# Homework Help: A problem that has force, friction on an incline

1. Sep 24, 2008

### musicfairy

A block of mass m, acted on by a force of magnitude F directed horizontally to the right as shown above, slides up an inclined plane that makes an angle $$\theta$$ with the horizontal. The coefficient of sliding friction between the block and the plane is $$\mu$$.

The picture is that of an incline, $$\theta$$ on the left of the incline, mass m on the incline, and a horizontal force is applied to m directed right.

Develop an expression for the magnitude of the force F that will allow the block to slide up the plane with constant velocity. What relation must $$\theta$$ and $$\mu$$ satisfy in order for this solution to be physically meaningful.

This is what I did. (wrong according to the answer)

F = ma
mgsin$$\theta$$ + $$\mu$$mgcos$$\theta$$ - Fcos$$\theta$$ = 0
Fcos$$\theta$$ = mg(sin$$\theta$$ + $$\mu$$cos$$\theta$$)
F = mg(sin$$\theta$$ + cos$$\theta$$) / cos$$\theta$$

This is how I reasoned: If it's moving up at a constant velocity, Fcos$$\theta$$ should equal mgsin$$\theta$$ + friction ($$\mu$$gcos$$\theta$$)

This is what the answer is supposed to be:

F = mg($$\mu$$cos$$\theta$$ + sin$$\theta$$) / (cos$$\theta$$ - $$\mu$$sin$$\theta$$)

F > 0 => cos$$\theta$$ > sin$$\theta$$
tan$$\theta$$ = 1/$$\mu$$

What mistake did I make in my reasoning and where did the answer come from?

2. Sep 24, 2008

### Staff: Mentor

Since the applied force has a component perpendicular to the plane, the normal force is not simply $mg \cos\theta$.

(Also: It's much easier if you write your entire equation using Latex, not just some pieces.)

3. Sep 24, 2008

### musicfairy

Ok, I see. Thanks for the tip. It worked.

I have a general/stupid question. How do you know which component of F is || or perpendicular to the incline. I tried several problems like this one (with numbers) and had trouble figuring out the axis

For the last part of the problem, how do I know that cos$$\theta$$ > sin$$\theta$$?

I'm still getting used to latex. It's coming to me.

4. Sep 24, 2008

### Staff: Mentor

Always draw yourself a diagram and mark the angles. (That's the only way I know.) Since we know the incline makes an angle $\theta$ with the horizontal, it must make the same angle with the applied force F.
Since $\cos\theta - \mu\sin\theta$ (the denominator) must be positive, you can deduce that $\cos\theta > \mu\sin\theta$. (Not $\cos\theta > \sin\theta$.)

5. Sep 24, 2008

### musicfairy

I see now. Thanks for all the help.