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Acceleration of slinding crate pushed and pulled at the same time

  1. Oct 18, 2009 #1
    Two workers must slide a crate designed to be pushed and pulled at the same time as shown in the figure below. Joe can always exert twice as much force as Paul, and Paul can exert 155 N of force. The crate has a mass m = 41 kg, the angle θ = 20° and the coefficient of kinetic friction between the floor and crate is 0.55.

    http://s783.photobucket.com/albums/yy113/eandronic/?action=view&current=4-p-073.gif

    If we want to move the crate as fast as possible, is it better to have Paul push and Joe pull, or vice-versa?
    Intuitively, I consider it's better to have Paul push, and Joe pull.

    Calculate the (horizontal) acceleration for both cases to find out.

    Paul pushing m/s2 ?
    Joe pushing m/s2 ?

    I am not sure which are all the forces on the x axis acting on the crate and how to calculate them at the given angle?
     

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    Last edited: Oct 19, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    Delphi51

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    What exactly is the question?
    It will take a long time to get your attachment approved so we can see it. Suggest you upload the diagram to a free photo site like photobucket.com and post a link here.
     
  4. Oct 18, 2009 #3
    http://s783.photobucket.com/albums/yy113/eandronic/

    If we want to move the crate as fast as possible, is it better to have Paul push and Joe pull, or vice-versa?
    1 It's better to have Paul push, and Joe pull.

    Calculate the (horizontal) acceleration for both cases to find out.

    Paul pushing m/s2 ?
    Joe pushing m/s2 ?
     
  5. Oct 18, 2009 #4

    Delphi51

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    Okay, so you'll have to use trig to find the horizontal and vertical parts of each force in each of the two scenarios. And find the friction force. Then you'll be in a position to do "sum of forces = ma".
     
  6. Oct 18, 2009 #5
    what i tried is
    when paul pushes:
    a(x)=fpush_paul*cos(180-20)+fpull_joe*cos20 -Ff

    when joe pushes:
    a(x)=fpush_joe*cos160+f pull_paul*cos20 - Ff

    and I am not sure of this and I also don't know the correct expression for Fn, therefore I don't have the good expression for Ff.

    in both cases i get negative accelerations.
     
  7. Oct 18, 2009 #6

    Delphi51

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    You have forgotten to include friction.
    Careful with cos(180-20). It is negative. Better to use cos(20) I'd say.
     
  8. Oct 19, 2009 #7
    I have included and subtracted friction. Just that I don't know the formula for it.

    Ff=Fn*miu(k)

    miu of kinetic friction is given, Fn is unknown and I am not sure how to express it.

    One option is Fn - m*g*sin_20 + Fpull_sin_20+ Fpush_sin_20 = 0 the other option is Fn=mgsin20.

    Regarding the acceleration on the x axis:
    If the angle is 20 in both cases would this be a right formula then? I am not sure what it is that quantifies the difference that I intuitively understand should give me a greater magnitude when the stronger guy pulls.

    when paul pushes:
    a(x)=(fpush_paul*cos20+fpull_joe*cos20 -Ffriction) / mass

    when joe pushes:
    a(x)=(fpush_joe*cos20+f pull_paul*cos20 - Ffriction) / mass

    Please advise.
     
  9. Oct 19, 2009 #8

    Delphi51

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    Yes, less friction when Joe pulls because the stronger force is partly upward and it therefore reduces Fn.
    Fn is the total vertical (downward) force pushing the block against the ground. Add mg, the component of Joe's force that is upward and the component of Paul's force that is downward.
     
  10. Oct 24, 2009 #9
    ok..note :paul=155N and joe=310N
    case 1:when paul pushes and joe pulls..
    ma(y)=N+F_joe sin20-F_paul sin20 - mg
    a(y)=0; solve for N
    ma(x)=F_joe cos20+F_paul cos20-miu N

    case 2:when joe pushes and paul pulls
    ma(y)=N+F_paul sin20-F_joe sin20 - mg
    a(y)=0 ;solve for N
    ma(x)=F_joe cos20+F_paul cos20-miu N
     
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