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