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Pulling a sled, work, and tension

  1. Feb 26, 2006 #1
    The problem is: A sledge loaded with bricks has a total mass of 17.0 kg and is pulled at constant speed by a rope inclined at 19.5 degrees above the horizontal. The sledge moves a distance of 21.0 m on a horizontal surface. The coefiicient of kinetic friction between the sledge and surface is 0.500.

    a) what is the tension of the rope?
    b) How much work is done by the rope on the sledge?
    c) What is the mechanical energy lost due to friction?

    So for (a) i figured to use the equation F(net)=ma which computes out to T=83.3/(sin19.5+cos19.5) but I came out with the wrong answer.
    And in order to work out b and c I need to have the correct tension. Any help would be apreciated :biggrin:
     
  2. jcsd
  3. Feb 26, 2006 #2

    Hootenanny

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    You cant use [itex]F = ma [/itex] for (a) because there is no acceleration.
     
  4. Feb 26, 2006 #3
    I thought I could use that equation to get the sum of all forces and solve for T (tension)
     
  5. Feb 26, 2006 #4

    Hootenanny

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    You cant use it because like I said, there is not acceleration [itex] \Rightarrow F = ma = m \times 0 = 0 [/itex]. Therefore, you know that all forces are balanced.
     
  6. Feb 26, 2006 #5

    Hootenanny

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    Show me how you've summed the forces.
     
  7. Feb 26, 2006 #6
    So what equation would I use to go about finding the tension?
     
  8. Feb 26, 2006 #7

    Hootenanny

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    Just think about what forces are acting. There is a component of gravity pulling it down the slope together with a frictional force. And the tension in the rope is pulling up the slope.
     
  9. Feb 26, 2006 #8
    But the sledge is horizontal to the ground while the rope is the only thing inclined at 19.5 degrees
     
  10. Feb 26, 2006 #9

    Hootenanny

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    Sorry I'd miss read the question. In that case there is only the frictional force opposing the tension. You need to resolve the tension using trig so that it is horiztonal.
     
  11. Feb 26, 2006 #10
    It's cool... so I came up with f(kinetic)= .5*17*9.8 and then made that to equal Tsin19.5+Tcos19.5 and then solved for T... am i on the right track?
     
  12. Feb 26, 2006 #11

    Hootenanny

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    The equation for frictional force is [itex]F = \mu R [/itex] where R is the normal reaction force. However,R will be reduced becuase there is a component of tension acting in the horizontal plane. [itex]T\sin 19.5 [/itex]. You don't need to put it in later on.
     
  13. Feb 26, 2006 #12
    I have a similar question in my book, but they use 20 degrees, 18 kg and a distance with 20 m with the same kinetic coefficient.... I solved for T=18/cos20 and came up with the wrong answer... the answer they have in the back of the book is 79.4N
     
  14. Feb 26, 2006 #13
    i am still coming up with the wrong answer...
     
  15. Feb 26, 2006 #14
    I thought that the natural reaction force is just mass*gravity?
     
  16. Feb 26, 2006 #15
    We note that there is no horizontal acceleration, so the horizontal component of the tension ([tex]Tcos\theta[/tex]) must equal the frictional foce. The frictional force is [tex]\mu N[/tex], where N is the normal force. Many times the normal force is just equal to the weight (mg) of the block, but there is an upward component of the tension that reduces the normal force ([tex]Tsin\theta[/tex]).

    [tex]\mu N = Tcos\theta[/tex]
    [tex]0.500(17*9.8 - Tsin 19.5) = Tcos 19.5[/tex]

    Solve for T and you get T is about 75 N. Is this correct?
     
  17. Mar 1, 2006 #16
    Yeah... I didn't understand that there was an upward componenet of tension that reduces the normal force. Thanks!!! :)
     
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