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Help with Friction and newtons laws

  1. Nov 21, 2004 #1
    I have this physics problem that has simply been driving me crazy...heres the problem:
    Jenn coasts down the hill on a sled, reaching the bottom with a speed of 7.0m/s. The coefficient of kinetic friction between the sled's runners and the snow is .05, Jenn has a mass of 62 kg, and the sled has a mass of 3.82 kg. How far do Jenn and the sled travel on the flat surface at the bottom of the hill before coming to rest?
    first i set up my diagram
    and i set the positives to be down (with gravity) and the direction the sled is going.
    then i set up my free body diagram which has mg going down, normal force up, the force of the motion going in the direction of the thread, and kinetic force opposing the motion.
    Then i added up the masses and got a total of 65.82kg. Then i continued to find my net forces which i got
    Fy = mg - Fn (normal) = ma = 0
    Fx = Fm (motion w/ the sled) - Fk.
    Since i know mg = 645.04 i plugged that in.
    Then i know that Fk = mue * mg so i got that Fk = .05.
    From here i don't know where to go. Help would be greatly appreciated! Thanks a ton!
  2. jcsd
  3. Nov 22, 2004 #2


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    "Then i know that Fk = mue * mg so i got that Fk = .05."

    How in the world did you get that? mu itself (the coefficient of friction) is 0.05.
    The friction force (which is the only force on the sled) is 0.05*645.04 and is negative since it is taken back opposite to the direction of motion.
    Once you know the force, F= ma gives you acceleration.

    The initial speed was 7.0 m/s so the speed at any time t is v(t)= 7- at (where a is the acceleration found above. I included the "negative" explicitely here so a is positive). The sled stops when v(t)= 7- at= 0, of course, so you can find the time until it stops.

    You know, I presume, that for any acceleration a and initial speed v0, the distance moved is given by x(t)= -(a/2)t2+ v0t (a is the acceleration, again taken positive, and v0 is the intial speed, which was 7).
    Since you solved the equation above for the time t until the sled stopped, you can plug it in here and find the distance the sled moved in that time.
  4. Nov 22, 2004 #3
    hey! Thanks a billion i did that! i came out with an answer of 50 meters...is that what you got? Also, i have another problem, which i'm finding more difficult-
    The Problem:
    Philip pushes a book up against a wall, holding the book motionless. Given that Philip is pushing with a force P at an angle z, the coefficient of static friction betweeen the book and the wall is h, and that the book has a mass m, find the magnitude of the force Pp such that the book is motionless.

    What i did:
    I set up my free body diagram, which is Fn (normal force) to the left, mg down, and Fp to the south west (at my angle of z) from here i am completely lost....how would you begin to tackle this problem...
    The answer must be in terms of h, z, m, or using sin cos, tan, etc.

    I'm not very good when it comes to variables so thats part of the problem...What would you do? Is my free body diagram correct?
  5. Nov 22, 2004 #4


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    1) The normal force from the wall must balance the normal component of P; otherwise, the book would be pushed into the wall.
    2) The forces in the vertical direction must also sum to zero..
  6. Nov 22, 2004 #5
    so are you saying that there would only be three forces affecting the book, P, mh down, and the wall into the book? If this is the case, would the force pushing into the wall be Fn (normal force?)
  7. Nov 22, 2004 #6
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