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Physics question- man in lift

Hi, please can i have help on this question? I thought the answer is D because if he is accelerating downwards, the non contact force will be decreasing and his weight will remain constant but is greater than the non contact force so he can accelerate downwards however the answer is D? Isn’t the weight of the man the ’force exerted by the man on the floor?’

Here is the question: https://app.gemoo.com/share/image-annotation/587433089591721984?codeId=M0Ge83YaVpW8o&origin=imageurlgenerator

Thank you!
Reply 1
Original post by anonymous294
Hi, please can i have help on this question? I thought the answer is D because if he is accelerating downwards, the non contact force will be decreasing and his weight will remain constant but is greater than the non contact force so he can accelerate downwards however the answer is D? Isn’t the weight of the man the ’force exerted by the man on the floor?’

Here is the question: https://app.gemoo.com/share/image-annotation/587433089591721984?codeId=M0Ge83YaVpW8o&origin=imageurlgenerator

Thank you!

The answer is D, similar to the reasons given in your other post. If the lift was stationary, the reaction force would be equal to the weight. If the lift was accelerating down at g, then the reaction force would be zero. Accelerating down < g, means the reaction is < weight.

What did you think it was?
(edited 3 months ago)
Reply 2
Original post by mqb2766
The answer is D, similar to the reasons given in your other post. If the lift was stationary, the reaction force would be equal to the weight. If the lift was accelerating down at g, then the reaction force would be zero. Accelerating down < g, means the reaction is < weight.

What did you think it was?

Sorry, i thought the answer was b using the reasoning i said in the post above..
Reply 3
Original post by anonymous294
Sorry, i thought the answer was b using the reasoning i said in the post above..

B is wrong as by newton 3, they are equal and opposite. They form a force pair.
https://www.physicsclassroom.com/class/newtlaws/Lesson-4/Identifying-Action-and-Reaction-Force-Pairs
Reply 4
Original post by mqb2766
B is wrong as by newton 3, they are equal and opposite. They form a force pair.
https://www.physicsclassroom.com/class/newtlaws/Lesson-4/Identifying-Action-and-Reaction-Force-Pairs

But isn’t the force exerted by the man on the floor the same as his weight?
Reply 5
Original post by anonymous294
But isn’t the force exerted by the man on the floor the same as his weight?

It would help to draw a proper diagram with the forces on as well as apply newton consistently.

So by newton 3, as its a force pair
force by man on floor = force by floor on man
Then by newton 2 on the man
ma = net force = weight - force by floor on man
Or
force by floor on man = weight - ma
The force by floor on man is only equal to the weight if a=0 (newton 1, in equilibrium). Otherwise its less than the weight and if a=g, then there is zero force by floor on man
Reply 6
Original post by mqb2766
It would help to draw a proper diagram with the forces on as well as apply newton consistently.

So by newton 3, as its a force pair
force by man on floor = force by floor on man
Then by newton 2 on the man
ma = net force = weight - force by floor on man
Or
force by floor on man = weight - ma
The force by floor on man is only equal to the weight if a=0 (newton 1, in equilibrium). Otherwise its less than the weight and if a=g, then there is zero force by floor on man

I almost get it but i just want to clarify what is the difference between the weight of the man and the force exerted by the man on the floor? Thanks!
Reply 7
Original post by anonymous294
I almost get it but i just want to clarify what is the difference between the weight of the man and the force exerted by the man on the floor? Thanks!

In the vomit comet lift, the reaction of the lift on the man is zero and the weight is mg. Theyre two completely different forces. The reaction forms a force pair at your feet and the weight is applied through your COM so effectively your stomach. Drawing a diagram of the person, the lift applies R upwards at your feet and gravity applies a force downwards (weight) at your COM. You resolve vertically and difference the two (opposite directions) to get the net force.

If youre in a stationary lift (or moving with constant velocity), gravity is exerting a force mg on the person (their weight). There is zero acceleration so newton 2 on the person gives
m*0 = weight - reaction = mg - reaction
so
reaction = mg = weight. Which is what most people think of as the "usual" reaction.

However, if both the lift and the person was accelerating downwards at g/2, then youd have for the person
ma = weight - reaction
so
reaction = mg - mg/2 = mg/2
so the reaction is half the previous value.

