Question

A firefighter who weighs 712 N slides down a vertical pole with an acceleration of 3.00 m/s 2 ,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) di- rection of the vertical force on the pole from the firefighter

Answer 1

Answer:

494.262996942 N

Upward

494.262996942 N

Downward

Explanation:

W = Weight of the firefighter = 712 N

g = Acceleration due to gravity = 9.81 m/s²

Mass is given by

$m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{712}{9.81}\\\Rightarrow m=72.5790010194\ kg$

Force is given by

$T-W=-ma\\\Rightarrow T=W-ma\\\Rightarrow T=712-72.5790010194\times 3\\\Rightarrow T=494.262996942\ N$

The force on the firefighter is 494.262996942 N

directed upward

On the pole the force will be the same 494.262996942 N

But the direction will be downward