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Rotational mechanics question

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data

    A cubical block of mass 'm' and edge length 'a' slides down a rough inclined plane of inclination alpha radian with a uniform speed. Find the torque of the normal force acting on the block about its center.

    On the back of the book answer is [itex]\frac{1}{2}mga sin(alpha)[/itex]
    2. The attempt at a solution

    I think the question is wrong. Am i correct????
    The torque due to normal should zero since line of action of normal force makes zero angle with center.
    this torque(as in answer) is of torque due to friction.

    Is all the things i write here are correct
     
  2. jcsd
  3. Sep 11, 2011 #2

    kuruman

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    The question is not wrong. Here is why.

    I am sure you have drawn a free body diagram. Look at the (components of) forces along the incline: There is friction uphill and gsinθ downhill. They are equal and opposite because the speed is uniform and the acceleration is zero. This means that they form a couple. Couples exert the same torque regardless of reference point. So if the normal force exerts no counter-torque, how come the block is sliding and not rolling downhill like a die?
     
  4. Sep 11, 2011 #3
    You are correct but the only force i can think of responsible for non rolling of this box is normal reaction. it is always perpendicular to plane so it must be in left with respect to center of mass of block(See attachemtn i draw a rough figure). am i correct?
     

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  5. Sep 11, 2011 #4

    kuruman

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    Correct. The normal force shifts off center just enough to make the net torque zero. If it didn't, the block would tumble which is not what's happening. So now you want to balance the torque generated by the couple with the counter-torque generated by the normal force.
     
  6. Sep 11, 2011 #5
    Oh wow what a question. This is first time when i found that normal is not passing through center.
    Is the free body diagram i draw(attachment in my last post) is correct?
    In fbd ,backward force is friction ,normal to plane is normal reaction and forward one is due to gravity.
     
  7. Sep 11, 2011 #6

    kuruman

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    The FBD needs to be drawn a bit more carefully. The block needs to be a cube of side a, gsinθ must act at the center of mass and static friction must be drawn on the surface of the incline directed uphill. Call x the offset of the normal force from the center and find what x should be so that the torques are balanced.
     
  8. Sep 11, 2011 #7
    what static friction. I think it should be kinetic friction.
    I have done question thanks yaar. I take lowest point on the cube as axis of rotation and did it.
     
  9. Sep 12, 2011 #8

    kuruman

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    Yes, kinetic friction of course. "Static" was a typo.
     
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