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Newton law problem?

  1. Dec 6, 2003 #1
    It is almost time for the end of term exam and I am looking over my old exams and am confused by one problem. Masses m1,m2,m3,m4 are cnnected together(by a "string") on an incline [tex]theta[/tex] = 22 degrees. m1 and tension = T1(connects m1 and m2) start nearest the bottem of the ramp, and a force F = 30N is pulling on m4 parallel to the ramp. We had to find the acceleration of m2 and the m3 tensions. The acceleration of course is the same for all of them.

    I got a = F cos[tex]theta[/tex] / (m4 + m3 + m2 + m1) This got marked wrong, if anyone knows why I would greatly appreciate it. Thank you
     
    Last edited: Dec 6, 2003
  2. jcsd
  3. Dec 6, 2003 #2

    Doc Al

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    Staff: Mentor

    Hint: the force F is parallel to the ramp.
     
  4. Dec 6, 2003 #3
    I am just really confused on this one, is it not needed to take the force in the "x" direction because since the acceleration and force are on the same "line", which we could use as the "x" axis? When finding the tension I had, for one example, m4*a = F cos(theta) - T3 cos(theta) and this was wrong too. I am really confused here if anyone could help with this type of problem, thank you very much.
     
  5. Dec 6, 2003 #4

    Doc Al

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    I guess my "hint" wasn't that helpful. :smile: I'll try again.

    To find the overall acceleration, first find all the external forces acting on the masses (taking them as a system), taking their components parallel to the plane. The weight acts down, so you need to find the component of the weights acting down the plane. (Careful with the angles!) The only other force is the applied force F. Now add these forces, then apply
    Fnet = mtotal a

    To find the tension in the strings, apply F=ma to each mass separately and solve the set of equations.

    Edit: Just to be clear, the first comment is a "short cut". By treating the masses as a single system, you can ignore the tensions in the strings--since they are all internal forces. If this is confusing you, forget it. Just apply F=ma to each mass separately: you'll get the same answer.
     
    Last edited: Dec 6, 2003
  6. Dec 7, 2003 #5
    So the force on the individual pieces, and the object as a whole would just be the applied force to the x plane and the gravitational force to the x plane?

    a = F + mg sin(Theta) / m

    Thank you very much for the help, I have no idea why I was having so much trouble.
     
    Last edited: Dec 7, 2003
  7. Dec 7, 2003 #6

    Doc Al

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    Right! Treating the four masses as a single object, that's how you can find the acceleration. (I assume you meant a = (F + mg sin(Theta)) / m )

    When treating each mass separately (which you must do to find the tension each string), don't forget to include the tension in any string touching the piece as well as the weight.
     
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