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Mechanics question (moments)

  1. Dec 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A uniform sphere rests on a horizontal plane. The sphere has centre O, radius 0.6m and weight 36N. A uniform rod AB, of weight 14N and length 1 m, rests with A in contact with the plane and B in contact with the sphere at the end of a horizontal diameter. The point of contact of the sphere with the plane is C, and A, B, C and O lie in the same vertical plane (see diagram). The contacts at A, B and C are rough and the system is in equilibrium. By taking moments about C for the system,
    (i) show that the magnitude of the normal contact force at A is 10N.

    (ii) Show that the magnitudes of the frictional forces at A, B and C are equal.

    http://papers.xtremepapers.com/CIE/...athematics - Further (9231)/9231_s09_qp_2.pdf
    The diagram can be found in this link, question no.4.

    2. Relevant equations



    3. The attempt at a solution
    I know that if the system is in equilibrium, then the resultant force in any direction and the sum of moments about any point is zero. What I don't get about sub-question (i) is: there is a reaction force acting at point B, and friction as well acting at the same point (due to the surface being rough), but why are we not required to take moments about point C for those forces? Why do we only take moments about point C for the weight of the rod and the reaction force at A? As for question (ii), the worked solution I have is that by taking moments about point O, we can prove that the frictional force at B is equals to that at C, but why isn't the frictional force at A taken into account when taking moments about C - the line of force for that friction does not go through C though!
    Here is the link to the worked solutions for both sub-questions:
    http://papers.xtremepapers.com/CIE/...athematics - Further (9231)/9231_s09_ms_2.pdf
    (Question 4)

    Thanks! I'd really appreciate if you could explain the worked solution for me! :)
     
  2. jcsd
  3. Dec 14, 2012 #2

    mfb

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    Everything apart from the weight of the rod and the normal component of the force of A has a matching force in the opposite direction at the same point (B) of the force points towards C (so its corresponding torque is 0 independent of the force - happens at A and C).

    It does, it is in the plane.
     
  4. Dec 14, 2012 #3
    Hi there, thanks for replying.
    So what you're trying to say is that at B, there is frictional force acting both downwards and upwards (upwards on the rod to prevent it from sliding down and downwards on the sphere presumably?). And there is a reaction force acting on the rod because the rod is leaning against the sphere, so everything cancels out in a way? By that logic, can't we say that at A, there's a reaction force on the plane by the rod as well? Or is that represented by the weight already? Sorry, I'm just trying to verify my understanding - I've been stuck on this question for quite some time now. :p

    As for question (ii), sorry I didn't mean moments about C, I meant moments about O instead. The frictional force at D doesn't go through O though right? But why is not taken into account as suggested by the worked solution?

    Thanks for your patience!
     
  5. Dec 14, 2012 #4

    mfb

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    Right.

    The floor is not included in your system., so you shouldn't consider forces acting on the floor if you want to calculate torque in your system.

    Where is D? Do you mean A?
    The moments for O are considered to exclude a rotating sphere only.
     
  6. Dec 14, 2012 #5
    Hi I'm so sorry, yes I do mean at A. What do you mean by that last statement? Doesn't the frictional force act perpendicularly to the line connecting O to C?
     
  7. Dec 14, 2012 #6

    mfb

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    You want that the sphere does not rotate around its center. Therefore, the sum of torques acting on the sphere has to be 0. There are two: Friction of the rod and friction on the ground.
     
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