G 27. We choose up as the +y direction, so a = (− 3.00 m/s 2 )ˆj (which, without the unitvector, we denote as a since this is a 1-dimensional problem in which Table 2-1 applies). From Eq. 5-12, we obtain the firefighter’s mass: m = W/g = 72.7 kg. G (a) We denote the force exerted by the pole on the firefighter Ff p = Ffp ˆj and apply Eq. G G 5-1. Since Fnet = ma , we have Ffp − Fg = ma
Ffp − 712 N = (72.7 kg)(−3.00 m/s 2 )
which yields Ffp = 494 N.
G (b) The fact that the result is positive means Ffp points up. G G (c) Newton’s third law indicates Ff p = − Fpf , which leads to the conclusion that G | Fpf | = 494 N . G (d) The direction of Fpf is down.