Finishing off a force equation

  • #1

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 [Broken]

Homework Equations





The Attempt at a Solution


[tex]\sum\vec{F}=ma[/tex]

[tex]\sum\vec{F}=0[/tex]

[tex]\sum\ F_{x}=0 => F_{x}-W_{x}=0[/tex]

[tex]\sum\ F_{y}=0[/tex]

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
[tex]F_{x}=F\cos{B}[/tex]
[tex]W_{x}=W\sin\theta[/tex]
[tex]F_{y}=F[/tex][tex]\sin[/tex][tex]B[/tex]
[tex]W_{y}=W\cos\theta[/tex]

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

  • #2
Doc Al
Mentor
45,011
1,288
[tex]\sum\ F_{x}=0 => F_{x}-W_{x}=0[/tex]
Good.

w=mg
[tex]F_{x}=F\cos{B}[/tex]
[tex]W_{x}=W\sin\theta[/tex]
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?
 
  • #3
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 [tex]43.2149=Fcos(25.1)[/tex]
 
Last edited:
  • #4
Doc Al
Mentor
45,011
1,288
Set up the equation Fx = Wx and solve for F. (Using your equations for Fx and Wx.)
 
  • #5
Alright, great! I got 43.2149=F*.9056

so 43.2149/.9056 = 47.7196N

: )
 

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