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Force to raise a cylinder over an edge

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    The cylinder with mass 200[kg] and radius 0.8[m] tries to climb a 0.3[m] high edge with a force F exerted at an angle of 300. what's the force and what's the minimum force for the cylinder to climb

    2. Relevant equations
    Moments: F1L1=F2L2

    3. The attempt at a solution
    The weight balances the the force, i made moments around the edge and got a true answer F=1670[N], but i don't understand the second question, about the minimum force, and it's about 766[N]
     

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  2. jcsd
  3. Jan 31, 2015 #2

    mfb

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    "Minimal force" might mean the force needed if the angle can be adjusted. But then the two values should not be so different. Maybe you can adjust the point where it acts on as well. If that is the full problem statement and 766 N is the given answer, it is poorly written.
     
  4. Jan 31, 2015 #3

    haruspex

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    To get an answer so much lower, it must be allowed to apply the force further from the step.
     
  5. Jan 31, 2015 #4

    TSny

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    Ah! Yes. That's how you can get about 766 N. o0)
     
    Last edited: Jan 31, 2015
  6. Feb 1, 2015 #5
    Hi Karol.

    I consider that the ball pivots about point of contact (O). So the CoM of the ball will perform circular motion about point O.

    Now - work energy theorem states that 'Work done by all the forces on the system is equal to the change in the kinetic energy'.
    As the required force is minimum so the ball the velocity of the CoM of the ball at the final position (i.e when the CoM is just above the point O) should be zero.

    Now there are three force acting on the ball -- gravity, F and Normal force (N). Work done by N is zero.

    Work done by F=work done by mg........(1)

    Work done by mg can be find easily.
    BALL.png

    The black curve shows the trajectory of the CoM. Angle ##\phi## can be found easily. Let the Com be at any arbitrary position such that it has rotated by angle ##\theta## about O.

    Work done by F is ##\int _{ 0 }^{ (90-\phi ) }{ FRcos(30-(\theta +\phi ))d\theta } ##.

    Just put the value in eq. (1). I think you will get the answers.:rolleyes:
     
    Last edited: Feb 1, 2015
  7. Feb 1, 2015 #6
    Work is either torque times angular displacement of force times displacement and in both ways i get an other equation than: ##\int _{ 0 }^{ (90-\phi ) }{ FRcos(\theta +\phi )d\theta }##
    According to the new drawing, for the initial position: α=90-(30+Φ) so W=FR cos[90-(30+Φ)]
    Of course we add θ for an arbitrary position and take an integral at the same boundaries
     

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    Last edited: Feb 1, 2015
  8. Feb 1, 2015 #7
    I made a mistake in #post 5.
    The work done by F is ##\int _{ 0 }^{ (90-\phi ) }{ FRcos((\theta +\phi)-60 )d\theta }##

    BALL.png

    If you consider the AO as the initial condition and BO as the final condition you will find that initially ##\theta=0## and final ##\theta=90-\phi##.

    What would you get after evaluating the integral?
     
  9. Feb 1, 2015 #8

    mfb

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    @Satvik Pandey: That does not work. What you calculate would be the average force needed, but the force depends on the angle, so you underestimate the force needed to start the lifting process.
     
  10. Feb 1, 2015 #9
    The geometry i have drawn is to scale, the force is above the tangent, at the initial position A and further, not like you drew, so the integral is:
    $$W=\int_0^{51.32} F\cdot 0.8 \cdot \cos (60-\phi)d\phi=-0.8F\cdot\sin(60-\phi)\vert^{51.32}_0=-0.8F(\sin 8.68-\sin 60)=0.57F$$
    The gain in potential enrgy equals W: ##0.57F=2000[N]\cdot 0.3[m]\rightarrow F=1053##
    But that is only theoretical since it's smaller than the force needed to lift it at the beginning
     
  11. Feb 1, 2015 #10

    haruspex

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    As mfb has pointed out, it is not only unnecessary to consider energy or perform integrals to answer this question, it will give the wrong answer.
    All you have to do is figure out where and at what angle to apply the force to get the cylinder moving. Beyond that, you could consider whether a larger force may be required at some later point in lifting it over the step (if the direction and point of application are not allowed to change?).
     
  12. Feb 2, 2015 #11
    Thank you mfb for pointing it. Sorry Karol, I lead you on the wrong path.:s
     
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