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Homework Help: Tension in the cord problem

  1. Jul 7, 2010 #1
    1. The problem statement, all variables and given/known data

    A worker pulls 3 carts connected with 2 horizontal cords. The mass of the last cart is 16.1kg, the middle cart's is 13.2kg and the first cart's is 15.0kg. Friction is negligible. A third cord which pulls on the first cart is at an angle of 21.0 degrees above the horizontal, has a tension of magnitude 35.3N. Determine the magnitude of
    a) the acceleration of the carts
    b)the tension in the last cord
    c) the tension in the middle cord.

    2. Relevant equations

    sigma F= ma

    3. The attempt at a solution
    a) I tried using Newton's 2nd law equation to find acceleration by using the tension in the cord as the force, but I don't get the write answer.
    35.3N = (16.1kg + 13.2kg + 15.0kg)a
    0.797m/s squared
    I then tried finding the horizontal force pulling on the carts instead of the force of the tension of the rope which was at an angle of 21 degrees above the horizontal.
    cos 21 = Fb/35.3N
    Fb = 32.955 N
    32.955N = (44.3kg)a
    0.744m/s squared= a
    I still get it wrong.
    The correct answer is 0.744 m/s squared.
    b) and c) I have no idea what I'm doing there.
     
    Last edited: Jul 7, 2010
  2. jcsd
  3. Jul 7, 2010 #2
    Re: Tension

    Can you draw free body diagrams of the forces that act on each cart?

    a) does not actually require this, but it is helpful for the others to help you see how to solve them.

    You got a) correct, what do you not understand on a?
     
    Last edited: Jul 7, 2010
  4. Jul 7, 2010 #3
    Re: Tension

    Do i just write it out here? Well,
    35.3N
    /
    | Fn | Fn | Fn /
    | | | / 21 degrees
    16.1 kg -------- 13.2kg --------- 16.1kg ______________________
    | | | 32.955N
    |Fg |Fg |Fg
     
  5. Jul 7, 2010 #4
    Re: Tension

    Well... that didn't end up too well...
    But it's just the horizontal force we're looking at isn't it? Do the normal and gravity forces affect the acceleration?
     
  6. Jul 7, 2010 #5
    Re: Tension

    Not in this case. Its all the horizontal forces, correct. Lets start here: Why do you think you got a) wrong, it looks correct?
     
  7. Jul 7, 2010 #6
    Re: Tension

    Oh... wait a minute a) is right >_> I just looked at the wrong answer.
    so to find the tension of the last rope do i just do this:? F= 16.1kg(0.744) = 11.97N
    and for the middle rope do i just do the same thing, but use the mass of bot the second cart and the last cart?
     
  8. Jul 7, 2010 #7
    Re: Tension

    Yep.
    Let me ask you this. How many forces are acting on the last cart in the horizontal?
     
  9. Jul 7, 2010 #8
    Re: Tension

    3? Gravity, normal force and the force applied?
     
  10. Jul 7, 2010 #9
    Re: Tension

    Ok. But only one in the horizontal and its Tension, what you are calling the applied force and you reasoned that the last cart accelerates the same as all the other carts so you got it.

    So, how many forces and in what direction do these forces act on the second cart, the 13.2 kg cart?
     
  11. Jul 7, 2010 #10
    Re: Tension

    Are there not two ropes attached to the 2nd cart and they "pull" opposite each other, yes?
     
  12. Jul 7, 2010 #11
    Re: Tension

    The tension from the first cord and the tension of the cord connecting the last cart to it?
     
  13. Jul 7, 2010 #12
    Re: Tension

    So yes, the tension from the cord attached to the last cart, and the tension of the cord attached to the 1st cart, right? Are there not 3 cords?
     
  14. Jul 7, 2010 #13
    Re: Tension

    yes.
     
  15. Jul 7, 2010 #14
    Re: Tension

    So a picture would look like this....

    carts.JPG

    If so lets now deal with the 2nd cart and the two ropes attached and the net force on the 2nd cart.
     
    Last edited: Jul 7, 2010
  16. Jul 7, 2010 #15
    Re: Tension

    Yeah
     
  17. Jul 7, 2010 #16
    Re: Tension

    So what is the net force on the middle cart (2nd cart) since it is accelerating? F = ma

    What number do you get in Newtons?
     
    Last edited: Jul 7, 2010
  18. Jul 7, 2010 #17
    Re: Tension

    F = (16.1 + 13.2) (0.744) = 11.98N
    Is that how you do it?
     
  19. Jul 7, 2010 #18
    Re: Tension

    Hold on. What is the mass of just the 2nd cart and what is its acceleration? You know both of these, one was given (mass,) and one you already determined, you just thought you got it wrong. ALL the carts must have the same acceleration, yes?
     
  20. Jul 7, 2010 #19
    Re: Tension

    The mass of the 2nd cart is 13.2kg and its accleration is 0.744m/s squared.
     
  21. Jul 7, 2010 #20
    Re: Tension

    Yes.
    So you know the net force. 13.2 * 0.744

    So what leads to that net force. There are two forces acting. You already calculated one of them. These two forces are both tension. Look at the picture. You know the force to the left and you know the net force which is to the right. So the force of tension to the right, which is the answer to the last part of your question must be what?
    Take a look at the picture, maybe you can see it better visually.
    carts.JPG
     
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