A 690 N physics student stands on a bathroom scale in an 850-kg (including the s

Question

A 690 N physics student stands on a bathroom scale in an 850-kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 590 N. Find the acceleration of the elevator (magnitude and direction -- take up to be the positive direction). What is the acceleration if the scale reads 810 N? If the scale reads zero, what is the acceleration of the elevator? Should the student worry? Explain your answer. What is the tension in the cable in parts (a) and (c)?

Explanation / Answer

on a weighing scale, we read the normal force as the reading of the scale.

when the elevator is going down with an acceleration of a m/s^2,

mass*g-normal force=mass*acceleration

==>normal force=mass*(g-a)

here if a is positive , the elevator is moving down and speed is increasing in downward direction (that is positive acceleration in downward direction)

part a:

given that m*g=690 N

m=690/9.8=70.408 kg

reading of the scale=normal force=590 N

==>590=70.408*(9.8-a)

==>9.8-a=8.3797 m/s^2

==>a=1.4203 m/s^2

as the acceleration is in downward direction and the question considers the acceleration in upward direction to be positive,

answer will be acceleration=-1.4203 m/s^2

part b:

810=70.408*(9.8-a)

==>a=-1.7044 m/s^2

as upward is positive, answer will be 1.7044 m/s^2.

part c:

for scale reading to be zero, a=g=-9.8 m/s^2 (-ve sign because g is directed in downward direction)

part d:

student has to be worried, because the elevator is in free fall and there is no normal force from the ground to support him.

part e:

in part a:

let tension in the cable be T.

as the elevator is moving down , writing force balance equation in downward direction:

mass of the elevator (including the student)*g-T=mass of the elevator*acceleration of the elevator

==>850*9.8-T=850*1.4203

==>T=850*(9.8-1.4203)=7122.7 N

in part c,

850*9.8-T=850*9.8

==>T=0

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