Alternatively if you were accelerating upwards at g/2 so a=-g/2 (positive downwards), then youd have (for the person) similar to above
reaction = mg + mg/2 = 3mg/2
so the floor is pushing you up harder than gravity is pulling you down and the net force and acceleration is upwards. If youre in a rapidly accelerating lift (up or down) you physically feel this effect.
(edited 3 months ago)
Reply 8
Original post by mqb2766
In the vomit comet lift, the reaction of the lift on the man is zero and the weight is mg. Theyre two completely different forces. The reaction forms a force pair at your feet and the weight is applied through your COM so effectively your stomach. Drawing a diagram of the person, the lift applies R upwards at your feet and gravity applies a force downwards (weight) at your COM. You resolve vertically and difference the two (opposite directions) to get the net force.

If youre in a stationary lift (or moving with constant velocity), gravity is exerting a force mg on the person (their weight). There is zero acceleration so newton 2 on the person gives
m*0 = weight - reaction = mg - reaction
so
reaction = mg = weight. Which is what most people think of as the "usual" reaction.

However, if both the lift and the person was accelerating downwards at g/2, then youd have for the person
ma = weight - reaction
so
reaction = mg - mg/2 = mg/2
so the reaction is half the previous value.

Alternatively if you were accelerating upwards at g/2 so a=-g/2 (positive downwards), then youd have (for the person) similar to above
reaction = mg + mg/2 = 3mg/2
so the floor is pushing you up harder than gravity is pulling you down and the net force and acceleration is upwards. If youre in a rapidly accelerating lift (up or down) you physically feel this effect.

Ohh ok but is the mans weight and the force exerted on the floor opposite to each other because don’t they need to be opposite to calculate the resultant force? Thank you for taking the time to explain this sorry :smile:
Reply 9
Original post by anonymous294
Ohh ok but is the mans weight and the force exerted on the floor opposite to each other because don’t they need to be opposite to calculate the resultant force? Thank you for taking the time to explain this sorry :smile:

The previous post was considering the forces acting on the person.

As in #5, there is a force pair (newton 3) at their feet as they remain in contact with the lift. So the force exerted by the person on the lift is equal and opposite to the force exerted by the lift on the person. This force pair and the persons weight are two completely different forces. The former is a contact force (pair) and the latter is contactless (gravity). You analyse the forces acting on the person, then make the argument about the force pair to get the force the person applies to the lift.

The weight acts downwards and the lift pushes upwards on the person. If the reaction force becomes zero (vomit comet), the lift loses contact with the persons feet and the force model is wrong.

If youre still unsure, draw a force diagram of what you think and write the (newton, simple) equations on it and upload.
(edited 3 months ago)
Original post by mqb2766
The previous post was considering the forces acting on the person.

As in #5, there is a force pair (newton 3) at their feet as they remain in contact with the lift. So the force exerted by the person on the lift is equal and opposite to the force exerted by the lift on the person. This force pair and the persons weight are two completely different forces. The former is a contact force (pair) and the latter is contactless (gravity). You analyse the forces acting on the person, then make the argument about the force pair to get the force the person applies to the lift.

The weight acts downwards and the lift pushes upwards on the person. If the reaction force becomes zero (vomit comet), the lift loses contact with the persons feet and the force model is wrong.

If youre still unsure, draw a force diagram of what you think and write the (newton, simple) equations on it and upload.

No that makes a lot more sense, thank you very much :smile:
Reply 11
Original post by anonymous294
Hi, please can i have help on this question? I thought the answer is D because if he is accelerating downwards, the non contact force will be decreasing and his weight will remain constant but is greater than the non contact force so he can accelerate downwards however the answer is D? Isn’t the weight of the man the ’force exerted by the man on the floor?’

Here is the question: https://app.gemoo.com/share/image-annotation/587433089591721984?codeId=M0Ge83YaVpW8o&origin=imageurlgenerator

Thank you!

The question asks about the force on the floor, not the man's weight. The lift is accelerating downward, if the downward acceleration was g, there would be no weight on the floor at all.
Looking at your discussion about "non contact force" I assume you mean his weight due to gravity. I really think the disccusion of "non-contact" force is causing you to over think the problem.
In general, in a question like this, think about yourself in a lift and consider a more extreme example, (e.g a lift accelerating downward at g) to get the feel of the problem.
Reply 12
Original post by anonymous294
But isn’t the force exerted by the man on the floor the same as his weight?

it is the same as his "experienced weight". if the acceleration is down then he feels lighter. this is because the upwards force from the floor decreases.
if the acceleration is up then he feels heavier. this is because the upwards force from the floor increases.

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