A climber with mass m=54.4 kg has reached the top of the l=15.70 m climbing wall, and then is lowered straight down by the belaying person without touching the wall. Assume that at the top the rope is going through the pulley, while to the belaying person the rope is inclined by the angle of θ=44.0° from the vertical. If the landing speed of the climber should not exceed 2.66 m/s, what is the magnitude of the minimum average force that is exerted by the belaying person? Express your answer in Newtons.
The Attempt at a Solution
•found the acceleration to be 0.2253..m/s2 using equation 2
•Have another equation for Fnet=FT-mg
unsure where the angle comes into play, and whether finding FT is the force exerted by the belayer.