Elevator Physics- Newton's Law of Motion

tyrantboy

1. The problem statement, all variables and given/known data
why is it that you sometimes feel 'light' or 'heavy' while moving in a fast moving lift? A lift has a total maximum mass, including occupants of 1000kg. it is supported by steel ables. the maximum acceleration or deceleration of the lift in operation is 2.0ms^-2. How big is the maximum tension (force) that needs to be supported by the cables from these figures?


2. Relevant equations
F=ma
.'. F= 1000x2= 2000 Newtons


3. The attempt at a solution
Well i think that you feel sometimes heavy because when you are going up, you are moving away from gravity. You are opposing gravity and it causes a downward feeling like you are being pulled down. It causes a drag as we are speeding away from gravity.

Now i think you feel light sometimes when you are traveling in a very fast lift down and since you are going with gravity and towards the earth, you have less drag.

Please help me out and could you also check my formula

Thank You

OwenMc

Your thinking is along the right lines. If you stand in a lift that isn't moving, you experience gravity, that is an acceleration of g. If the lift goes up, you now experience an additional acceleration (in your case 2ms-^2 more), so you feel heavier. The opposite is true if you go down.
When calculating the tension you need to remember that even when the lift isn't moving it has to support its own weight (mg). When it starts to accelerate upwards you have to add this in seperately. So:

F = m * (g + a)

dynamicsolo

If you are working with free-body diagrams in your course, draw the force diagram acting on the passenger in the elevator: there is their own weight, W = Mg, downward and the normal force, N, from the floor acting upward. So the net force on the passenger when the elevator is accelerating upward at rate a is

N - W = Ma .

To deal with the first question, when you "feel your weight", what you actually sense, in Newton's mechanics, is the magnitude of the normal force, N, pushing against the soles of your feet. (A scale measures the magnitude of the force its mechanism must supply to push back against your weight, so it displays N, which you read as "your weight".)

So the weight you "feel" in the elevator is found by solving the equation above for N:

N = W + Ma = Mg + Ma = M·(g+a)

The cases for you to consider then are: 1) elevator stationary or moving at constant speed (a = 0); 2) elevator accelerating upward (a > 0); 3) elevator accelerating downward (a < 0) -- the extreme case for which is free-fall (a = -g).

A cable towing the elevator car must supply a tension found in a similar fashion, as OwenMc described, so you obtain the same result for T.

tyrantboy

thx guys

i'll get back at you with my new answer afterwards