Force Needed to Push a Box on an Inclined Plane

[Tastosis]

Homework Statement

A man is pushing a box on top of a plane that is inclined 20 degrees above the horizontal, with a distance of 10 m. The weight of the box is 20 kg and the coefficient of kinetic friction is 0.30. The velocity is constant.
How much force is exerted by the man? Find the work done by force and the work done by gravity.

Homework Equations

Summation of forces along x-axis = F - f = 0
Summation of forces along y-axis = n - Wy = 0
f = uk * n
W of Force = Fdcos
W of weight = mgdcos

The Attempt at a Solution

[PLAIN]http://img96.imageshack.us/img96/4928/17512380.jpg
Is my free body diagram correct?

If someone can give a clue on how to get Wy, I think I can solve this. Thanks!

 

[PhanthomJay]

Your first FBD is correct...then you apparently rotated it in your second FBD, which is OK, as long as it doesn't confuse you. You'll need the x and y components of the weight force, which you can find from basic geometry and trig once you realize what is the interior angle in between the Wy and W forces.

 

[Doc Al]

Summation of forces along x-axis = F - f = 0

You forgot the x-component of the weight.

If someone can give a clue on how to get Wy, I think I can solve this.

See: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm"

 

[Tastosis]

You forgot the x-component of the weight.

Wait, what does this mean? What should the summation of forces in the x-axis should be, then?

 

[Doc Al]

Wait, what does this mean? What should the summation of forces in the x-axis should be, then?

Well, when you sum the forces in the x-direction you must include all forces that have an x-component. That includes the weight.

 

[Tastosis]

So sum of forces in the x-axis is: F - f - Wx?

Now I'm really confused. How do I get Wx and Wy?

 

[Doc Al]

So sum of forces in the x-axis is: F - f - Wx?

Yes.

Now I'm really confused. How do I get Wx and Wy?

Go to the link I gave in post #3.

 

[Tastosis]

Oh, forgot about the link. Here's what I got...

Wy = mgcos
Wy = 20 kg * 9.8 m/s^2 * cos(20)
Wy = 184.18 N

Wx = mgsin
Wx = 67.04 N

normal force = Wy = 184.18 N

f = uk * n = 0.30 * 184.18 N = 55.25 N

F = f + Wx
F = 55.25 N + 67.04 N
F = 122.29 N

Before I proceed with the work, did I get everything right?

 

[Doc Al]

Looks good to me.

 

[Tastosis]

Work of force would be = Fdcos
WF = 122.29 N * 10 m * cos 20
WF = 1149.15 J?

Work of weight would be = mgcos? How do I get work of the weight?

 

[Doc Al]

Work of force would be = Fdcos
WF = 122.29 N * 10 m * cos 20
WF = 1149.15 J?

No. In the formula W = Fd cosθ, θ is the angle between the force and the displacement. What's that angle?

Work of weight would be = mgcos? How do I get work of the weight?

Use the same method as above. What's the angle between the displacement and the weight?

 

[Tastosis]

For force, is the angle 0?

For weight...250? >.<

 

[Doc Al]

For force, is the angle 0?

Right. The applied force is up the incline and so is the displacement.

For weight...250?

OK. 110° also works.

 

[Tastosis]

OK. 110° also works.

So it's cos 250 or cos 110?

Also, just a quick question, the method on how I got Wx and Wy (in your link):
[URL]http://www.physicsclassroom.com/Class/vectors/u3l3e5.gif[/URL]

Does this only apply to inclined planes and never to horizontal planes?

Thanks!

 

[Doc Al]

So it's cos 250 or cos 110?

Those are equivalent.

Also, just a quick question, the method on how I got Wx and Wy (in your link):
[URL]http://www.physicsclassroom.com/Class/vectors/u3l3e5.gif[/URL]

Does this only apply to inclined planes and never to horizontal planes?

It works fine for horizontal planes. (But you shouldn't need it.) If the plane is horizontal, the angle is 0.