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

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Homework Help Overview

The problem involves two masses, one on an incline and the other on a horizontal surface, connected by a cord. The incline is at an angle of 35.0°, and the system is subject to a force of 17.5 N. The objective is to determine the tension in the connecting cord.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of free body diagrams and the forces acting on each mass. There are attempts to set up equations based on Newton's second law, but confusion arises regarding the correct forces to include, particularly the components of gravitational force on the incline.

Discussion Status

Some participants have offered guidance on setting up the equations, while others express confusion about specific terms and the relationship between tension and acceleration. There is an ongoing exploration of the problem, with varying levels of understanding among participants.

Contextual Notes

Participants are grappling with the definitions and implications of forces acting on the masses, particularly in relation to the incline. There is a mention of a potential mistake in calculations, and some participants are encouraged to share their work for further assistance.

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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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