# Homework Help: Applying Newtons Laws

1. Jun 10, 2009

### coco87

1. The problem statement, all variables and given/known data
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 [Broken]

2. Relevant equations
F = ma
Fg = mg (g=9.8)
Fnet = Fg + F

3. 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

Last edited by a moderator: May 4, 2017
2. Jun 10, 2009

### rock.freak667

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

3. Jun 10, 2009

### coco87

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

4. Jun 10, 2009

### PhanthomJay

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

5. Jun 10, 2009

### coco87

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

6. Jun 10, 2009

### PhanthomJay

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???

7. Jun 10, 2009

### coco87

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