# Homework Help: Forces - finding minimum mass given coefficients of friction

1. Jul 10, 2010

### shawli

1. The problem statement, all variables and given/known data

A rope exerts a force of magnitude of 21 N, at an angle 31 degrees above the horizontal, on a box at rest on a horizontal floor. The coefficients of friction between the box and the floor are s=0.55 and k=0.50 . The box remains at rest. Determine the smallest possible mass of the box.

2. Relevant equations

Fs= s*Fnormal , Fk=k*Fnormal

F=ma

3. The attempt at a solution

I've attached my system diagram to the post.
Using vector components, I find that Fapp-x=18 N and Fapp-y=10.8N. That's all I know how to do here :/.

I have to solve for mass, but I'm not sure how the physics is working. Can someone start me off on how to solve this question?

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2. Jul 10, 2010

### pgardn

So you start off by looking at Fs compared to F applied in the horizontal. If they are exactly equal in magnitude, the mass will not accelerate. Does this make sense?

3. Jul 10, 2010

### pgardn

You have calculated Fx, or the applied horizontal force, so thats done. How are you going to calculate Fs? In other words, can you write an equation to calculate Fs? I bet the equation you write will have one unknown, m...

Last edited: Jul 10, 2010
4. Jul 10, 2010

### shawli

Fs = Fapp in order for the object to be at rest. Now, since the question says that force is being applied to move it, I'm not sure how to use Fk with Fs.

Fn = Fg+Fapp-y (you have to consider the upward component of the applied force, right?)

Fs = s*Fn
Fs = 0.55 * (9.8m+10.8) ...I think this is it?

I'm lost with how to manipulate Fk though.

5. Jul 10, 2010

### pgardn

I thought it said minimum mass so it stays at rest. Why do you need the coefficient of kinetic friction?

6. Jul 10, 2010

### pgardn

What does the normal for Fn equal in this situation? I think you may have a sign problem.

7. Jul 10, 2010

### shawli

If the object is at rest then Fs= Fapp, like you said before.
and Fapp = Fapp-x (horizontal component of the applied force).

So then

18 = 0.55 * (9.8m+10.8)
m = 2.03 kg

8. Jul 10, 2010

### shawli

Did I do something wrong?

9. Jul 10, 2010

### shawli

This is what I have as the normal force:

Fn = 9.8m+10.8

It should be a positive value, no?

10. Jul 10, 2010

### pgardn

The normal force is equal to Mg - F(applied)y... or Mg - F(vertical)yes?

11. Jul 10, 2010

### shawli

Yup, the F vertical/F applied y is equal to 10.8N (calculated with vector components).
So ...the sign is flipped! I guess that makes sense because the F vertical is pointing downwards. But then, I didn't make 9.8m/s^2 negative, even though its the downward acceleration of gravity. I get confused about sign flipping all the time, but okay, thank you!

12. Jul 10, 2010

### pgardn

Well the push back from the table on the block has to be less than mg if you are also pulling up on the block with a string. The Normal must be less. The normal force would be more than mg if you pushed down on a block with your finger. The table would push back with mg + force of finger. I hope this makes sense.