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Newton's 3rd Law - Firefighter slides down a pole

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    A firefighter who weighs 712 N slides down a vertical pole with an acceleration 3 m/s2, directed downward. What are the (a) magnitude and (b) direction (up or down) of the vertical force on the firefighter from the pole and the (c) magnitude and (d) direction (up or down) of the vertical force on the pole from the firefighter?

    2. Relevant equations
    Newton's Second Law: Fnet = ma
    Newton's Third Law: F1 + F2 = 0

    3. The attempt at a solution

    I know that the magnitude of the force on the pole from the firefighter will be the same as the force on the firefighter from the pole, but I'm having trouble implementing this to Newton's Second and Third Laws.


    Let [itex]\textit{F}_{P}[/itex] be the vertical force on the firefighter from the pole.
    Let [itex]\textit{W}_{F}[/itex] be the weight of the firefighter.
    Let [itex]\textit{m}_{F}[/itex] be the mass of the firefighter.

    [itex]\textit{m}_{F} = \frac{W_F}{9.8}\textit{ = 72.6531 kg}[/itex]

    [itex]\sum\textit{F}_{y} = \textit{F}_{P}\textit{ - W}_{F}\textit{ = m}_{F}\textit{ a}_{y}\textit{ = }\textit{(72.6531 kg)(-3}\frac{m}{s^2}\textit{) + (712 N) = }[/itex][itex]\textit{494.0408 N}[/itex]

    This answer is correct according to the text.

    (b) Up, to resist the net downward acceleration.

    Also correct according to the text.

    (c) Now this is where I don't know how to implement the equations properly and visualize the diagram for forces.

    First, I tried identifying all the forces on the body of interest; the pole. Since in part (a) I identified all the forces acting on the firefighter, I chose to identify all the forces acting on the pole.

    With this, I reasoned that [itex]\textit{W}{_F}[/itex] acts on the pole in the downward direction, while the frictional force [itex]\textit{F}{_f}[/itex] acts upward on the pole.

    Since the net acceleration of the body (the pole) was 0, I put [itex]\sum\textit{F}{_y}\textit{ = F}{_f}\textit{ - W}{_F = 0}[/itex]

    This is wrong according to the text, but I also feel that I'm missing a key idea for this problem. I just don't know what :/
  2. jcsd
  3. Sep 29, 2011 #2
    steve , have you drawn free body diagram for the firefighter and the pole separately ?
    it does helps to understand the situation. its better than imagining in your mind.

    now remember in newton's third law, the action and reaction always act on different bodies. also all forces in nature come in action-reaction pairs. so [itex]W_F[/itex]doesn't act on the pole. thats wrong kind of language. [itex]W_F[/itex] is force exerted by the earth on the man and one should ask instantly , what is its reaction according to third law ? well, that is force exerted by the man on the earth. these two forces form action-reaction pair. now what you found in part a is force exerted on the man by the pole. so its reaction is the force exerted by the man on the pole and by newton's third law , their magnitudes are equal....... does it help
  4. Sep 29, 2011 #3
    Yes I did also draw a free-body diagram to go along with this problem.

    Well I did already know the answer to part (c) because of Newton's Third Law but I didn't know if it had anything to do with the free-body diagram and the affecting forces.

    I guess what I'm confused about is figuring how to apply Newton's Second Law in finding (c).

    Or is it that you CAN'T just use Newton's Second Law to find (c)?
  5. Sep 30, 2011 #4
    yes you can use second law, but the intermediate step would be newton's third law anyway.
  6. Sep 30, 2011 #5
    Wait. Does the firefighter also exert a frictional force on the pole?

    I always assumed friction to be be made from only one body to another interacting body. Not both bodies.
  7. Sep 30, 2011 #6
    the force of interaction between the man and the pole would be, yes , frictional force.
    frictional force is just one type of force.... since its interactive force, you need two bodies
    to interact
  8. Sep 30, 2011 #7
    Ahh I see. Okay, so since there's a frictional force exerted by both the pole and firefighter, you'd have to reuse the force found in (a) to use Newton's Second Law in part (c) correct?
  9. Oct 1, 2011 #8
    yes, correct...... but we can directly use third law... since every force has its reaction paired with it (unless its pseudo force, which doesn't have reaction pair... thats why its called 'pseudo' force.......), and in part a), you found one of these pairs, part c) is asking for the reaction associated with the force in part a)......
  10. Oct 1, 2011 #9
    Okay thanks much :]
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