# Force and Static Friction

1. Feb 3, 2012

### zaper

I've worked through this problem for a while now and can't quite come up with an answer. The problem is as follows:

A block with mass 5 kg sits on an inclined plane of 30°. The coefficient of static friction is 0.1 between the plane and block. A force F is applied in the horizontal direction on the block. (I have attached a picture of the problem)

What is the minimum magnitude of F to keep the block from slipping?

What I have so far is this:

I assume then that since we're talking about static friction the block is not moving so there is no acceleration other than gravity anywhere.

The weight of the block in the x direction (Wx) = mg*cos(30) and Wy = mg*sin(30)
The force of F in the x direction (Fx) = F*cos(30) and Fy = mg*sin(30)

The Normal force (N) = Wy + Fy

The maximum static friction (fs) = μ*N which I solved down to .1*(F+mg)*sin(30)

I then set Fx = fs to find the F for the maximum static friction and got F = 3.

However I believe that this is the maximum force such that the block will not move not the minimum.

Have I gone horribly awry here? What should I do next?

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2. Feb 3, 2012

### BruceW

I think you started off good. But you've got Wy and Wx the wrong way around. You've got the right idea to calculate the normal force and the maximum static friction. (But the calculation has gone wrong because you got Wx and Wy the wrong way). And at the end, I don't get what you mean by 'Fx=fs' ? Fx should be zero, and you should use that fact to calculate F.

3. Feb 3, 2012

### SammyS

Staff Emeritus
I think your only mistake (besides leaving units off of F) was the direction you assumed for the force of friction. You want the minimum force, F, needed to keep the block from sliding down the incline. To find that, the friction should be up the incline to help keep the block from sliding down.

What you found is the maximum force that can be applied to the block, without it sliding up the ramp.

Added in Edit: Never mind. See the following Post by BruceW

Last edited: Feb 3, 2012
4. Feb 3, 2012

### BruceW

I don't think that was his mistake, Sammy. If you look through his working, he has Wy=mg*sin(30) But this is not correct. (A fast way to check is to imagine the angle got close to zero, then we should have the weight purely in the y direction, but his equation says none of it would be in the y direction).

5. Feb 3, 2012

### zaper

I think Bruce is correct here. I attached a picture of Wy and Wx drawn out in a triangle and Wx is obviously the sin side of the angle. I apologize that I mixed this up in my original post.

#### Attached Files:

• ###### 4.2-2.png
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Last edited: Feb 3, 2012
6. Feb 3, 2012

### zaper

In that case would I set Wx = fs + Fx?

7. Feb 3, 2012

### BruceW

Yep, I think that is right. fs is in the opposite direction to Wx, so that it effectively 'cancels out' some of the weight, so that Fx can be as small as possible.

When I said "Fx should be zero, and you should use that fact to calculate F." I was wrong to say this, because you were using Fx to mean the applied force, but I was thinking of Fx as the net force. Sorry if I caused confusion there.

8. Feb 3, 2012

### zaper

Ok so I've gone back through all my steps and I keep getting 24N as F. This seems high to me so if you wouldn't mind going through my steps to check here's what I did:

1) fs = μ*N = .1*(Fy+Wy) = .1*(F*sin(30) + 5*9.8*sin(30))

2) Wx = fs + Fx => 5*9.8*cos(30) = .1*(F*sin(30) + 5*9.8*sin(30)) + F*cos(30)

3) F = 24N

Last edited: Feb 3, 2012
9. Feb 3, 2012

### SammyS

Staff Emeritus
It looks to me like your normal force is wrong.

You still have Wx as (mg)cos(30°), and Wy as (mg)sin(30°).

10. Feb 3, 2012

### zaper

Sorry. I fixed all the W's and changed the steps to the ones below:000

1) fs = μ*N = .1*(Fy+Wy) = .1*(F*sin(30) + mg*cos(30))

2) Wx = fs + Fx => mg*sin(30) = .1*(F*sin(30) + mg*cos(30)) + F*cos(30)

3) F = 22.1N

Anything wrong still?

11. Feb 3, 2012

### BruceW

Your working all looks good, and I got the same answer. So I reckon you've done it right. And your working is pretty neat and understandable (which is always a good thing).

The answer does look large, but it would take almost 50N to hold the block without using the wedge. So in view of this, maybe 22.1N is not unexpectedly large.