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Magnitude of tenstion required to get chest moving.

  1. Mar 18, 2017 #1
    1. The problem statement, all variables and given/known data
    A filled treasure chest (Mass = 375 kg) with a long rope tied along its center lies in the middle of a room. Dirk wishes to drag the chest, but there is friction between the chest and the floor, with the coeffecient of static friction = .52

    If the angle between the rope and the floor is 30 degrees, what is the magnitude of the tension required just to get the chest moving?

    2. Relevant equations
    [itex]f=ma[/itex]
    [itex] F_static = MkFn [/itex]


    3. The attempt at a solution
    First I would like for everyone to see this diagram to see if it is correct:

    http://i.imgur.com/mCpPKIR.png

    So is it true, that the Force of static friction is going to be opposite to the tension force? Meaning the force of friction isn't going to be pointed in the -j hat direction? Because the rope is being lifted at an angle of 30 degrees, does that mean that the force of static friction has i hat and j hat components?

    Working this problem out... I did:

    [itex] F_g = 375 kg * 9.81 = 3678.75 N [/itex]
    [itex] F_n = F_g [/itex]
    [itex] F_static = F_n * .52 = 1912.95 N [/itex]

    so finding the x component of [itex] F_static [/itex] :

    [itex] 1912.95cos(30) = 1655.840572 [/itex]

    and now y component of static :
    [itex] 1912.95sin(30) = 956.475 [/itex]

    Okay now this is where I'm stumped...

    My books says the answer is 1700 N, but the force of static friction is 1912.95 N! wouldn't that mean that the tension has to be at least this much just to get the thing moving?
     
  2. jcsd
  3. Mar 18, 2017 #2

    gneill

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    Staff: Mentor

    No. Friction is directed opposite to the direction of motion (or the potential direction of motion in the case of static friction). Friction parallels the surfaces between the sliding objects. In this case the chest is going to slide along the horizontal floor, so the friction force vector will be parallel to the floor.

    Note that the weight of the chest is not the only force component contributing in the vertical direction.
     
  4. Mar 18, 2017 #3
    Okay so this brings me to further questions:

    Wouldn't the direction of motion = tension in this case? So wouldn't the friction be opposite to this?

    If so, since the rope is being pulled at an angle of 30 degrees, that means it would be moving in the x direction AND y direction? I picture this as the right side of the chest being slightly lifted while the left side remains on the ground.
     
  5. Mar 18, 2017 #4
    Anyways, if the force of friction is only going to the -i hat direction, and tension is at an angle 30 degrees from the floor, that means xcos(30) = static friction,

    or x = static friction / cos(30)

    so x = 2208 N, which isn't the answer either..
     
  6. Mar 18, 2017 #5

    gneill

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    Staff: Mentor

    No, the motion will be along the floor. The chest will slide along the floor and friction occurs between the chest and the floor.
    No. Even if the chest tipped it would still not lose contact with the floor. Try an experiment: tie a long string around a book lying on a flat surface and pull it at about a 30° angle so that it just starts moving. Does the book rise in the air at the end of the string?

    It's best to imagine the applied force acting through the center of mass of the object (chest). So no tipping involved.

    Hint: The vertical component of the applied force counters some of the weight of the chest.
     
  7. Mar 18, 2017 #6
    I see. So this means that [itex] F_g =/= F_n [/itex], correct? so [itex] F_g = F_n + F_{tension(y)} [/itex], right?
     
  8. Mar 18, 2017 #7

    gneill

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    Staff: Mentor

    While it's true that x = static friction / cos(30), note that the value of static friction is not what you're claiming. As I stated previously, some of the weight of the chest is being countered by the vertical component of x.
     
  9. Mar 18, 2017 #8

    gneill

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    Staff: Mentor

    Right.
     
  10. Mar 18, 2017 #9
    First I would like to say thanks for helping me through this problem Gneill, I think it would be fair when I get proficient enough to land a job to compensate you monetarily for the help haha.

    Anyways, using the knowledge you gave me,

    I realized that
    [itex] F_g = F_n + xsin(30) [/itex]
    [itex] xcos(30) = F_n c [/itex] with c = .52

    solved these two equations for x, plugged that in to the opposite equation, then solved for x one last time, and reached answer of:

    [itex] 1698.85 N [/itex], which is very close to the books answer, I'm sure they rounded. Pretty sure my method is now correct, thanks a lot Gneill.
     
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