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Help wit a force question please

  1. Sep 21, 2004 #1
    Hi, can somebody help me out with this question:

    A car is stuck on the side of the road and the driver has only a long piece of non-stretchable rope. Seeing a tree 20 m away from the car he ties the rope to the tree and then to the car such that it is quite tight and any sag is negligible. The driver then pushes the rope at the midpoint of the rope with a push prependicular yet horizontal to the line made by the rope. What force does the rope exert on the car if the driver can push the rope to a distance of 2 m with a force of 600 N?

    I know (or atleast think) that I must use the similar triangles of the force and distance vectors. I have tried it many different times and i'm getting 3000 N being exerted on the car. Can anyone show me if its right. Thanks!
     
  2. jcsd
  3. Sep 21, 2004 #2

    Doc Al

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    You are almost there. Draw a diagram of the forces acting on the piece of rope that the driver pushes on. Call the force that he pushes with F, and the tension in the rope T. The forces must balance, since there is equilibrium. (The force on the car will be the tension generated in the rope.)
     
  4. Sep 21, 2004 #3
    Yea I drew the diagram and put 600 N perpendicular to the rope. Are the forces acting on the car and tree equal? I ask this because the car is moving while the tree sits still. What effect does the tree have on the tensions.
     
  5. Sep 21, 2004 #4

    Doc Al

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    If the person pushes on the rope with 600 N, then the rope must push back on the person with 600 N. What must the tension be for the rope to exert a sideways force of 600 N? Draw the picture.
     
  6. Sep 21, 2004 #5
    I can't seem to understand what u mean by the rope having to exert 600 N sidways. Why must it have to exert 600 N when the force is perpendicular to it. I drew the diagram with the 600 N force being exerted onto the rope and the tensions in the rope directed towards both the car and tree. Sorry if i'm being dumb lol.
     
  7. Sep 21, 2004 #6

    Doc Al

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    OK, but do you agree that the rope must be pushing back on the driver? (Consider Newton's 3rd law.) What's the component of the rope tension pushing back on the driver?

    Call the line between car and tree the y-axis. The driver pushes the rope in the x-direction. The rope is now pushed sideways (it is no longer straight), so the tension now has a component in the x-direction. The net force that the rope exerts in the x-direction must balance the force that the man exerts (600 N). Set up the equilibrium equation and you can solve for T.
     
  8. Sep 21, 2004 #7
    Oh, now I understand what u meant by the rope exerting a force sideways, its a different component. And yea I agree with the rope pushing back on the driver of course. Ok ill try to solve it now and I'll tell you what I get. Thanks for the help.
     
  9. Sep 21, 2004 #8
    Ok, I've tried it again many times yet I'm still getting 3000 N. Is this right and if not can u tell me what answer you get. If I have the answer I most of the time find a way to get there and I better understand. The answer is useless unless I know how to get there in terms of marks anyway.
     
  10. Sep 21, 2004 #9

    Doc Al

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    Tell me how you get 3000 and I'll point out what you're doing wrong. :smile: (I think you are making a simple mistake.)

    Hint: both sides of the rope pull back on the driver.
     
  11. Sep 21, 2004 #10
    The way I'm getting 3000 N is by using similar triangles. One triangle consists of the the distances. One side is 10 m and the other is 2 m. The other triangle consists of the forces. The side corresponding to the 2 m is 600 N and the one corresponding to the 10 m is x. When I solve for x I get 3000 N. Is this wrong?
     
  12. Sep 21, 2004 #11

    Doc Al

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    What you need to do is find the sideways component (the x-component) of the tension from each side of the rope. Since the rope (and thus the tension) is at an angle given by that right triangle, you can use it to find the x-component of the force exerted by each side of the rope. If the rope tension is T, the x-component is [itex]T_x = T cos\theta = T (2/10) = T/5[/itex]. But two sides of the rope are being pulled by the driver, so his force of 600 N is equal to [itex]2 T_x[/itex], thus [itex]T_x = 300[/itex]. Now solve for the tension T: [itex]T = 5 T_x = 1500[/itex] N.

    Let me know if this makes sense.
     
  13. Sep 22, 2004 #12
    OK I knew the answer was 1500 N from before but I was still getting 3000 N. I thought to myself that since the force of 600 N is being dispersed between two objects it must be halved to isolate the car. And if I use 300 N on my similar triangles I do end up getting 1500 N. Tell me if this is a logical way to get the answer. Thanks for the help.
     
  14. Sep 22, 2004 #13

    Doc Al

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    I don't know what you mean by "isolate the car". I would say that the applied force of 600 N is being spread over both sides of the rope. So each side must be contributing an x-component of 300 N of force.
     
  15. Sep 22, 2004 #14
    Ok so the x direction is the direction the 600 N force is being applied to right?
     
  16. Sep 22, 2004 #15

    Doc Al

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    No, the 600 N is in the x-direction. The line between tree and car is the y-direction; sideways to that (where the rope is being pushed) is the x-direction.
     
  17. Sep 22, 2004 #16

    Doc Al

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    That's right.
     
  18. Sep 22, 2004 #17
    Oh ok I think I finally get it... but one last thing... the "T" in your equation is the resultant of the x and y force right? And since this is true you didnt need to solve for the y force because its right triangle.
     
  19. Sep 22, 2004 #18

    Doc Al

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    T represents the tension in the rope created by the sideways force. That's the force that the rope will exert on the car (and on the tree), which is what you were asked to find.
     
  20. Sep 22, 2004 #19
    Ok I finally solved it while understanding it...It seriously is all about the way you draw the vector diagrams and the way you picture it (I seem to have trouble with that). Thanks a lot for the help... but can I ask you one last thing? Is there any mathematical way to show that the 600 N must be halved for the x-components? Or is it just something you have to know by reading the question and state it in the answer?
     
  21. Sep 22, 2004 #20

    Doc Al

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    You should be able to see it from the symmetry of the problem. Each side of the rope is pulling back on the man, so each pulls the same amount, which must be half of the total.
     
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