Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 368: 31

$F_2=3318N$ $F_1=-2.2KN$

We know that $F_2=\frac{\frac{1}{2}W_bL+m_dgL}{d}$ We plug in the known values to obtain: $F_2=\frac{\frac{1}{2}(225N)(5.00m)+(90.0Kg)(9.81m/s^2)(5.00m)}{1.50m}$ $\implies F_2=3318N$ Now $F_1=m_dg+W_b-F_2$ We plug in the known values to obtain: $F_1=(90.0Kg)(9.81m/s^2)+225N-3318N$ $\implies F_1=-2210N=-2.2KN$