# Finishing off a force equation

1. Oct 11, 2009

### neutron star

1. The problem statement, all variables and given/known data
A 18.6-kg box rests on a frictionless ramp with a 13.7° slope. The mover pulls on a rope attached to the box to pull it up the incline. If the rope makes an angle of 38.8° with the horizontal, what is the smallest force F the mover will have to exert to move the box up the ramp?

____ N
http://img93.imageshack.us/img93/7287/4figure44alt.gif [Broken]

2. Relevant equations

3. The attempt at a solution
$$\sum\vec{F}=ma$$

$$\sum\vec{F}=0$$

$$\sum\ F_{x}=0 => F_{x}-W_{x}=0$$

$$\sum\ F_{y}=0$$

I made a graph from the ramp model and found:

a=0
m=18.6 kg
θ=13.7°
σ=38.8°
B=25.1°
F=?

w=mg
$$F_{x}=F\cos{B}$$
$$W_{x}=W\sin\theta$$
$$F_{y}=F$$$$\sin$$$$B$$
$$W_{y}=W\cos\theta$$

I thought 18.6(9.81)sinθ would give me the answer but that wasn't correct.

Last edited by a moderator: May 4, 2017
2. Oct 11, 2009

### Staff: Mentor

Good.

Good.

How did you solve for F from the equations above?

3. Oct 11, 2009

### neutron star

Oh, I think I see what I was missing. I need to find w=mg and plug that into Wx and solve right?

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Ok, I got $$43.2149=Fcos(25.1)$$

Last edited: Oct 11, 2009
4. Oct 11, 2009

### Staff: Mentor

Set up the equation Fx = Wx and solve for F. (Using your equations for Fx and Wx.)

5. Oct 11, 2009

### neutron star

Alright, great! I got 43.2149=F*.9056

so 43.2149/.9056 = 47.7196N

: )