How Does the Angle Theta Affect Frictional Force in a Mop Sliding Problem?

[WarpDrive]

The handle of a floor mop of mass m makes an angle theta with the vertical direction. Let muK be the coefficient of kinetic friction between the mop and the floor and muS be the coefficient of static friction. Neglect the mass of the handle.

a) Find the magnitude of the force F directed along the handle required to slide the mop with uniform velocity across the floor

b) Show that if theta is smaller than a certain angle theta0, the mop cannot be made to slide across the flor no matter how great a force is directed along the handle. What is the angle theta0?

Answers:

So far, I have been able to get a. It's an odd question, so the answers are there in the back of the book. I got the answer for part a:
|F| = MuK *mg / (sin(theta) - MuK *cos(theta))

But I'm still stuck on B. I know that the horizontal component of the force F must be smaller than friction, so |F|sin(theta0) < MuS(mg+|F|cos(theta0).

Now, the answer in the back of the book is simply Tan-1(MuS). But the extra MuS*mg is getting in the way. The only solution I see is the claim that because |F|sin(theta0) is so much larger than MuS*m*g, that the mg part of the normal force can be ignored. This allows you to simplify the problem to the answer, but is this right? It seems odd to simply get rid of one of the components.

 

[Doc Al]

Think of it this way: As long as the angle of the mop handle is less than \theta_0, where \tan \theta_0 = \mu, the mop won't slide regardless of its mass. Thus that angle works for any mop with the same \mu. (Of course, as you realize, if you know the mass you can increase that angle a bit.)