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Friction problem that's really bugging me

  1. Oct 19, 2005 #1
    A man drags a 150-lb crate across a floor by pulling on a rope inclined 15 degrees above the horizontal. a) If the coefficient of static friction is 0.50, what tension in the rope is required to start the crate moving?
    The answer is supposedly 68 lb.
    I start by assuming the normal force = weight, so Fn = 150 lb. The force of friction operating against the direction of motion is Ff = Fn*coefficient of static friction, which is at its maximum.
    So 150*.5 = 75 lb = Ff

    Now the x component of the force required to move the crate should be equal to the frictional force (Ff). So Cos 15 = 75/Frope, and Frope turns out to be 77.65 lb. This is wrong, but oddly it is 9.8 off from the correct answer (though I realize this is irrelevant since the problem is in lbs). What am I doing wrong? Thanks :smile:
     
  2. jcsd
  3. Oct 19, 2005 #2

    Tide

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    Nice start but the normal force is not 150 lb. The tension has a vertical component.
     
  4. Oct 19, 2005 #3
    I suspected that, but have no clue how to find it.. it seems like in order to solve for it, I need to find the frictional force, but the frictional force is calculated using the normal force, so I don't know where to start. :(
     
  5. Oct 19, 2005 #4

    Tide

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    Sure you do! You know what the tension is and you calculated its horizontal component so surely you can also calculate the vertical component.
     
  6. Oct 19, 2005 #5
    Er, but the only reason I know the tension is 68 lb is because I checked the answer, it's not given at the start of the problem. And I can't calculate the normal force since I don't know the vertical component, and since I don't know the normal force I can't calculate the frictional force, and thus the horizontal component. Obviously I'm missing something simple, but I'm not sure what. :(
     
  7. Oct 19, 2005 #6

    Tide

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    If T is the tension in the rope then the horizontal component is [itex]T \cos \theta[/itex] where [itex]\theta[/itex] is the angle with respect to the horizontal. Therefore, the vertical component will be ...??
     
  8. Oct 19, 2005 #7
    I knew Tsintheta was the vertical component, I was just afraid to leave the normal force as 150 - Fropesin15 and try to solve from there. Anyway, I plugged it in and got the right answer. Thanks a lot :)
     
  9. Oct 20, 2005 #8

    Tide

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    You're welcome!
     
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