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Dynamics problem: determine maximum angle for equilibrium

  1. Sep 11, 2016 #1
    1. The problem statement, all variables and given/known data
    A uniform bar with lenght c can slide a cylindrical surface with a radius of r. Determine the maximum angle theta that guarantees the equilibrium of the bar if the friction coefficient at the points of contact is u.
    exercise.png

    2. Relevant equations
    Friction force = u*N (where N is the normal force, and u is the friction coefficient).


    3. The attempt at a solution
    Okay, I'm really lost with this exercise. I didn't understand the statement. The bar is at the top of a cylindrical surface, and can slide within it, or is the cylinder hollow and the bar is supported on both ends by the friction force (touching the inside walls of the cylinder)? For me, the second option makes more sense, but how would the bar slide then?

    I would appreciate any help.
     
  2. jcsd
  3. Sep 11, 2016 #2

    Nidum

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    Small print says inside in the original problem statement . So it is a bar inside a tube .
     
  4. Sep 11, 2016 #3
    So it's the bar supported on both ends by the tube walls? I didn't understand what the question means by equilibrium... Does the bar rotates around some point? I'm really not sure about how does the angle changes.
     
  5. Sep 11, 2016 #4

    Nidum

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    Equilibrium just means not moving in this case . This is essentially a static analysis problem .

    The bar does not start from some unknown initial angle and slide round until it comes to a halt .

    Think of it more like a lab experiment where you place the bar in the tube by hand at a range of test angles and note for which angles the bar stays put and for which angles it slides(1) .

    This problem is really only a variation of the 'ladder against the wall ' problem .

    Note (1) :There is a range of angles where the bar could tip over or fall down rather than just slip .
     
    Last edited: Sep 12, 2016
  6. Sep 11, 2016 #5
    Okay, now I understand the problem. So I will have the gravitational force acting on the center of mass, friction forces on both ends of the bar plus two normal forces to balance the gravitational force.

    Can I make the assumption that those normal forces have only perpendicular components?
     
  7. Sep 11, 2016 #6

    Nidum

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    I have to break off for tonight - it is getting very late here in UK . I expect someone else will jump in to help you but if not I'll be back tomorrow am .
     
    Last edited: Sep 11, 2016
  8. Sep 11, 2016 #7
    The rod would tip over if the angle was 90º (completely vertical), and fall down if it was 0º (completely horizontal).

    Hm... So the normal force would only have purely perpendicular components in the first case, right? In all the other cases, it would have components in other direction that I should take into account.
     
  9. Sep 12, 2016 #8

    Nidum

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    Let's start by having a look at the basic problem as given in the question .

    We need the free body diagram(s) for the bar .

    You may find it useful to draw separate diagrams for the different cases where the higher bar end is above or below the tube horizontal centreline and where the lower bar end is left or right of the tube vertical centreline .
     
    Last edited: Sep 12, 2016
  10. Sep 12, 2016 #9
    Ok. The gravity force acting at the center of mass, pointing down. The normal forces, pointing towards the center of the circle, and the friction forces opposing the sliding motion of the rod?
     
  11. Sep 12, 2016 #10

    Nidum

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  12. Sep 12, 2016 #11
    Ok, I'm working again on the problem, and I'm having trouble to figure out relationships between the angles.
    Here's a rough sketch (showing only the normal forces):
    20160912_110539.jpg
    Where Fn and Fm are the normal forces.

    I'm trying to express phi and alpha in therms of theta, so I can decompose the x and y components and write the equilbrium equations.
     
    Last edited: Sep 12, 2016
  13. Sep 13, 2016 #12

    Nidum

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    Have you made any progress ?
     
  14. Sep 13, 2016 #13
    Not much, I fear.I think I established a relationship between the angles (but I'm not sure if I made it correctly). I found out that sin(phi) = cos(theta).

    That's what I did:
    20160913_214126.jpg

    Then I came up with those two equilibrium equations in x and y, but got stuck there (note: my professor said that we dont have to account for the gravity force, but he hasn't been very helpful beyond that. Anyway, I left it out there). Solving the equations led me nowhere.

    EDIT: Also, I've considered that the friction forces are perpendicular to the radius of the circle. Is that right?
     
  15. Sep 14, 2016 #14
    Now I see that I've made incorrect assumptions in my prior attempt, but I'm still stuck with those angles.
     
  16. Sep 14, 2016 #15
    Okay, I've struggled a little bit with those angles, but if I'm not mistake again, then phi = theta.

    Also, I've added the gravity force again (which was absent in my last picture).

    Now I have three equilibrium equations:

    x: Fg*sin(theta) + Fn*cos(theta) - Fm*cos(theta) - uFn*sin(theta) - uFm*sin(theta) = 0

    y: -Fgcos(theta) + Fn*sin(theta) + Fm*sin(theta) + u*Fn*cos(theta) - u*Fm*cos(theta) = 0

    Moment: -Fg*cos(theta)*(c/2) + Fm*sin(theta)*c - u*Fm*cos(theta)*c = 0

    Where u is the friction constant, and c is the length of the rod. Fn and Fm are normal forces, and Fg is the gravity force.
    Solving those equations should give me expressions for Fg, Fm and Fn, right? After that, I should isolate theta in one of those equations, and that would give me the answer?
     
  17. Sep 16, 2016 #16

    Nidum

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    The method you are using is not exactly incorrect but it is more complicated than needs be and does not really reflect the essential nature of the problem .

    Think about the problem this way :

    The bar when set at any particular angle can only be in one of two conditions - static or in motion .

    There are driving forces trying to cause motion of the bar and retarding forces trying to prevent motion .

    Depending on the relative magnitude of these forces the bar will move or it will not .

    So can you see a way of analysing this problem in terms of driving forces and retarding forces .?
     
  18. Sep 16, 2016 #17

    Nidum

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    Removed pro tem .
     
    Last edited: Sep 16, 2016
  19. Sep 16, 2016 #18
    Let me guess... My driving forces will be the gravity force and the normal forces, while the retarding forces will be the friction forces.

    The conditions for equilbrium are R(resultant) = 0 and M(torque with respect to some point) = 0.

    Anyway, I will have to decompose the forces using the angles the way I was doing before, right?
     
  20. Sep 16, 2016 #19
    Ok, think I've got something...

    If I calculate the distance from the center of mass of the rod (where the gravity force is being applied) to the center of the tube, I will get:
    $$ \sqrt{r^2 - \frac{c^2}{4}}$$

    The cross product of this position vector with the gravity force will be the torque of this point around the center of the tube.

    Now I have to find a way to express the gravity force in terms of u (the friction constant), and r...
     
    Last edited: Sep 16, 2016
  21. Sep 21, 2016 #20

    Nidum

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    Have you got your final answer now or do you want to discuss this problem further ?
     
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