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Chain sliding off the edge of a table

  1. Oct 21, 2004 #1
    A uniform chain of length 8.00m initially lies stretched out on a horizontal table.

    If the coefficient of static friction between the chain and the table is 0.600, show that the chain will begin to slide off the table if at least 3.00m of it hangs over the edge of the table.

    Determine the speed of the chain as all of it leaves the table, given that the coefficient of the kinetic friction between the chain and the table is 0.400.

    I do not have a clue on how to draw up an FBD and start working on the problem. I kind of think it has to do with partial derivatives since the length of the chain, its mass hanging off the edge and its speed and acceleration varies with time.

    Can someone please, help me out to get started by giving a direction or a generic setup which I can start to work with.

    Thank you in advance!
  2. jcsd
  3. Oct 21, 2004 #2

    Doc Al

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    Staff: Mentor

    Treat the chain as having two segments: one piece on top of the table, the other hanging. What are the forces on each?
  4. Oct 21, 2004 #3
    The first one is just setting the friction force equal to the force of gravity. The second one is trickier, however. We know that [tex]F_f = \mu mg[/tex] and that [tex]F_g=mg[/tex] but the m's are changing over time. Let [tex]m=m_f+m_g[/tex] to denote the portion of the mass being affected by friction and gravity, respectively. The net force acting on the chain is then [tex]F_g-F_f = m_g g - \mu m_f g = m a[/tex]. I may have stopped my explanation too early so tell me if you need me to extend it a little further. Perhaps someone knows a less complicated way as well.
  5. Oct 21, 2004 #4
    Does either have any tension forces acting on it?

    If not the 5m segment has a negative F(friction) and it equals to the gravity acting on the 3m segment.

    Is this correct?


    0=F(friction)-F(gravity) (equlibrium, chain does not slide if ratio is 5/3)

    Also, how do I account for the lenght of the chain segments in the force equation since I do not have the density to calculate the mass? Do I just use ratios?

    Thank you
  6. Oct 21, 2004 #5
    Thank you for your reply. Yes, that would be great if you could expend on it a bit.

  7. Oct 21, 2004 #6

    Doc Al

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    They better--or else how can the weight of the hanging piece affect the piece on the table? In equilibrium, the tension at the top of the hanging piece will equal its weight. And that tension exerts a force on the tabled piece tending to pull it towards the edge.

    Essentially correct, but how would you get the answer if it wasn't given? (See below.)

    The way I'd do it is to call the hanging length "x". Its mass is then (M/L)x, where M is the mass of the chain and L the length of the chain. You'll find that M drops out.

    To solve the first part, write the mass of the tabled piece and hanging piece using the above trick. Then set up your equilibrium equation and solve for x.
  8. Oct 21, 2004 #7
    Here is how I solved the first part:


    Where x is the length of the chain

    So m1=5ro, m2=3ro


    Since the g*ro will drop off we are left with 3-(5*0.6)=0 and that is true!!

    Now onto the second part....

    Can you please start me out on how to get the V out of the above equations? Do I work backwards from F=ma, then solve for a, substitute dv/dt and do a partial derivative??

    Thank you
  9. Oct 21, 2004 #8

    Doc Al

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    Sounds like a lot of work to me! :smile: To find the speed of the chain, I would use energy methods. An interesting problem!
  10. Oct 21, 2004 #9
    Can you please get me started on the energy method? I understand that the chain as it slides down has both kinetic and potential energy. Do I start with the chain hanging as initial condition and just as it passed the edge of the table as the final condition and use Ui+Ki+W=Uf+Kf? How would I account for the Friction Force? Static or Kinetic?
  11. Oct 21, 2004 #10

    Doc Al

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    Yes, I would start at the point it begins to fall. As you know, it has an initial potential energy, and ends up with some kinetic energy. Along the way it transforms some of its mechanical energy to friction. The challenge is to figure out the work done against friction. (And you must use kinetic friction--otherwise no displacement and no work! :smile: ) Hint: the chain is stretched out for a reason.
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