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Coefficient of sliding friction homework

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data
    A large package weighing 517 N is accelerated across a gym floor at 3 m/sec^2 because it is being pushed at a force of 285 N at 50 degrees to the floor. What is the normal force of the package? What is the coefficient of sliding friction between the package and the floor? Approximately what percentage of the 285 N force is being wasted due to the excessive angle produced by the pusher?

    2. Relevant equations
    F=MA
    HORIZONTALLY
    FORCE APPLIED, X= (285)(cos 50)= 183.194

    VERTICALLY
    FORCE OF Y=0 (not rising off the ground)
    FORCE APPLIED, Y= (285)(sin 50)= 218.323

    3. The attempt at a solution
    I have filled in as much as I know, please help!
    Thanks

    1. The problem statement, all variables and given/known data
    To slide a 400 N box up a 25 degree ramp at a constant speed, a force of 250 N is required. What is the net force of the box? What is the frictional force on the box? What is the coefficient of friction between the box and the plane?

    2. Relevant equations
    Horizontally
    Fg,X=(9.81)(cos 25)= 8.89 N
    FX= Fg,X-Force of Friction

    Vertically
    Fg, Y= (9.81)(sin 25)= 4.145 N

    3. The attempt at a solution
    I am still unsure how to put these products into a formula.

    Thanks for your help!
     
  2. jcsd
  3. Oct 7, 2007 #2

    Kurdt

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    So you've resolved the force horizontally and vertically. Lets address the question of kinetic friction. The friction when two surfaces slide over another is constan and is the product of the coefficient of friction and the normal force. You will have to work out the force due to friction which is where the acceleration of the package comes in, and then the normal force. This will allow you to work out the coefficient of friction.

    For the percentage waste compare the force actually going into accelerating the box here with the force that would be going into accelerating the box had the force been applied completely horizontally.

    The weight of the box is 400N which is given by W = mg. Work out the component of weight parallel to and perpendicular to the ramp. The component of weight parallel must be balanced with the friction by the applied force. The friction is found similarly from the normal as above.
     
  4. Oct 8, 2007 #3
    So,
    NET FORCE
    =250 N
    VERTICAL
    Y Component= (250)(sin 25)=105.654

    HORIZONTAL
    X Component= (250)(cos 25)=226.576

    For problem 2, how do I finish finding friction?
     
  5. Oct 8, 2007 #4

    Kurdt

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    You want to resolve the components of the weight of the box which is 400 N not the net force. The net force will be the sum of the frictional force and the component of weight parallel to the ramp.

    To find the friction force you will need to find the weight of the box parallel to the ramp. Once you have that you know that friction plus weight is equal to the 250 N force acting on the box. You will also know the normal force from resolving the weight and you can then find the coefficient of friction.
     
  6. Oct 8, 2007 #5
    Net Force = Frictional Force + Weight Parallel
    250 = Frictional Force + 226.576
    23.424 = Frictional Force

    Is this correct?
    Once Again, THANKS FOR YOUR HELP AND QUICK RESPONE!
     
  7. Oct 8, 2007 #6

    Kurdt

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    The weight is 400N not the 250 newtons net force. So you need 400 sin (25) to find the weight parallel to the ramp. The method however is correct.
     
  8. Oct 8, 2007 #7
    THANK YOU SO MUCH!
    Is the first problem solved in the same way?
     
  9. Oct 8, 2007 #8

    Kurdt

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    The first question is similar. You have resolved the forces correctly but be careful because the normal force will be the weight plus the extra component of the force acting veritically downward on the package. To work out the friction wou need to find the frictional force by subtracting the component of force horizontally from the force needed to accelerate the package at 3m/s2. Then to find the final piece of information proceed as I suggested in my first post.
     
  10. Oct 8, 2007 #9
    So,
    Normal Force = 517 - (517x9.81)

    Normal Force = 4554.77

    Force of Friction = 183.194 - 285

    Force of Friction = 101.806 N

    Is this correct?

    THANK YOU SO MUCH!
     
  11. Oct 8, 2007 #10

    Kurdt

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    Normal force is the weight (517 N) + the vertical component of the applied force which you worked out as 218.323N.

    Now: Frictional force = force applied - force required for 3m/s{sup]2[/sup] acceleration.

    Can you take it from there?
     
  12. Oct 8, 2007 #11
    I think I got it. THANK YOU for your help. I appreciate it!
     
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