Static Friction: Mop Handle Critical Angle

[amcavoy]

A handle of a floor mop makes an angle of θ with the vertical direction. Let mu be the coefficient of static friction. Show that there is a critical angle in which the mop will not move no matter how much force is applied on the handle.

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

 

[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.

 

[amcavoy]

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.

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
Thanks again,
Alex

 

[Pyrrhus]

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

 

[amcavoy]

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

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

Alex

 

[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

 

[hotvette]

Deleted because it leads in a misleading direction.

 

[amcavoy]

Actually, your approach is correct. However, there is some reasoning that you have to go through.
1. The

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

 

[hotvette]

Look at my edit.

 

[amcavoy]

This is a good problem. Actually, your approach is correct. However, there is some reasoning that you have to go through.
1. The objective is to find the angle such that the mop can't move no matter how big F is
2. Friction force > 0 only if there is a force opposing it. In other words it isn't mu*N under all circumstances. Think of a mass with no external forces (except gravity) resting on a perfectly horizontal plane.

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.

 

[hotvette]

Deleted because it goes in a misleading direction.

 

[amcavoy]

That's the point. If you don't want the object to move in the x direction, the only way to do that is to have zero force. How do you get zero force in the x direction?

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. ?

 

[hotvette]

Deleted because it goes in a misleading direction.

 

[amcavoy]

I'd place my bet on the first one. Think about it...

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

 

[hotvette]

Yep!

Hold on, I had another thought.

 

[hotvette]

What I should have said is the following:

Go back to your inequality:

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

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?)

 

[amcavoy]

what happens if $\mu = tan \theta$?

Hmm. Where did you come up with that?

Thanks again hotvette