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Pulley physics help

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data
    ok so box 1 (5kg) is ontop of box 2 (10kg) and this guy has to pull the box with an unknown force, but the boxes are on a pulley conected to a wall (not a celling).
    The coefficient of static friction between the box and the floor is .9. The coefficient of kinetic friction between the box and the floor is .7. The coefficient of static friction between the boxes is .7. The coefficient of kinetic friction between the boxes is .5. (2 points)

    2. Relevant equations

    Last week in class my teacher gave us all some equations.
    Ffs=Ms*Fn
    Ffk=Mk*Fn
    Where Ffs and Ffk is friction do to static and kanetic, Fn is friction normal.
    but i dont rember what M is, is it somthing as simple as mass?
     
    Last edited: Nov 18, 2008
  2. jcsd
  3. Nov 18, 2008 #2
    Re: pulleys

    so im kinda going off of what i think but am sure its not the correct way to solve it. Friction can be a deseleration and to find force you do F=MA. so for each box moving in a given direction with friction their is a force acting in the opposet direction, so i put 2.5N for box1 and 7N for box2.
     
  4. Nov 18, 2008 #3
    Re: pulleys

    o also our teacher said that the rope is massless and has no friction between the pullie thingy.
     
  5. Nov 18, 2008 #4
    Re: pulleys

    If F is the force of friction and N is the normal force, then [tex] F = \mu N [/tex], where the constant [tex] \mu [/tex] is the friction coefficient.

    Learn to love to read your textbook.
     
  6. Nov 18, 2008 #5
    Re: pulleys

    yea but i dont have the normal force. the force of both of the boxes down ward is 147N but thats down wards. how do i find the normal force, and in what dirrection
     
  7. Nov 18, 2008 #6
    Re: pulleys

    is the norml force equal to the mass times gravity?
     
  8. Nov 19, 2008 #7
    Re: pulleys

    If a box on a table is not accelerating vertically, the normal force is the force of the table on the box that cancels the net downward force on the box (excluding N). Otherwise, the box would accelerate vertically, wouldn't it?
     
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