 0

or

T = 15000 N

for t = 8 to 9.5, a = 2.67 m/ss

or

T = 1500 (10-2.67) = 10995 N

and for last phase, t = 9.5 onwards, a = 0

or T = 15000 N

the curve for tention in the cable would be:

b)

Downward velocity, u = 12 m/s

it would be braked to a stop in a distance of s = 7.5 m

the final velocity after stopping will be, u = 0 m/s

using the third equation of motion:

v^2 = u^2 - 2as

thus a = u^2/(2a) = 9.6 m/ss

SInce it will do work against gravity thus net acceleration = 9.6 + 10 = 19.6 m/ss

Hence force applied = 19.6*1500 = 29400 N

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