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Friction problem

  1. Nov 15, 2005 #1
    Block A is on top of Block B. They are connected by a light, flexible cord passing through a fixed, frictionless pulley. Block A weighs 3.60 N and block B weighs 5.40 N. The coefficient of kinetic friction between all surfaces is 0.25. Find the magnitude of the force F necessary to drag Block B to the left at a constant speed.
    Ok so the key here is constant speed. So that means [tex] F = 0 [/tex]. So all forces must be equal and opposite one another in the same direction. Since [tex] w_{total} = 9.0 N [/tex] does that mean that the force required is [tex] f_{k} = \mu_{k}N = 0.25(9.0 N) = 2.5 N [/tex]?
    Last edited: Nov 15, 2005
  2. jcsd
  3. Nov 15, 2005 #2


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    You have part of the answer.

    Block A is on top of block B, and they are connected by a pulley. If B moves left, A moves right, and the friction force of A's motion resists B's motion.

    Block B is on a surface, also with [itex]\mu[/itex] = 0.25, but the downward force arises from the masses of A and B.

    Now at constant velocity, there is not acceleration, so the net force F = 0, so F = Friction force of block A on B + Friction force of block B on the surface underneath.
  4. Nov 15, 2005 #3
    so it would be [tex] 0.25(3.6 N) + 0.25(5.40 N) = 2.25 N [/tex]. I forgot the fact that when you pull on block A, block B goes in the opposite direction.

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