Newton's Second Law and an elevator

[Peter G.]

So I had this problem regarding an elevator with a man inside.

In the first part of the question we had to calculate the tension in the cable, which I managed to do alright.

The Second Part however, I am having difficulty and it asks for the following: When the lift is decelerating by 1 m/s², what is the normal reaction force of the lift on the man? (The man weights 90kg and take the acceleration due to gravity to be 10 m/s)

I know the answer because he did it on the board. But I don't understand why (I know I am wrong but I can't get my head around this) the reaction force is not 900N.

Can anyone try and explain to me why?

Thanks in advance,
PeterG

 

[cupid.callin]

left is going up or down?

 

[Doc Al]

I know the answer because he did it on the board. But I don't understand why (I know I am wrong but I can't get my head around this) the reaction force is not 900N.

If the normal force were just 900N, the man would not accelerate. That's the normal force if the elevator is not accelerating.

If the elevator accelerates downward, then it exerts less force on the man. Thus the man has a net downward force on him, so he can accelerate along with the elevator. You figure out the normal force by using Newton's 2nd law. Only two forces act on the man: gravity, which doesn't change, and the normal force, which adjusts itself with the movement of the elevator.

 

[Peter G.]

The reaction force varies according to the tension of the cable pulling on the elevator? So if it is going very fast upwards, the reaction force on the man would be much greater than his weight? Your apparent weight going up an elevator would be your weight + normal reaction?

 

[Doc Al]

The reaction force varies according to the tension of the cable pulling on the elevator?

What it directly depends on is the elevator's acceleration, which varies with the tension in the cable.

So if it is going very fast upwards, the reaction force on the man would be much greater than his weight?

Yes, but it's not speed but acceleration that's important. If the elevator accelerates upward, the normal force on the man will exceed his weight. (So his apparent weight will be greater than his true weight.)

Your apparent weight going up an elevator would be your weight + normal reaction?

No. The normal force on you is your apparent weight. (When the elevator is stationary--or just not accelerating--the normal force on you will equal your weight. In that case your apparent weight will be equal to your true weight.)

 

[Peter G.]

Ok, thanks, sorry for the fast/speed/acceleration misconceptions and the last question was just downright stupid I got it: Apparent Weight = Normal Force, what I really meant was Normal Force = Your Weight + Resultant Force (in the case of Lift going upwards)

Thanks once again Doc Al, for being very didactic.