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Spring in Elevator

A spring of unstretched length 1 m is suspended inside a stationary lift and is extended by 5 cm when a mass M hangs on it. If the lift now accelerates upwards at 4 times the acceleration of free fall, what will be the length of the spring?

Is the net force 4Mg during acceleration? Can someone explain a free body diagram with the question?

Reply 1

Morbo

No need for a free body diagram really.

In the stationary lift, normal gravity acts on the spring. Assuming the spring's mass is negligible compared to M, then the force on the spring is:

F = Mg

Invoking Hooke's law, tells us the spring constant, k:
Mg = kx --> k = Mg/(5 cm)

Now when the lift accelerates upwards at 4g, this means there is an extra fictious force (in the accelerating frame) of 4Mg acting upon the spring:
F_new = mg + 4Mg = 5Mg

And again using Hooke's law, we find the new extension:
5Mg = kx_new ---> x_new = 5Mg/k = 5Mg/(Mg/[5 cm]) = 25cm.

Thus the new length is 1m + 25cm = 1.25m

Reply 2

wickeddude12

I don't understand this part:

Now when the lift accelerates upwards at 4g, this means there is an extra fictious force (in the accelerating frame) of 4Mg acting upon the spring:
Fnew = mg + 4Mg = 5Mg

isn't Mg in the opposite direction of 4Mg?

Reply 3

Morbo

Remember/imagine what happens when you are standing in a lift that accelerates upwards. Does it feel as though there is additional gravity, or less gravity? What does this mean for the direction of the ficticious force?

Reply 4

wickeddude12

Would this idea of a FBD be correct?

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Reply 5

Morbo

No. That's not right. An upwards accelerating elevator introduces a downwards ficticious force in the elevator. That's why you feel heavier when the elevator starts to go up.

Mg acts downwards on the mass from gravity, whether the elevator is stationary or otherwise. Then, if the elevator accelerates upwards at 4g, this makes the effective feeling of gravity in the elevator increase by 4g.

g + 4g = 5g