What is the Tension in a Connecting Cord for Two Masses on an Incline?

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The discussion centers on calculating the tension in a connecting cord between two masses, one on an incline and the other on a horizontal surface. The masses involved are 1.1 kg on a 35° incline and 6.0 kg on a flat surface, with a force of 17.5 N applied. Participants suggest using free body diagrams and the equation F=ma to analyze the forces acting on each mass. The correct approach involves setting up equations for net forces, incorporating gravitational components, and recognizing that both masses accelerate at the same rate. The conversation highlights confusion over the calculations, but ultimately leads to a resolution for one participant.
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Homework Statement


A mass m2 = 1.1 kg, on a 35.0° incline, is connected to a mass m1 = 6.0 kg, on a horizontal surface. The surfaces and the pulley are frictionless. If F = 17.5 N, what is the magnitude of the tension in the connecting cord?
http://img717.imageshack.us/img717/1943/hmwork1.gif


Homework Equations


F=ma


The Attempt at a Solution


Realy stumped on this one, i drew free body diagrams for each of the objects and attempted to eliminate the forces, i believe that F-(tension+m*g*cos(35)) = net force. Then i add the masses together and divide by M to get acceleration, and then plug acceleration back into f=ma but instead of the total mass i use 1.1 for m2.
 
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Hi juggalomike, welcome to PF.
F-(tension+m*g*cos(35)) = net force
This step should be
F-(tension+m*g*sin(35)) = m2*a ...(1)
Next T = m1*a...(2)
Solve these two equation to find a and T.
 
i don't understand what mgsin is... are you talking about m1 or m2? help i am trying to figure this problem out and i still don't get it...
 
mgsinθ is the resultant of normal force and gravity on the mass on the ramp.
 
i still get it wrong.. i don't understand how to get "a" when tension is in the eq... F-(tension+m*g*sin(35)) = m2*a ...(1) :( helppp
 
You can us ma in place of tension. Both objects will be accelerating at the same rate, so both a's will be the same. You now only have 1 variable, and can solve for acceleration.
 
thank you for trying it seems i am doing something wrong bc it tells me its wrong. :(
 
Well, tell us the answer, tell us what you are getting and show your work. I'm sure that we can help you figure out what you are doing wrong.
 
wait! i was doing a silly mistake! i got it! thank you thank you thank you! :)
 
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I still don't understand how to solve this problem.. please explain?
 
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