A firefighter of mass 120 kg slides down a vertical pole with an acceleration of 4 m/s per second. What is the friction force that acts on the firefighters?

Question

In this case, the friction force is the force that is acting against gravity.

Use F=M*a
If g=9.8m/s^2 and the firefighter's acceleration is 4m/s^2 then (9.8-4)m/s^2 is the deceleration suppied by the firefighter's hands (or legs, whatever part of the body he/she is using to decelerate). So the force is
F=M*(9.8-4)m/s^2 where M is the firefighter's mass.

Answer 1

480kgm/s^2

Answer 2

Thank you