# Weight of man in a lift when lift is ascending/descending.

1. Jun 11, 2010

### equilibrum

I guess this is really basic physics to all of you but it got me thinking abit.

1. The problem statement, all variables and given/known data
A man of mass 72kg tries to weigh himself in the lift using a weighing machine
I am required to find his weight when the lift is stationary ,accelerating upwards(3m/s^2),travelling upwards at a constant speed(2m/s) and when the lift is accelerating downwards(4m/s^2)

2. Relevant equations
W=mg
F=ma

3. The attempt at a solution
If i am not wrong,
the weight of man when lift is stationary , W=mg = 72x10(at my level we don't require 9.81 yet) = 720N

I got stuck when it came to accelerating up/downwards and moving at constant speeds cause I notice when I take the lifts down(descend) I seem to feel like something is pushing me upwards(g-force?) and when im taking the lift up I fee like something is pressing down on me. When the lift is coming to a halt I always feel a short sensation of weightlessness(well,a little)

This got me to think that shouldn't the weight of the man fluctuate then when the lift is moving? If so,then there should be other forces causing the fluctuation. Can someone enlighten me on what other forces are acting? I'm guessing really small amounts of g-force,like when you jam the breaks in the car you come to a quick halt and you feel like something is pressing down on you(I assume it's the same feeling in F1 cars)

Thanks in advance for anyone who would take time to clarify my doubts. :)

Last edited: Jun 11, 2010
2. Jun 11, 2010

### aim1732

You can count on other forces when you are in a non-inertial ,accelerated that is, frame of reference. These are pseudo forces and have no real existence. Weight of a man for us is actually the normal force that the floor of the lift would exert to balance other forces. Now if your observations are with respect to the earth you would agree that the man is accelerated. Apply the second law and calculate the normal force - that would be the apparent weight.

3. Jun 11, 2010

### equilibrum

With my current knowledge I do not really understand what it means by 'non-inertial frame of reference' as I have not learnt it yet. Did a quick Wikipedia check,correct me if I am wrong but a frame of reference that is accelerating or decelerating is considered 'non-inertial' where we can neglect the fictitious forces of 'pressing' and 'weightlessness' in the lift? Is the lift a 'inertial frame of reference' then when it is moving at constant speed(zero acceleration)? Also,if the lift is a inertial frame of reference when it is moving at a constant speed,do we take into consideration the 'fictitious forces'(if any) as mentioned? Cause if we don't, I would get F=ma to be F=72 x 0 and thus F=0 which contradicts the F=720N that we would get when the man measures his weight inside a stationary lift.

Last edited: Jun 11, 2010
4. Jun 11, 2010

### hikaru1221

Since you haven't learned the non-inertial frame yet, I think the only way is to approach it in the reference frame of the earth (or the ground, the building, etc, as long as it's an inertial frame).
The "weight" the man experiences is actually the normal force N the floor exerts on him (note: this is what the man experiences).
1 - When you're at rest: N - mg = 0
2 - When you're accelerating upwards: N - mg = ma, so N = 72 x (10+3) = 936 N
3 - When you're traveling at constant speed: Since a=0, N - mg = 0.
4 - When you're accelerating downwards: mg - N = ma, so N = 72 x (10-4) = 432 N

1 - The non-inertial reference frame is the frame which accelerates or decelerates (in the view point of an inertial frame). Thus, moving at constant speed means it's an inertial frame.
2 - You seem to not understand the fictitious forces mentioned in wikipedia. I think you should forget what wiki tells you, because sometimes they mention some complicated concepts (such as relativity) which may confuse you, and wait for the next lesson in class or read the book at your level if you want.

Last edited: Jun 11, 2010
5. Jun 11, 2010

### equilibrum

I get it now. Thanks for clarifying. Although not tested at my level,I will find out more about the concepts mentioned here that I did not really understand from my teacher. :)