Applying Newton's Laws: Crate on an Inclined Ramp

[coco87]

Homework Statement

In the figure, a crate of mass m = 96 kg is pushed at a constant speed up a frictionless ramp (θ = 27°) by a horizontal force F. The positive direction of an x-axis is up the ramp, and the positive direction of a y-axis is perpendicular to the ramp. (a) What is the magnitude of F? (b) What is the magnitude of the normal force on the crate?

m = 96 kg
θ = 27°

http://lcphr3ak.is-a-geek.com/figpf1.png

Homework Equations

F = ma
Fg = mg (g=9.8)
Fnet = Fg + F

The Attempt at a Solution

Currently I'm trying to map out the situation, but am running into an issue. Considering we are dealing with constant speed, we can assume that the acceleration is 0. First, I figured out Fg, which is (96)*(9.8), so Fg = (940.8). Using a little trig, we can derive a triangle with a hypotenuse Fg, and a base angle of 63°. This means that Fgx = (940.8)(cos 63°) = -427.1143 (negative because it lies in the third quadrant); and Fgy = (940.8)(sin 63°) = -838.2589. Ultimately (I think I am understanding this properly), Fnetx will be 0, because the acceleration is 0. So, Fnetx = Fgx + Fx, which is 0 = -427.1143 + Fx; so Fx = 427.1143. I've verified all of this is correct. However, When trying to find Fy, I'm (at least I think so) stuck with Fnety = Fgy + Fy; which is Fnety = -838.2589 + Fy. How am I suppose to figure out Fy from here?

Thanks for any help

 

[rock.freak667]

Since the crate is not moving the direction of the y-axis, the net force in that direction is zero.

 

[coco87]

Since the crate is not moving the direction of the y-axis, the net force in that direction is zero.

I apologize, I left out a part. The answer that is given for that part is: Fy = -217.8476. I did set Fnety to 0 and solve that way, but when I do that I get: Fy = 838.2589. So there is something I'm missing here

 

[PhanthomJay]

...and don't forget the normal force, Fn, in the y direction. That's what you need. Fy you can get with trig.

 

[coco87]

...and don't forget the normal force, Fn, in the y direction. That's what you need. Fy you can get with trig.

Well, Fn = mg (since there's no acceleration), so that's not hard to find. I'm assuming Fy isn't either, and I figured it was only trig, but whenever I try with trig, I can't seem to get anywhere near -217.8476

 

[PhanthomJay]

Well, Fn = mg (since there's no acceleration), so that's not hard to find.

Fn does not equal mg. You are not looking correctly at all components of the forces in the y direction, which, as Rockfreak points out, must equal zero when summed. You already calculated the y component of Fg. You now need Fy (the y component of F) to solve for Fn. Note that Fn lies along the y axis.

I'm assuming Fy isn't either, and I figured it was only trig, but whenever I try with trig, I can't seem to get anywhere near -217.8476

Well, you got the answer for Fx ok, so what's the trig relationship between Fx and Fy, knowing that theta is 27 degrees?

 

[coco87]

Fn does not equal mg. You are not looking correctly at all components of the forces in the y direction, which, as Rockfreak points out, must equal zero when summed. You already calculated the y component of Fg. You now need Fy (the y component of F) to solve for Fn. Note that Fn lies along the y axis. Well, you got the answer for Fx ok, so what's the trig relationship between Fx and Fy, knowing that theta is 27 degrees?

Thank you for your response, it seems I did not understand the relationship between forces correctly. I've got it now :)