Calculating Tension in a Pulley System on a Sloped Surface

[SilentBlade91]

Homework Statement

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

Homework Equations

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

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?

 

[PhanthomJay]

I found the net force for m2 to be 386.643-the friction force which is 87.201=299.442N.

That's NOT[/color] the net force, because you did not include the tension force, T, in that equation. The net force is then equal to m2a

subtracted from the Fg of m1=1764N. So 1764-299.442=1464.56N so the tension=1464.56N did I do this right?

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.

 

[SilentBlade91]

Ummm I am 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.

 

[PhanthomJay]

Ummm I am 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.

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

 

[SilentBlade91]

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

 

[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...)

 

[SilentBlade91]

Turns out my answer was right.