ch05-p027 - j 27. We choose up as the +y direction, so a =...

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27. We choose up as the + y direction, so 2 ˆ (3 . 0 0m / s) j a =− G (which, without the unit- vector, 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. (a) We denote the force exerted by the pole on the firefighter fp ˆ j FF = G and apply Eq. 5-1. Since net Fm a = G G , 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.

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