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Cylinder and Torque

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data

    Figure P8.74 shows a vertical force applied tangentially to a uniform cylinder of weight w. The coefficient of static friction between the cylinder and all surfaces is 0.500. Find, in terms of w, the maximum force F that can be applied without causing the cylinder to rotate. [Hint: When the cylinder is on the verge of slipping, both friction forces are at their maximum values. Why?]

    2. Relevant equations

    Torque = F * d

    F = Mg

    Friction = coefficient * normal force

    3. The attempt at a solution

    There are a few force: weight force, F, <right wall>: normal force to the left, friction force, <lower wall>: friction force to the right, normal force upward. However, I don't know how to combine these forces to solve the problem.
  2. jcsd
  3. Feb 16, 2015 #2
    Have you started properly by drawing a free body diagram showing all the forces acting on the cylinder, or do you feel that you have progressed beyond the point where you need to use free body diagrams?

  4. Feb 16, 2015 #3


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    You must have heard about Newton's laws. There is one that says ##\ \vec F = m\; \vec a\ ## (*) .
    If the thing don't move, the forces must add up to zero. That's your combination !

    (*) Wrote it as vectors, so you may use it for horizontal non-movement and for vertical idem as well.

    And if you're still short on equations, something similar for non-rotation also holds. You know what I mean ? :wink:

    [edit] bedtime for me; leave you to Chet !
  5. Feb 16, 2015 #4
    @chet, Yes, I have drawn free body diagrams, but I cannot upload here. The forces on the FBD were written in the "attempt at a solution" section.

    From the FBD, I have n1 = coefficient * n2, coefficient * n1 + F + n2 = w. Is that correct?
  6. Feb 16, 2015 #5
    Yes. Sorry. I didn't read your post carefully enough.

    You have 2 equations here and 3 unknowns. The third equation would be a moment balance.

  7. Feb 16, 2015 #6
    Can you explain more about how to set up the moment balance?

  8. Feb 16, 2015 #7
    Well, first of all, do you know what a moment (torque) is? Do you know how to calculate the moment of a force about an axis of rotation?

  9. Feb 16, 2015 #8
    Yes, torque = F * d, to calculate moment, is mr^2.
  10. Feb 16, 2015 #9
    Not exactly. mr^2 is what you use to calculate the so called "moment of inertia." That's different from the moment of a force around an axis. A moment is the same thing as a torque. They're different words for the same entity. Your equation torque = F*d needs a little more explanation. How do you determine what value of d to use?

  11. Feb 16, 2015 #10
    It's radius, but no radius is given
  12. Feb 16, 2015 #11
    There's a rule for determining the correct distance d to use. You drop a normal from the axis of rotation to the line of action of the force. d is the distance from the axis to the line of action of the force. Does this ring a bell? If not, go back and review your textbook or notes.

  13. Feb 16, 2015 #12
    It's ## Fr \sin \theta ##, but I'm still not sure how to apply it. Can you give a start?
  14. Feb 16, 2015 #13
    OK. Make a list of the moments of each of your forces around the axis of the cylinder. Let's see what you get.

  15. Feb 16, 2015 #14

    Torque caused by f1: ## f_1 \cdot R##
    Torque caused by n1, W, n2: 0
    Torque caused by f2: ##f_2 \cdot R##
    Torque caused by F: ##F \cdot -R##

    Therefore, ##F = f_1 + f_2##?

    Now we have that equation and ## n_1 + f_2 +F= W## and ##f_1 = n_2##?
  16. Feb 16, 2015 #15


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    That all looks right.
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