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Magnitude of the Force of Friction

  1. Jan 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A wooden crate with a mass of 20.0 kg is dragged across a floor at a constant speed by a force of 85.0N. The magnitude of the force of friction here is ____ N.




    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 11, 2010 #2

    rl.bhat

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    According to Newton's first law, when a body moves with constant velocity?
     
  4. Jan 11, 2010 #3
    ? no acceleration?
     
  5. Jan 11, 2010 #4

    rl.bhat

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    Can you state the Newton's first law?
     
  6. Jan 11, 2010 #5
    Correct, now expand on this using Newtons 2nd law.
     
  7. Jan 11, 2010 #6
    I'm confused, constant velocity should mean that there is 0 acceleration?
     
  8. Jan 11, 2010 #7
    newtons first law is pretty much intertia, an object in motion tends to stay in motion, its 11:00pm my time and ive been struggling with it for a while, kinda frustrated so i no its rude but if u dont mind i need a lil fast help
     
  9. Jan 11, 2010 #8
    In the absence of a net force, a body either is at rest or moves in a straight line with constant speed.

    Apply this knowledge with Newton's 2nd law to draw your conclusion.

    [tex]\sum F = ma [/tex]
     
  10. Jan 11, 2010 #9
    k heres what i got

    F=ma
    85N=20kg(a)
    a=4.25 m/s2

    I dont no how to get force of friction
     
  11. Jan 11, 2010 #10
    You told me before that with constant velocity, acceleration was 0. You should be able to conclude that,

    [tex]\sum F = 0[/tex]

    Now simply choose a coordinate system for your x axis (which way is positive which way is negative) and assign each horizontal force the proper sign and solve for the force of friction.

    You need to review your understanding of Newtons laws.
     
  12. Jan 11, 2010 #11
    And mind you that friction force acts the opposite direction from the force acting on the object. i.e., if you move a block to the right, friction force act to the left. So, friction force would have minus sign.
     
  13. Jan 11, 2010 #12
    well could the answer be a negative number then? I need a 3 digit answer

    the second part of the question is

    The coefficient of friction here is a.bc x 10 ^-d
    The values of a, b, c, and d respectivly are ____
     
  14. Jan 11, 2010 #13

    Sorry, I might have confused you. What I want to say is that find all force act on the object, use net force equation(f=ma=0, which I told you in another post). net force will be
    total force-friction force because friction acts the opposite. That is why I mentioned minus sign. I hope I am going right. Could someone check this for me?
    Net force=total force-friction force = 0
     
  15. Jan 11, 2010 #14
    wow so confused so my final answer for the first part will be zero? and for the seconds part it will be zero??
     
  16. Jan 11, 2010 #15
    No. Now take a closer and careful look at the eqation I gave it to you.
    total force-force of friction=0.
    This does not imply force of friction is zero.
    Equate this eqation we get
    force of friction=total force.
    now find WHat force acts on the object as I told you before?
    we find gravity and pulling force.
    so, mg+pulling force=total force
    (weight of the object)(9.8)+85n=total force.

    And I am so sorry. Here I am staying right now is 12:30 in the evening. And visit this website. It has specific information that will help you to find answer. There it tells you how to find the total force too.
    http://www.physics247.com/physics-homework-help/friction-and-weight.php
     
    Last edited: Jan 11, 2010
  17. Jan 11, 2010 #16
    so 281.2 is the total force but where do i go from there?
     
  18. Jan 11, 2010 #17
    This is incorrect. We're applying Newton's 2nd law in a 1 dimensional motion.
     
  19. Jan 11, 2010 #18
    lets clarify i only have physics 20-1, and i really need help with this questions, can someone please show me the formulas to use and how to fill them in? it would be greatly appreciated
     
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