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Forces and Torque

  1. Nov 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A 4.5-kg bowling ball is perched on a concrete ledge directly below your dorm room window, with the side of the ball opposite the holes touching the wall. Wanting to hold the ball in place so that it doesn't roll off and land on somebody, you manage to hook one of the holes with a wire and exert a purely tangential (and vertical) force on the ball. The coefficient of static friction between ball and ledge is the same as that between ball and wall, μs = 0.43. What is the maximum upward force you can exert so that the ball does not rotate and you lose your hold? Even though the ball has holes drilled in it, assume a uniform distribution of inertia.

    2. Relevant equations
    τ = Fr
    F = ma
    ƒ = μN

    3. The attempt at a solution
    I drew a diagram and wrote equations for net torque, net force in x direction, and net force in y direction.

    τnet=T-ƒwallledge=0
    Fxledge-Nwall=0
    Fy=T-mg+Nledgewall=0

    To find the maximum force, I set the frictional force from the wall = μ times the normal force from the ledge. Also, I set the frictional force from the ledge = μ times the normal force from the wall.

    Substituting values:
    T-.43Nledge-.43Nwall=0
    .43Nwall-Nwall=0
    T-44.1+Nledge+.43Nledge=0

    Solving for Nwall gave a value of 0, which doesn't seem correct. Continuing through regardless, I finished with an answer of 10.2 N for the tension, which wasn't correct.
     
  2. jcsd
  3. Nov 10, 2015 #2

    haruspex

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    Did you really do that? A little unconventional perhaps?
     
  4. Nov 10, 2015 #3
    I was told to do that from a tutor I was seeking help from. Clearly there was some misthought or miscommunication.
     
  5. Nov 10, 2015 #4

    haruspex

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    What's the usual relationship between maximum frictional force and normal force?
     
  6. Nov 10, 2015 #5
    ƒs = μs x FN
     
  7. Nov 10, 2015 #6

    haruspex

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    Yes, but what are those two forces exactly? Where do they act and in what directions in relation to each other?
     
  8. Nov 11, 2015 #7
    Since friction is along a surface, and normal force is perpendicular to the surface, the forces should be perpendicular to one another. So in this case, the max friction force (along wall) would equal the μs times the normal force from the wall? And same thing for the ledge?
     
  9. Nov 11, 2015 #8

    haruspex

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    Yes.
     
  10. Nov 11, 2015 #9
    I got 15 N for the tension, and this answer is correct. Thanks for all your help!
     
  11. Feb 24, 2017 #10
    Care to explain further? I'm stumped on this one too.
     
  12. Feb 24, 2017 #11

    haruspex

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    As with the initial post of a thread, if you have the same homework you need to post your attempt.
    Have you drawn a diagram? What forces act on the ball?
     
  13. Feb 24, 2017 #12
    Fx=Normalforce (wall)-Friction (ledge)=0
    Fy=Normal(ledge)+Tension-mg-friction(wall)=0

    Solved for Y direction for Normal Force Fn= (mg-T)/0.57
    Then plugged into T=mg-fs-Fn
     
  14. Feb 24, 2017 #13

    haruspex

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    You have two equations but three unknowns. What do you need to do?
     
  15. Feb 24, 2017 #14
    Do I need a third equation? f=μN?
     
  16. Feb 24, 2017 #15

    haruspex

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    You need a third equation, but I was assuming you were already using f=μN. (If not, you had five unknowns.)
    In 2D statics problems, there are always three equations available. Exactly which three is a matter of choice, but the usual choice is horizontal linear force balance, vertical linear force balance, and ....?
     
  17. Feb 24, 2017 #16
    Rotational force? Torque perhaps? Torque=rF
     
  18. Feb 24, 2017 #17

    haruspex

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    Yes.
     
  19. Feb 24, 2017 #18
    I'm not sure how to use torque in this situation, is the Net torque from the tension and the 2 frictions?
     
  20. Feb 24, 2017 #19

    haruspex

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    All the forces contribute to torque in principle, but you can eliminate some by choosing an axis through which the forces pass. E.g. if you choose the centre of the ball then you can ignore the normal forces because they have no torque about that point.
     
  21. Feb 24, 2017 #20
    Στ=τT-2τf
    Tr=2(μFn)r
    T=2μ[(mg-T)/(1-μ)]
    This is what I got using Net Torque and ΣFy
     
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