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What is Tension of the rope?

  1. Apr 14, 2005 #1
    A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at a contstant speed by a rope. The rope is inclined at 20.00degrees about the horizontal, and the sledge moves a distance of 20.0m on a horizontal surface. the coefficient of kinetic friction between the sledge and surface is 0.500. What is Tension of the rope?

    For the free body diagram...I have tension on the +x axis and Force of friction on the -x axis. There is no resultant force on the y axis so i did not need that. BUT, tension needs to be broken up into component form because of 20 degrees incline, and thats what I dont know how to do because so far my resultant x force is T=Fk..which is not right.
    Last edited: Apr 14, 2005
  2. jcsd
  3. Apr 14, 2005 #2
    You would do well in drawing the free body diagram for this problem. Please explain your reasoning to us first... we want it to be worth both your time and our time.

    Ahh I see you edited your post... sorry about that disregard my response... Let me work it out and tell you a strategy.
  4. Apr 14, 2005 #3


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    I think the key phrase here is,
    "is pulled at a contstant speed by a rope",
    meaning that there is no acceleration (in either direction), which means that the new force = 0.
    when you think of the force diagram, there would be 5 forces you might want to draw, the force of wieght (mg acting straight down), the force perpendicular to the place/the normal force (mg*cos(angle)), the force parrallel to the place (mg*sin(angle)),the force of friction (normal force*coefficient of friction), and the force of tension.

    since the box is moving up the ramp at a constant speed, the we know that the forces down the ramp = the forces up the ramp.
    what forces are acting down the ramp, and which are acting up?
    the force of friction acts in the opposite direction of the motion, so it is acting down, and obviosly, the parralell force is acting down too.
    the only force acting up the ramp is the tention.
    set these 3 forces (2 on one side and one on the other side of the equal sign), and then solve for the force of tension.

    mg*sin(angle) + mg*cos(angle)*coefficient of friction = Force of tension
  5. Apr 14, 2005 #4
    But the sledge is not being pulled up a ramp thus it is not on an inline itself. the rope is the only thing on the incline and the sledge is only being moved horizontally. I tried plugging in the numbers and i got 113.05N but I know the answer is supposed to be 79.4 N i just don't know how to get there.
  6. Apr 14, 2005 #5


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    I misread the question, I took it that the object was going up a ramp, but in reality, it is the rope that is at the angle.

    OK, now you know that the force in the X direction of the tension is equal to the frictional force. the force in the Y direction of the tension is in the opposite direction of the weight, thus reducing the "felt" force of weight and reducing friction.

    for is friction = the normal force * the coefficient of friction,
    we need to find the normal force, which is now equal to
    normal force = mg - T*sin(angle)
    so friction = (mg - T*sin(angle))*coefficient of friction

    so set that equal to the X direction foce of tension,
    T*cos(angle) = (mg-T*sin(angle))*coefficient
    and solve for T, which is the tension force.
  7. Apr 15, 2005 #6
    79.4N is the correct answer
    Suppose the force we look for is F

    along the y-direction [tex]N = mg - Fsin(20)[/tex] N is the normal force
    along the x-direction [tex]0 = Fcos(20)-0.5mg+0.5Fsin(20)[/tex]

    Solve for F
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