Question

A firefighter with a weight of 718 N slides down a vertical pole with an acceleration of 3.22 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? 1 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?

Answer 1

a) 482N Upward

b) 482N Downward

Step-by-step explanation:

According to Newton's second law:

`\sum F=m.a\\F_p-m.g=m.a\\F_p=m(g+a)\\m=(F)/(g)\\m=(718N)/(9.81m/s^2)\\m=73.2kg\\\\F_p=73.2kg(9.81m/s^2-3.22m/s^2)\\F_p=482N`

We substract the acceleration of 3.22m/s^2 because it is going downwards.

The direction of the force is upward because it is opposite to the movement, that's why the acceleration is smaller than the gravity's acceleration.

According to Newton's third law, The force of the firefighter must be the same but in the opposite direction, so:

`F_f=482N\\`

going downward.