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Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a Pole

  1. Sep 27, 2008 #1
    The alarm at a fire station rings and a 96.0-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 3.55 m). Just before landing, his speed is 1.79 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?



    Relevant equations:
    I'm not sure, but I believe this has something to do with Newton's second law of motion? So this may be helpful {[tex]\Sigma[/tex]Fnet }/mass = acceleration , although I couldn't see a way to use it myself

    I mainly used: y = (v2y - v2oy)/(2ay), where ay can be substituted with ay= (-fk)/mass



    3. The attempt at a solution

    Sooo, my messy and essentially incorrect attempt:

    y = -3.55 m
    v0 = 0 m/s
    voy = 1.79 m/s
    mass = 96.0 kg

    using the equation for the y-displacement, I attempted to find fk:

    -3.55 m = (1.792 - 0 m/s) / (2(-fk)/96kg)

    multiplying both sides by -(2/96) fk ----> .07396 fk = 1.79^2

    fk = 43.3 N


    if someone could point me in the right direction i would be terribly grateful!
     
  2. jcsd
  3. Sep 27, 2008 #2

    Doc Al

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    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    Solve this in two stages. First find the acceleration, using the kinematic formula that you stated. Then apply Newton's 2nd law to find the unknown friction force. What forces act on the fireman?
     
  4. Sep 27, 2008 #3
    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    i found the acceleration to be -.04512, but i'm confused right now about [tex]\Sigma[tex]
    Fy and -fk.

    if [tex]\Sigma[tex]
    Fy = ma_y and a_y = -fk/mass, then are these two force values the same?
     
  5. Sep 27, 2008 #4

    Doc Al

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    Staff: Mentor

    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    Recheck that calculation. (Pay attention to the decimal point.)

    When using [itex]\Sigma F = m a[/itex], [itex]\Sigma F[/itex] represents the net force on the man. What forces act on the man?
     
  6. Sep 27, 2008 #5
    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    he is being acted on by a friction force from the pole... which is slowing down his acceleration
     
  7. Sep 27, 2008 #6
    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    oh sorry! i meant to type -0.4512 m/s^2!!

    in that case, would it be correct of me to say that the net force and not the kinetic frictional force is 43.3 N ?
     
  8. Sep 27, 2008 #7
    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    forces which act on the man:
    gravity, kinetic frictional force
     
  9. Sep 27, 2008 #8

    Doc Al

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    Staff: Mentor

    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    Good. That's one of the forces acting on him. What's the other?

    Good. (I assume you made it negative to indicate that the acceleration acts down.)

    Correct.
     
  10. Sep 27, 2008 #9

    Doc Al

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    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    Good!

    Now write an equation. Add up the forces to find the net force (direction counts!) and set it equal to "ma".
     
  11. Sep 27, 2008 #10
    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    So, Fnet = -W + f_k = ma = 43.3 N?

    = -(mg) + f_k = ma = 43.3 N
    = -(96 kg)((9.8 m/s) + f_k = 43.3 N
    = f_k = 940.8 N

    does this look right? it doesn't seem right to me...
     
  12. Sep 27, 2008 #11

    Doc Al

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    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    Careful with signs. Which way does the net force point?
     
  13. Sep 27, 2008 #12
    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    downward.
    so, would it be
    Fnet = -W + f_k = ma = -43.3 N?

    so... f_k = (96 kg)(9.8 m/s) -43.3 N

    and f_k = 897.5 N ?

    i hope its right :x
     
  14. Sep 27, 2008 #13

    Doc Al

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    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    Looks good.
     
  15. Sep 27, 2008 #14
    Re: Newton's Second Law of Motion, Kinetic Friction Force of Fireman Sliding Down a P

    hoorayy~! and yet i would never have expected the number to come out so big... thank you for your help!
     
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