Question

A 557-N physics student stands on a bathroom scale in an 859-kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 496 N

Find the direction of the acceleration of the elevator.

What is the acceleration if the scale reads 682 N ?

If the scale reads zero, should the student worry? Explain.

Essay answers are limited to about 500 words (3800 characters maximum, including spaces).

3800 Character(s) remaining

What is the tension in the cable in part A?

What is the tension in the cable in part D?

Answer 1

Part A.

Using Net force balance on student:

Fy_net = m*ay

N - W = m*ay

N = normal Force on student OR apparent weight

W = weight of student = 557 N = m*g

m = mass of student = W/g = 557/9.81 = 56.78 kg

Now

ay = (N - W)/m = (496 - 557)/56.78

ay = -1.07 m/sec^2

So negative sign means acceleration will be downward.

Part B.

When N = apparent weight = 682 N

ay = (N - W)/m = (682 - 557)/56.78

ay = 2.20 m/sec^2

Part C.

If the scale reads zero, which means N = 0

ay = (0 - W)/m = -W/m

Since W = m*g

ay = -m*g/m

ay = -g

Which means that student should worry, as the elevator is in free fall.

Part D.

I think your part A was 'the acceleration when scale reads 496 N'

So in that case Using force balance in vertical direction

Fnet = T - M*g = M*ay

T = M*(ay + g) = Tension in cable

M = mass of elevator + person = 859 kg

T = 859*(-1.07 + 9.81)

T = 7507.66 N

Part E.

when ay = -g

T = M*(ay + g)

T = M*(-g + g) = M*0

T = 0 N

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