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Coefficient Kenetic Friction and Kenetic Force

  1. Oct 14, 2006 #1
    Hey, can you guys help me out with this...

    A 75 kg box slides down a 25 degree ramp with an acceleration of 3.6 m/s^2 ...what is the coefficient of kenetic friction?

    I know how to do the basic equation problems from what our teacher told us, but I don't know how to manipulate the equations here to find Fk (kenetic force).

    Also just double checking, does Fn (normal force) = (75)x(3.6)?

    Can someone please confirm and point me in the right direction?
  2. jcsd
  3. Oct 14, 2006 #2
    Kinetic friction is given by=> Coefficient of kinetic friction X Normal force. To find the coefficient. You need to first know wat's the friction on the box sliding down the ramp and the normal force.

    Nope. Draw out the free body diagram of the box first. Draw out the forces acting on the block.

    Mainly there is the weight of the box acting perpendicular downwards and the friction which is in oposite direction to the direction of the motion.

    Next draw out the vertical and horizontal components of the weight. The vertical component will be equals to ur normal force, according to ur newton's 3rd law. Hence, normal force is given by=>Weight of the box X cos 25.

    To find frctional force, take the horizontal component of the force minus the force causing the acceleration of the box down the ramp. Note, the horizontal component of the force will be given by (Weight of box X Sin 25) and the acceleration of the box will be given by (F=ma). After finding frictional force, you can hence use the above eqn to find kinetic friction.
  4. Oct 15, 2006 #3
    I used free body diagram and based on numbers that we have, I didn't get the right answer for the coefficient of kenetic friction (mu k), which should become 0.061. I can't figure out how to get to here.. it's so frusterating.. please help clearify the steps.
  5. Oct 15, 2006 #4


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    Staff: Mentor

    The weight of an object is due to gravity pulling on its mass m, producing an acceleration g, so weight W = mg.

    The box on the incline is being pulled downward (vertically) by gravity, but it is contrained by the ramp, so is must slide down the ramp.

    The weight vector is at an angle with respect to the ramp, so one must resolve the weight into the normal force applied on the ramp and the component which is parallel to the ramp. Opposing the gravitational force pulling the box down the ramp is friction, which is proportional to the normal force by the kinetic friction factor.

    Please refer to these -


  6. Oct 15, 2006 #5
    Alright, to make it simple, the steps you need to go through is this..
    1) Find kinetic friction.(mgsin(theta)-ma)
    2) Find normal force. given by( mg cos(theta))
    3) By using the formula for kinetic friction( kinetic friction=coefficient of friction X normal force), You should be able to find the coefficient.

    Yes. The answer you should get is 0.061.
  7. Oct 16, 2006 #6
    Ok, I am confused about how to derive the normal force (mgcos(25)) and frictional force
  8. Oct 16, 2006 #7
    Refer to post #2, draw out the free body diagram and break up the components of the weight. The horizontal component of the weight is the kinetic friction and the vertical component of the weight is the normal force.
  9. Oct 16, 2006 #8
    Why do you subtract ma when calculating frictional force?
  10. Oct 16, 2006 #9
    The horizontal component of the weight actually gives the value of the box when there is no frictional force. Note that ma= horizontal component of the weight-frictional force. Therefore, frictional force=horizontal component of weight-ma.

    Note that ma is ur resultant force on ur box. The force minus friction.
  11. Oct 16, 2006 #10
    hi guys.. could you please answer this?A 20 kilogram sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80 Newton and is directed at an angle of 30degrees above the horizontal. Determine the coeffecient of kinetic friction....
    please i badly need your help as soon as possible.. thanks
  12. Oct 16, 2006 #11
    Where's ur own working.? A little clue to get u started.. draw a free body diagram of the sled.
    Last edited: Oct 16, 2006
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