# Homework Help: Force and acceleration

1. Apr 30, 2006

### oooo

This is a question I keep getting wrong on my Physics homework:
An elevator is moving with an initial velocity of 10 m/s downward with a 47 kg rider standing inside. What should be the final velocity of the elevator after 15 seconds of constant acceleration if the rider is to feel twice as heavy as he does when at rest? Upward is the positive direction.

Step 1: What is the normal force on the rider during the 15-second acceleration period? What I did was: find the acceleration and multiplied that by the mass to get the net force. Then I subtracted the force of gravity from the net force to get the normal force. However there are two problems with my approach : (1) The steps I completed to find the normal force were seen further down the line of steps suggested for finding the final velocity (signifying that I did my process out of order) and (2) The answer was marked incorrect.

2. Apr 30, 2006

### God64bit

Confused :P

i thought about it and now im confused.
Considering the wieght of the perosn is (47)(9.8) = wieght
to get twice the wieght or 2(wieght)= (47)(9.8)(2)
since multiplication is culmative or the order don't matter we can say eithier their mass or 9.8 or gravity has doubled. there mass is probaly constant...
so why can't we say the elavator must move up at 9.8m/s in order for their wieght to double that of rest?

3. Apr 30, 2006

### oooo

and so if we assume the velocity is 9.8 m/s, where can we go from there to find the normal force?

4. Apr 30, 2006

### nrqed

the fact that he feels twice as heavy does not mean that his weight has changed. The weight is always mg downward. There are two forces acting on the person: the force of gravity downward and the normal force exerted by the floor acting upward. The fact that he feels twice as heavy is due to the fact that the *normal* force exerted by the floor is twice what he usually experiences, so the normal force has a magnitude of 2mg and is acting upward. So you get the equation $(F_g)_y + N_y = m a_y \rightarrow -mg + 2 mg = m a_ y$.
Therefore a_y is + 9.8 m/s^2.

You should always draw a free body diagram and identify clearly the forces. The weight never changes (unless you go to a very high altitude)

5. Apr 30, 2006

### God64bit

lets leave out everything we have done so far at rest what is Fn?
Fn=mg
so his normal force is (47)(9.8) or 460.6N now "if the rider is to feel twice as heavy as he does when at rest?" if thats how much he feels at rest how does he wieght is thats doubled?

6. Apr 30, 2006

### God64bit

So in english im wrong and what i said won't work?

7. Apr 30, 2006

### nrqed

I am not sure I understand the sentence. But the point is that the weight is actually unchanged. If he *feels* twice as heavy it is because the NORMAL force is twice as large as when there is no acceleration. You are right that the normal force when there is no acceleration is mg (as long as no other force beside gravity is in effect). So if he feels twice as heavy it means that the normal force will be 2mg (but the weight stays the same...mg downward)

8. Apr 30, 2006

### oooo

so the normal force is (2)(460.6)?

9. Apr 30, 2006

### nrqed

yes. And is acting upward. The weight is the same as always, mg acting downward

10. Apr 30, 2006

### oooo

thanks for clearing that up for me!

11. Apr 30, 2006

### oooo

a related question: If a block whose mass is 17 kg is attached to a rocket that exerts 169.88 N upward on the block, what is its acceleration?
I thought to just do a=F/m, so a= 169.88/17 where a is roughly 9.993 m/s^2. This is wrong. What is the matter with what I am doing?

12. Apr 30, 2006

### nrqed

Again, you should be thinking in terms of free body diagrams and thinking about all the forces involved. There is the force exerted by the rocket and there is still the force of gravity acting downward! (the weight if you will). SO the equation is +169.88 - mg = m a_y.

13. Apr 30, 2006

### oooo

oh, I understand now, thanks.