27. We choose up as the +ydirection, so 2ˆ(3.00m/s)ja=−G(which, without the unit-vector, we denote as asince 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. (a) We denote the force exerted by the pole on the firefighter fpˆjFF=Gand apply Eq.5-1. Since netFma=GG, we have
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.