A firefighter of mass 39 kg slides down a vertical pole with an acceleration of 3 m/s^2.

Question

The acceleration of gravity is 10 m/s^2.
What is the friction force that acts on him?
Answer in units of N.

Answer 1

F = m a

mg - Ffriction = m (3)
so
F friction = m(g-3)

= 39 (7)

= 273 Newtons