What is the Minimum Force Needed to Move a Box Up a Frictionless Ramp?

[neutron star]

Homework Statement

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

Homework Equations

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\sinB
W_{y}=W\cos\theta

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

 

[Doc Al]

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

Good.

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

Good.

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

How did you solve for F from the equations above?

 

[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?

-----

Ok, I got 43.2149=Fcos(25.1)

 

[Doc Al]

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

 

[neutron star]

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

so 43.2149/.9056 = 47.7196N

: )