Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 373: 101

a) $L=2.49m$ b) $F_2=1.49KN$

(a) We know that $L-d=\frac{Fd}{mg}$ We plug in the known values to obtain: $L-d=\frac{(828N)(1.10m)}{(67.0Kg)(9.81m/s^2)}$ $\implies L-d=1.39m$ Total length: $L=1.39+1.1=2.49 m$ (b) We know that $F_2=F_1+mg$ We plug in the known values to obtain: $F_2=828N+(67.0Kg)(9.81m/s^2)$ $\implies F_2=1485.27N=1.49KN$