A climber with mass m =67.2 kg has reached the top of the l =13.29 m climbing wa

Question

A climber with mass m=67.2 kg has reached the top of the l=13.29 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 =39.1° from the vertical. If the landing speed of the climber should not exceed 1.48 m/s, what is the magnitude of the minimum average force that is exerted by the belaying person? Express your answer in Newtons.

Explanation / Answer

vo = initial speed at the top = 0 m/s

vf = final speed at landing = 1.48 m/s

a = acceleration = ?

Y = vertical displacement = l = 13.29 m

using the equation

vf2 = vo2 + 2 a Y

1.482 = 02 + 2 (13.29) a

a = 0.0824 m/s2

T = tension force in the rope

m = mass of climber = 67.2 kg

force equation for the motion of climber is given as

mg - T = ma

(67.2) (9.8) - T = (67.2) (0.0824)

T = 653 N

so force exerted by the person = F = T = 653 N

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