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A firefighter with a weight of 722 N slides down a vertical pole with an acceleration of 3.11 m/s2, directed downward.
(a) What are the magnitude and direction of the vertical force (use up as the positive direction) exerted by the pole on the firefighter?
N

(b) What are the magnitude and direction of the vertical force (use up as the positive direction) exerted by the firefighter on the pole?
N

  • physics -

    a) Normally, al objects on earth fall with the same (nearly constant)acceleration (namely 9.81 m/s²).
    We see here that the firefighter only slides down the pole with an acceleration of 3.11 m/s². This means, that in order to slow down the firefighter, there must be an upward acceleration of 6.70 m/s² which results from the force of the pole acting on the firefighter.
    Now, according to Newtons first law F = m.a (Force equals mass times acceleration)
    So, since we have our upward acceleration and since we have the mass of the firefighter (m=722N/(9.81 m/s²) = 73.6 kg), we can calculate the force of the pole acting on the firefighter as follows:

    F= m.a = 73,6 kg . 6.70 m/s² = 493.12 N

    This force acts in the upward direction.

  • physics -

    b) According to Newton's 3rd law, for every action there is an equal but opposite reaction. This means that if the pole exerts a force of 493.12 N on the firefighter in the upward direction, the firefighter must also be exerting a force on the pole of 493.12 N in the downard direction.

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