# Friction and pullies at slope

1. Sep 28, 2010

1. The problem statement, all variables and given/known data
In the diagram, the pulley is frictionless and the
string is massless. Given: m2 = 90 kg, m1 = 2m2,  angle=
26o, and μk = 0.11. Determine the tension in the string.

The m2 is on a sloped hill, m1 is dangling down off of the pully

2. Relevant equations

Fg=mg
F=ma
Fk=uk(Fn)

3. The attempt at a solution

I found the net force for m2 to be 386.643-the friction force which is 87.201=299.442N. then I subtracted from the Fg of m1=1764N. So 1764-299.442=1464.56N so the tension=1464.56N did I do this right?

2. Sep 28, 2010

### PhanthomJay

That's NOT the net force, because you did not include the tension force, T, in that equation. The net force is then equal to m2a
No, you have to look at the hanging block separately using a free body diagram. What are the forces acting on m1, what's the net force? Then again the net force =m1a. Solve the 2 equations with 2 unknowns for T.

Last edited: Sep 28, 2010
3. Sep 28, 2010

Ummm im not sure exactly if I got what ur saying right, but I used the Fnet for m2 I found before that you said was right (299.442) and used the mass of m2 (90) to find a which equaled 3.32713m/s^2. Then i used that acceleration and used it to multiply with m1 (180) to get 598.883N. Idk if that is the T or what... I probably did none of that right hahaa.

4. Sep 28, 2010

### PhanthomJay

Sorry, i meant to say that was not the net force. Ill edit that response.

5. Sep 29, 2010

Okay so I made separate free body diagrams of each.

On m1 there would only be 2 forces correct? The FGm1-1764 going down and the T going up?

Then on m2 there would be a Force of 299.442 (after subtracting the friction force) but then going that same direction as F would be T right?

So I did T+F=m2a which would come out as T=m2a-F

Then for m1 i did Fgm1-T=m1a then replaced the T with (m2a-F)

so Fgm1-(m2a-F)=m1a which finding acceleration would come out as a=(Fgm1+F)/(m1+m2)

so a=(1764+299.442)/(180+90)=7.64238m/s^2

so then I plugged that acceleration into T=m2a-F so T=90(7.64238)-299.442= a tension of 388.372N. I hope that is right this has been a pain haaha

6. Sep 29, 2010

### jhae2.718

I get a different answer. (Edit: seems like I solved a different variation of the problem based on my interpretation of the words...)

Last edited: Sep 29, 2010
7. Sep 29, 2010