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

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SUMMARY

The forum discussion centers on calculating the tension in a connecting cord between two masses, m1 = 6.0 kg on a horizontal surface and m2 = 1.1 kg on a 35.0° incline, with a force F = 17.5 N applied. The correct equations derived include F - (tension + m2 * g * sin(35°)) = m2 * a and T = m1 * a, where 'g' represents gravitational acceleration. The solution involves solving these equations simultaneously to find both acceleration and tension. The user initially struggled with the application of forces but ultimately resolved their confusion with guidance from other forum members.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of free body diagrams for analyzing forces
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Basic principles of dynamics involving connected masses
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  • Study the derivation of tension in connected mass systems
  • Learn about free body diagram techniques for inclined planes
  • Explore the role of gravitational force components in dynamics
  • Practice solving problems involving multiple masses and pulleys
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This discussion is beneficial for physics students, educators, and anyone interested in understanding dynamics involving connected masses and forces on inclines.

juggalomike
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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! :)
 
  • #10
I still don't understand how to solve this problem.. please explain?
 

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