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Pulling object at an angle problem

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    A woman at an airport is dragging her 9.0 kg suitcase at a constant speed by pulling a strapa at an angle of θ above the horizontal. She pulls the streap with a 24.0 N force, and the friction force on the suitcase is 7.5 N.

    a)Draw a free body diagram of the suitcase
    b)What angle does the strap make with the horizontal?
    c)What normal force does the ground exert on the suitcase?

    2. Relevant equations
    Fa=cosθ-fk=0


    3. The attempt at a solution
    a) This is what I got. Should 7.5 N (friction) point directly to the left instead? If so, why? http://puu.sh/chxIH/b42d9f8b68.png [Broken]

    b) I think this might be it:
    cosθ-fk=0
    cosθ = 7.5N/24N
    θ = cos^-1 (7.5/24)
    θ = 71.79`

    c) This is what I am trying:

    n = mg sinθ
    n= 9.0kg*9.8m/s*sin71.79`
    n= 83.78N


    Please let me know if something is wrong/how to fix it. Thank you
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 18, 2014 #2

    NascentOxygen

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    Friction acts along the rough surface. So friction is always perpendicular to N.

    You then ignored your incorrect diagram, and got the right answer in (a). Though your first line in (a) is not right.

    You are nowhere near being right in (b). Look at it again, once you draw the right diagram.
     
  4. Oct 18, 2014 #3
    Ok so for a this is it then:

    a)http://puu.sh/chQHU/7023dd03b8.png [Broken]

    note: on the post above did you mean 'b' and 'c' , instead of 'a' and 'b' respectively? .
    if so:
    b)θ = 71.79`

    and

    c) would normal force = mg , since there is no movement in the vertical axis?
    if so: 9.0kg*9.8m/s = 88.2 N ?
     
    Last edited by a moderator: May 7, 2017
  5. Oct 19, 2014 #4

    haruspex

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    I feel sure that's what NascentO meant.
    What forces acting on the case have a vertical component? What is the vertical component of each?
     
  6. Oct 19, 2014 #5

    NascentOxygen

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    Weight acts vertically, so is normal to the surface, yes. But the rope conveys a vertical component of force, too, in addition to its horizontal component.
     
  7. Oct 21, 2014 #6
    Ok so I am thinking I use Pythagorean theorem to combine them

    therefore:

    sqrt(((mgcos71.79)^2)+((mgsin71.79)^2))
    = 88.20 N

    I just realized that is the same as mg.
    I am confused on this one, to me it makes sense that the normal force = mg, since there is no vertical movement. But I feel something is wrong.
     
  8. Oct 21, 2014 #7

    NascentOxygen

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    There is a vertical force acting downwards on the body, at the same time there is a force delivering an upwards action to the body. These will partly cancel.
     
  9. Oct 21, 2014 #8
    then:

    vertical component + normal force = mg?
    mgsin71.79 + n = 88.2N
    n = 4.42N
     
  10. Oct 22, 2014 #9

    NascentOxygen

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    That doesn't mean anything to me.

    You should draw the free body diagram showing forces in equilibrium. One force is weight, acting downwards. Another is friction. Etc. Once you have all the forces sketched, their vectors must all add to zero, showing equilibrium.
     
  11. Oct 22, 2014 #10

    haruspex

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    The normal force and mg are only two of the three forces acting on the case which have a vertical component. It is the combination of all three that produces equilibrium.
     
    Last edited: Oct 22, 2014
  12. Oct 23, 2014 #11
    I know that. In my previous post I attempted to combine the 3 of them (normal force, the 'y' component of the external force, and mg).
    I understand you are trying to make me figure it out on my own but this is a bit frustrating. Thank you all for your help but I'll call this one off.
     
  13. Oct 23, 2014 #12

    haruspex

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    Yes.
    mg sin θ for the vertical component of the strap tension? How do you arrive at that?
     
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