Kinetic Frictonal Forces

1. Sep 28, 2004

BlackMamba

Hello,

I have a problem that I am stuck with. The problem consists of a 2.60kg box sliding across the horizontal floor of an elevator. The coefficient of kinetic friction between the floor and the box is 0.490. I have to determine the kinetic frictional forces of different cases.

(a) when the elevator is stationary ------Which I have already done. That part was easy.

but I'm stuck on the following two.

(b) when the elevator is accelerating upward with an acceleration of 1.30 m/s^2

(c) when the elevator is accelerating downward with an acceleration of 1.30 m/s^2.

I assumed that I could use the equation, kinetic frictional force = kinetic friction times mass and acceleration. That however was not correct.

Any help or direction provided would be greatly appreciated.

2. Sep 28, 2004

Pyrrhus

Remember $$F_{f} = \mu N$$

Do a analysis on the y-axis of the forces, you know the mass and acceleration.

3. Sep 28, 2004

BlackMamba

That is the first time I've seen the equation you provided. Im not sure I understand it. Is that the Force of friction = the coefficient of the normal force?

I also don't understand what you mean by doing an analysis of the y axis.

Yes I know the mass (2.60kg) and the accleration is (1.30 m/s^2). By analysis, do you mean what direction the accleration is going? If so, when the elevator is moving upward the accleration is pointing downward....right? And vice versa with the opposite direction. Other than that, what else do I need to know about the y axis?

Oh.....there would be a Weight force acting on the block. Would I need to know that? Would that be in play?

Thanks for helping

4. Sep 28, 2004

Pyrrhus

$$F_{f} = \mu N$$

$$\mu$$ is the coefficient of friction, in this case kinetic, and N is the normal force.

I'm talking about a Free Body diagram of the forces acting on the block, you know the elevator is accelerating upwards/downwards, so the block will have a movement on y-axis.

Use Newton's 2nd Law

$$\sum^{n}_{i=1} \vec{F}_{i} = m \vec{a}$$

5. Sep 28, 2004

BlackMamba

Right, I drew a free body diagram and that did nothing. LOL

Oh and your initial equation is just a bit different from how it is in my book. That's why I didn't quite get it at first. I used that equation to find the frictional force when the elevator was stationary, that was easy. But I don't understand where acceleration comes into play.

Using Newton's second law, I understand that net force equals mass times acceleration. Oh...I think I get it now. I have to find the net force of the box when the elevator is travelling up and down then plug that into my equation to find the kinetic friction force. Is that correct?

So if this were the case then the number would be the same for up and down?

Last edited: Sep 28, 2004
6. Sep 28, 2004

Pyrrhus

Give it a try

Well what's your sign convention? if up and right is positive, then left and down must be negative.

7. Sep 28, 2004

BlackMamba

LOL, thanks. But I have to be somewhat sure before I post my answer considering my class uses webassign and we are only alloted so many submissions and this is only my second problem I'm working on. LOL But yes, I will give it a try.

Thank you for your help. :)

8. Sep 28, 2004

BlackMamba

Well I did it your way and still got the same answer I got before. LOL And I know that's wrong because I submitted it once already and it was wrong.

9. Sep 28, 2004

Pyrrhus

Show me what you did.

10. Sep 28, 2004

BlackMamba

I found the net force of the box when the elevator is going up and down using Newton's second law.

$\sum \vec{F} = m \vec{a}$
So that gave me a total -3.38 and 3.38

I pluged those figures into : $F_{f} = \mu N$

and got -1.66 and 1.66.

11. Sep 28, 2004

Pyrrhus

I will do b)

Y- Axis
Up and Right positive.

$$N - mg = ma$$

$$N = ma + mg$$

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

Now

$$F_{f} = \mu N$$

I got 14.14 N

12. Sep 28, 2004

BlackMamba

Just as you posted this I figured out that I was neglecting gravity. LOL Thanks again.

13. Sep 28, 2004

BlackMamba

Just double checking. For (c) then the answer would be: 10.83N correct?

14. Sep 28, 2004

Pyrrhus

I believe so.

15. Sep 28, 2004

BlackMamba

Thanks again for your help. It is much appreciated.

16. Sep 28, 2004

Pyrrhus

Well, It's always good to be of help

17. Sep 28, 2004

BlackMamba

Well perhaps you could help me with another problem then. :D

Here is a link to the picture: http://www.webassign.net/CJ/4-67alt.gif

The block on the table is 492N and the hanging block is 175N.

(a) find the acceleration of both blocks
(b) find the tension in the cord

I know the tension will be the same for both blocks and have managed to come up with these two equations for each block.

Block 1
$$\sum F_{x} = - W_{1} + T = m_{1}a$$

Block 2
$$\sum F_{y} = T - W_{2} = m_{2}(-a)$$

Now I'm supposed to solve for both T and a and I'm supposed to do this by solving simultaneously. My question being how on earth do I solve them simultaneously?

Thanks for any help.

18. Sep 28, 2004

Pyrrhus

if there's no friction, there's only tension on the first block.

19. Sep 28, 2004

BlackMamba

Right that's what I meant. But I'm still getting extremely frustrated in that I've tried solving this every which way and I don't appear to be getting any closer.

I tried by getting T alone for the block 2 equation and substituting that into the block 1 equation and still nothing.

20. Sep 28, 2004

Pyrrhus

I will solve it, if there's something you don't understand, let me know.

On x-axis:

$$T = m_{1}a$$

On the y-axis:

$$T - m_{2}g = -m_{2}a$$

$$m_{2}g - T = m_{2}a$$

Adding Equations on X-axis and Y-axis:

$$T = m_{1}a$$

$$m_{2}g - T = m_{2}a$$

$$m_{2}g = (m_{2}+m_{1})a$$

$$\frac{m_{2}g}{m_{2}+m_{1}} = a$$