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Find the force of friction provided by the floor

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  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Please see attached picture.

    2. Relevant equations
    Ff=uFn



    3. The attempt at a solution

    I'm really not sure where to start, since I don't know the coefficient of friction.
    Fn = ma
    Fn = 0.1kg(9.8m/s^2)
    Fn = 0.98N

    Does this mean that the force in the image is 0.98N as well?

    Any hints would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Mar 12, 2015 #2

    Bystander

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    First hint, "... on the verge of sliding ...." means the arrangement is static, nothing is moving. You'll be finding a coefficient of static friction.
     
  4. Mar 12, 2015 #3
    Thanks Bystander. So to find a coefficient of static friction, my formula would by Fsf = u(static)Fn. I know the normal force is 0.98N, but I don't know how to find either the coefficient or force of friction from here. What should I try next?
     
  5. Mar 12, 2015 #4

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    No. That would be the case if the stick were balanced vertically; there is a force being applied at the top end, at a right angle to the stick. It can be regarded as the resultant of a horizontal force to the right, and a vertical force that reduces the normal force against the floor. The force to the right is opposed by the friction force, and is also equal to the force the leaning stick exerts to the left.
     
  6. Mar 12, 2015 #5
    Okay. So how would I find the horizontal and vertical components of the force at the top of the stick?
     
  7. Mar 12, 2015 #6

    Bystander

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    Assume the mass of the meter stick is uniformly distributed along its length. It is static, and supported at two points against gravitational force.
     
  8. Mar 12, 2015 #7
    Okay. So the gravitational force, or the normal force is 0.98m. If the stick isn't moving, that means the force acting upwards on the stick is 0.98m as well?
     
  9. Mar 12, 2015 #8

    Bystander

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    Yes. Now what're the forces at the two support points?
     
  10. Mar 12, 2015 #9
    The net forces? or are you talking about the horizontal components?
     
  11. Mar 12, 2015 #10

    Bystander

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    Vertical components.
     
  12. Mar 12, 2015 #11
    Okay, I must be really confused. I thought 0.98N was the vertical components?
     
  13. Mar 12, 2015 #12

    Bystander

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    That's the sum of the vertical components, one acting at the left hand (upper) end, and the other acting at the right hand lower end, to support the stick against gravity.
     
  14. Mar 12, 2015 #13
    How do I get the separate vertical components then? Since the mass is uniform, do I just divide 0.98 by 2 to get 0.49N?
     
  15. Mar 12, 2015 #14

    Bystander

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    Two points of support, one at one end, the other at the other, assumed uniform distribution of mass, one carries half, and the other carries half.
     
  16. Mar 12, 2015 #15
    Okay, so then the vertical components are 0.49N. That means the force of friction by the floor is 0.49 N in the opposite direction?
     
  17. Mar 12, 2015 #16

    haruspex

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    That is not correct. The two will have a net torque about the mid point of the rod, balancing a net torque from the horizontal force components.
    MadMcB, there are in general three available equations for 2D statics: net horizontal force is zero; net vertical force is zero, net torque is zero.
    For the torque, you can in principle choose any point as the reference axis, but some are much more convenient than others.
    Since you don't really care about the normal force, the trick is to find two equations that do not involve it. It isn't involved in the horizontal force balance, so you can use that. Look for a torque axis that won't involve it.
     
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