How Does Newton's Third Law Apply to Forces in an Elevator?

[qazxsw11111]

Hi. Let's say I have a lift with a man in it. The lift is moving upwards, and the forces acting on the man if I am not wrong is R (Normal reaction force by floor on man) and weight (mg). Hence R-mg=ma. Correct me if I am wrong or anything.

But my teacher says that (this is a marking point for my mid-year exam),

From Newton’s third law,
When a man is standing in a descending lift (there could be deceleration, acceleration or constant speed i.e. zero acceleration) the magnitude of the force exerted on the man’s feet by the floor is always equal to the magnitude of the force exerted on the lift floor by his feet.

Does this mean that force exerted by man's feet is not the weight of man? I am so confused. My free-body diagram is starting to screw up if I can't even identify the forces

 

[arildno]

Hi. Let's say I have a lift with a man in it. The lift is moving upwards, and the forces acting on the man if I am not wrong is R (Normal reaction force by floor on man) and weight (mg). Hence R-mg=ma. Correct me if I am wrong or anything.

Why? You are thinking correctly already!

From Newton’s third law,
When a man is standing in a descending lift (there could be deceleration, acceleration or constant speed i.e. zero acceleration) the magnitude of the force exerted on the man’s feet by the floor is always equal to the magnitude of the force exerted on the lift floor by his feet.

True, but IRRELEVANT.

Does this mean that force exerted by man's feet is not the weight of man?

Correct. The force we call the weight of the man is the force from the Earth on the man, the force the man exerts upon the Earth equals this in magnitude, as follows from Newton's 3. law.

 

[uart]

Both are correct.

The statement of Newtons ?th law "the force exerted on the man’s feet by the floor is always equal to the magnitude of the force exerted on the lift floor by his feet" is correct, but so to is your reaction force calculation correct "R-mg=ma"

For the free body diagram of the man you don't need to worry about the "force exerted on the lift floor by his feet". You use the reaction force (same magnetude as the above but directed upward) acting upward on the man and the gravitational force (mg) acting downward to find the total force (and hence acceleration) of the man/lift.

That is, ma = R - mg.

You appear to have done this correctly but there may be some aspect of your working which your teacher thought to show a lack of understanding or perhaps the actual question being asked was something other than what you answered. Your answer however is certainly correct.

BTW. Note that in the above acceleration (a) is taken as directed upward. A positive value of "a" corresponds to either an ascending lift increasing in ascent speed or a decending lift decreasing in decent speed.

 

[jimvoit]

Does this mean that force exerted by man's feet is not the weight of man?

What would be the force on the mans feet if he and the elevator were in free fall? Zero ?

 

[qazxsw11111]

Correct. The force we call the weight of the man is the force from the Earth on the man, the force the man exerts upon the Earth equals this in magnitude, as follows from Newton's 3. law.

So, what exactly is the force exerted by the man on the floor? It isn't the weight?

 

[Doc Al]

So, what exactly is the force exerted by the man on the floor? It isn't the weight?

The normal force between the man and the floor depends on the acceleration of the elevator. If the acceleration is zero, then the magnitude of the normal force will equal the man's weight.

As arildno explained, the term "weight" refers to the gravitational force of the Earth on the man. Sometimes the normal force is called the apparent weight, because that force is what you feel to give you the sensation of having weight. If the elevator is in free fall, the normal force goes to zero and your apparent weight (not your real weight) goes to zero: that's what the term "weightlessness" means.

 

[uart]

So, what exactly is the force exerted by the man on the floor? It isn't the weight?

It's exactly the reaction force that you already correctly expressed when you wrote $ma = R -mg$

That is,

$$R = m (a + g)$$

Note that this is different from the mans "normal"* weight which is just mg

Though technically one could argue that his weight in the lift at that instant in time is indeed R, though it's not what someone would usually refer to as their true weight.