Non-Inertial Frame Homework: Find Lift Acceleration

[bon]

Homework Statement

A passenger holding a parcel mass m is standing in a lift which is being accelerated upwards by a constant force F. Total mass of life plus passenger is M.

First q asks: what is the accel of the lift?

Homework Equations

The Attempt at a Solution

So i know i use resultant force = ma

Is it: F - (M+m)g = (M+m)a i.e. do i use M+m for the mass of the lift? and for the m in mg?

thanks

 

[bon]

Because it later asks - If the passenger drops the parcel from height h, how long does it take to hit the floor?

So at this point does the accel of the lift change, since the mass is less? (i.e. M rather than M+m)?

 

[tiny-tim]

Because it later asks - If the passenger drops the parcel from height h, how long does it take to hit the floor?

So at this point does the accel of the lift change, since the mass is less? (i.e. M rather than M+m)?

Hi bon!

In these problems, you always want the system on which all the external forces are known.

At first, that's the lift pasenger and parcel … leave any out, and the reaction force becomes an external force, whose value you don't know.

When the parcel is dropped, the accelerations of course are different, and the reaction force to/from the parcel vanishes.

 

[bon]

Hi bon!

In these problems, you always want the system on which all the external forces are known.

At first, that's the lift pasenger and parcel … leave any out, and the reaction force becomes an external force, whose value you don't know.

When the parcel is dropped, the accelerations of course are different, and the reaction force to/from the parcel vanishes.

Ok thanks so are the results as follows:

Q1) Accel of lift: F/(M+m) - g

Q2) Apparent weight of parcel: mF/(M+m)

Q3) if the passenger drops the parcel from height h how long does it take to hit floor?

So accel changes to F/M - g thus it takes root(2Mh/F) seconds?

Thanks!

 

[tiny-tim]

Hi bon!

(have a square-root: √ )

Q2) Apparent weight of parcel: mF/(M+m)

No … if F was zero, the apparent weight would be mg, wouldn't it?

Q3) if the passenger drops the parcel from height h how long does it take to hit floor?

So accel changes to F/M - g

Yes, acceleration of the lift is F/M - g.

… thus it takes root(2Mh/F) seconds?

No (again, doesn't work for F = 0, does it?).

 

[bon]

Hi bon!

(have a square-root: √ )No … if F was zero, the apparent weight would be mg, wouldn't it?


Yes, acceleration of the lift is F/M - g.No (again, doesn't work for F = 0, does it?).

No If F was 0 the lift would be in free fall

therefore apparent weight would =0 and if it was let go it would never hit the ground..

I think my equations are right? Anyone else?

 

[tiny-tim]

No If F was 0 the lift would be in free fall

That's not the way I read the question …

"a lift which is being accelerated upwards by a constant force F" …

I read that as a pushing force, separate from gravity.

 

[bon]

That's not the way I read the question …

"a lift which is being accelerated upwards by a constant force F" …

I read that as a pushing force, separate from gravity.

But I think the point is that if that force was not there then the only force would be gravity, hence it would be free falling..

 

[tiny-tim]

If the lift was being hauled up on a rope, the tension force would be more than g, and the person/machine pulling would be pulling with that tension force.

If a question said "a box is being pulled up a smooth slope with a force F", would you assume that an mgsinθ had been subtracted from the tension to give F?

Perhaps you'd better check with whoever set the question?​

 

[ideasrule]

If F=mg, the elevator could be at a standstill. Try to make your equations so that if F=mg, the normal equations of motion (a=g, t=sqrt(2H/g)) are recovered.

You shouldn't expect your equations to work if the elevator is accelerating downwards because in your solution, you assumed the elevator was accelerating upwards. (In fact, most of the time they won't work.)