# Coefficient of friction Problem

1. Jul 20, 2006

### highc

A small box resting on a larger box, which in turn sits on a table. When a horizontal force is applied to the larger box, both boxes accelerate together. The small box does not slip on the larger box.

If the acceleration of the pair of boxes has a magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.

I'm don't really know where to start on this one, it seems that the only useful information that I have (so far as I can see) is the accleration.

2. Jul 20, 2006

### Staff: Mentor

Start with F=ma=uN. What is N? Hint -- the mass of the small box probably will cancel out....

3. Jul 21, 2006

### deepakalways

looking frm the ground frame of refernce, the smaller block seems to be moving bcoz of friction, as there is no other force on it. ..... (1)

k= coeff of friction,m=mass of smaller block, N=normal rxn on it,g=10 m/s2

balancing forces on small block in vertical direction,
mg=N

also, in horizontal direction,
ma=kN (due to (1))

where a = 2.5
therefore ma=kN
or ma=kmg
or a=kg
k= a/g = 0.25 .. answer

4. Jul 21, 2006

### Hootenanny

Staff Emeritus
Hi there deepakalways, please refrain from showing full solutions. It is better to guide a poster through the process than simply providing answers.

5. Jul 21, 2006

### Andrew Mason

The coefficient of static friction gives the maximum force that static friction can impart. The actual force depends on the applied force. What is the force that the lower box applies to the upper?

AM

6. Jul 21, 2006

### highc

The force that the lower box applies to the other is the normal force equal and opposite to the force of gravity. Would that be 9.8 N*kg?

7. Jul 21, 2006

### Staff: Mentor

The lower box exerts two forces on the upper box: One is the normal force, which acts vertically upward and happens to equal the weight of the upper box (w = mg). The other force is the friction force, which acts horizontally. I believe that's the force that Andrew was asking you about. Hint: You know the horizontal acceleration of the upper box, so what must be the force on it? (Answer in symbols, not numbers.)

8. Aug 30, 2008

### gonzalezphys

hi, new to the forums... am following along with you here. I also need help on this question.

Doc Al would the force be Kinetic Friction?

9. Aug 30, 2008

### Staff: Mentor

Since the boxes do not slip, the friction will be static, not kinetic.

10. Aug 30, 2008

### gonzalezphys

alright thanks! but I believe the original poster did not write the second part of the question.

c) If the acceleration of the pair of boxes has the magnitude of 2.5 m/s^2, determine the smallest coefficient of friction between the boxes that will prevent slippage.

how would i start this part?

11. Aug 30, 2008

### Staff: Mentor

Not only did the original poster write this part of the problem, it's word-for-word in the post that you quoted! In fact, that's the only part of the problem discussed in this thread. (Re-read the thread for tips.)

12. Aug 30, 2008

### gonzalezphys

LOL, I just realized, sorry I was talking on the phone while trying to post...

i'm going to sit down now before I hurt myself

13. Oct 31, 2008

### randomwinner

Excuse me, but don't you need the mass to calculate the coefficient of friction, since the friction between two objects depends on the force they apply against each other, and to know the force you need the mass. Am I wrong on both points? But to find the coefficient of friction you need to know the normal force. Do you need the mass to calculate the coefficient of friction?

14. Oct 31, 2008

### Staff: Mentor

To find the friction force, you'd need the mass. Luckily, this problem only asks for the coefficient of friction. It turns out that you don't need the mass to find that. (Try it and see.)

15. Oct 31, 2008

### randomwinner

I am sorry, but I still don't understand. The formula for the coefficient of static friction is us = FSmax/FN. FN is the normal force, which can only be calculated using the mass. Is this the way to conduct this problem solving?

16. Oct 31, 2008

### Staff: Mentor

Hint: The friction force is the only horizontal force acting on the smaller box. Relate that force to the acceleration of the box. (Note that you only need the ratio of the two forces, not their absolute measures.)

In general: When you don't have a needed quantity, give it a symbol and carry on. Maybe you won't really need it.

17. Nov 5, 2008

### randomwinner

Am I correct in using the formula us = Fs / Fn?

18. Nov 5, 2008

### randomwinner

To find the coefficient of friction using that formula, the mass is necessary. Since another variable's value is needed, either the normal force or the mass must be known. Incidentally, mass also needs to be known to calculate the normal force. I found that I could not answer this question using this formula, but it is the only formula the book gives, so I must be doing something wrong. I just can't see a solution to this problem.

19. Nov 5, 2008

### randomwinner

Okay. So I use the formula and plug in the acceleration value making this:

us = Fs / (2.5 m/s2) (mass)

In accord to my last post, no solution can be made without additional information.

20. Nov 5, 2008

### Staff: Mentor

You'll need that formula.

What's Fs? Hint: Use Newton's 2nd law.
What's Fn?

You do not need to know the mass. Just call it "m". Try it!