# Mop and Handle

1. Oct 13, 2005

### amcavoy

This is what I did:
$$F_x\leq\mu_s N=\mu_s\left(mg+F\cos{\theta}\right)$$
...and then I would simplify. But unfortunately, I was expecting the "F"s to cancel out, meaning the force didn't matter at that critical angle. They didn't obviously. Did I set this up correctly? If not, where did I go wrong?
Thanks, I appreciate it.
Alex

Last edited: Oct 13, 2005
2. Oct 13, 2005

### hotvette

Is the normal component of F really in the same direction as mg? You need to check your coordinate system and make sure you are consistent with + and - directions.

3. Oct 13, 2005

### amcavoy

It seems to be. The force F is broken down as follows:
$$F_x=\left<-F\sin{\theta},0\right>$$
$$F_y=\left<0,-F\cos{\theta}\right>$$
Now the normal force for Fy is -Fy:
$$N_{F_y}=\left<0,F\cos{\theta}\right>$$
Adding this to the normal from gravity clearly gives:
$$N=mg+F\cos{\theta}$$
I don't see where I'm going wrong :grumpy:
Thanks again,
Alex

4. Oct 13, 2005

### Pyrrhus

You got a sign error in Fx, it should be positive, isn't it pointing to +x?

5. Oct 13, 2005

### amcavoy

Sorry I should have mentioned: The force is being applied from the right. Any ideas?

Alex

Last edited: Oct 13, 2005
6. Oct 13, 2005

### amcavoy

Alright this is what I am down to (using the above equations I posted):

$$-F\sin{\theta}\leq\mu_smg+\mu_sF\cos{\theta}$$

Now I can solve for θ and come up with an inequality for that, but it still depends on F! I am confused as to why I still have an F here because it seems obvious that at that critical angle (or anything less), the applied force wouldn't make a difference; the mop would stay still. Is what I posted above correct, or have I gone wrong? I really appreciate it

Thanks again,

Alex

Last edited: Oct 13, 2005
7. Oct 13, 2005

### hotvette

Last edited: Oct 14, 2005
8. Oct 13, 2005

### amcavoy

I only got part of your message (as you can see in the quote) but I was thinking of breaking the equation down into:

$$0\leq\mu_smg$$

$$-F\sin{\theta}\leq\mu_sF\cos{\theta}$$

????

Alex

9. Oct 13, 2005

### hotvette

Look at my edit.

10. Oct 13, 2005

### amcavoy

I understand that, I just can't see how it applies to my equation. When I break it up like I did in my last post, I am able to solve for θ.

As for #2, it makes sense, but in this problem I do have external forces.

11. Oct 13, 2005

### hotvette

Deleted because it goes in a misleading direction.

Last edited: Oct 14, 2005
12. Oct 13, 2005

### amcavoy

By applying a vertical force only (θ=0) or if the force is less than or equal to the max. static friction. If it is less than or equal to the max. static friction, then the forces will cancel out and there won't be any x-force. ?

13. Oct 13, 2005

### hotvette

Deleted because it goes in a misleading direction.

Last edited: Oct 14, 2005
14. Oct 13, 2005

### amcavoy

Sorry for being difficult here. I see what you're saying. So if the angle is not zero, there will always be a force that is greater than the static friction. Because of this, the only angle that fits this problem is θ=0. Is this what you're saying?

Thanks again,

Alex

15. Oct 13, 2005

### hotvette

Yep!

Hold on, I had another thought.

16. Oct 13, 2005

### hotvette

What I should have said is the following:

$$-F\sin{\theta}\leq\mu_smg+\mu_sF\cos{\theta}$$

First of all, there is a sign mistake on the left side (like Cyclovenom pointed out). Gather all terms with F and theta on the left side and factor. Take a look at the left side and consider how to make the inequality valid (Hint: one possibility is that the left size is zero. How could that happen?)

Last edited: Oct 14, 2005
17. Oct 13, 2005

### amcavoy

Hmm. Where did you come up with that?

Thanks again hotvette