A problem that has force, friction on an incline

[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 + \mumgcos\theta - Fcos\theta = 0
Fcos\theta = mg(sin\theta + \mucos\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 (\mugcos\theta)




This is what the answer is supposed to be:

F = mg(\mucos\theta + sin\theta) / (cos\theta - \musin\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?

 

[Doc Al]

F = ma
mgsin\theta + \mumgcos\theta - Fcos\theta = 0

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.)

 

[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.

 

[Doc Al]

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

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.

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

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.)

 

[musicfairy]

I see now. Thanks for all the help.