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Packing Case Question: Friction

  1. Nov 15, 2005 #1
    A 30.0 kg packing case is initially at rest on the floor of a truck. The coefficient of static friction between the case and the floor is 0.30, and the coefficient of kinetic friction is 0.20. The truck is traveling due east at a constant speed. Find the magnitude and direction of the friction force acting on the case (a) when the truck accelerates at 2.20 m/s^2 eastward; (b) when it accelerates at 3.50 m/s^2 westward.

    Ok so initially, we have the force of gravity [tex] w = (30 kg)(9.8 \frac{m}{s^{2}}) [/tex], the static friction force, and the normal force (equal to the weight but opposite in direction). So for parts (a) and (b) since everything is moving, we have to find the magnitude and direction of the kinetic frictional force. So [tex] f_{k} = \mu_{k}\times N [/tex]. This is where I become stuck. How do you relate the acceleration to finding the frictional force?

    Any help is appreciated.

    Thanks
     
  2. jcsd
  3. Nov 15, 2005 #2

    Fermat

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    You only use kinetic friction when the case is moving over the surface it had been resting on. The truck (and case) may be moving, but that doesn't mean that the case is sliding over the floor. That only happens when the acceleration is so great that the accelerating force is greater then the force of static friction. In that case it becomes the force of kinetic friction that takes over.
     
  4. Nov 16, 2005 #3
    So would I use F = ma to find the magnitude of the accelerating force, and then find the friction?

    Thanks
     
  5. Nov 16, 2005 #4

    Fermat

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    Use Fr=mu*Fn to find the (maximum) friction force.

    Then use F = ma on the box and show that F is less tham (max) Fr, hence the box does not slide hence static friction rules, hence state what the friction force is.

    Part b) is little bit different (but not much) - can you do that ?
     
  6. Nov 16, 2005 #5
    Thanks a lot

    For part (B) you would use F = ma or F = (30 kg)(-3.50 m/s^2) = 105 N due west. The friction force must then act eastward. Since [itex] f_{s,max} = 90 N [/itex] the kinetic friction rules in this case. So [itex] f_{k} = 0.20(300 N) = 60 N [/itex].

    Is this correct?

    Thanks
     
  7. Nov 16, 2005 #6

    Fermat

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    All correct. Kinetic friction, 60N, acting east.
     
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