Answer
$$
\begin{aligned}
& F=75 \mathrm{~N} \\
& P=100 \mathrm{~N}
\end{aligned}
$$
Work Step by Step
$$
B A=0.5 \mathrm{~m}
$$
The couple created by the $150-\mathrm{N}$ forces is
$$
M_{C 1}=150(0.5)=75 \mathrm{~N} \cdot \mathrm{m}
$$
Then
$$
\begin{aligned}
\mathbf{M}_{C 1} & =75\left(\frac{3}{5}\right) \mathbf{j}+75\left(\frac{4}{5}\right) \mathbf{k} \\
& =45 \mathbf{j}+60 \mathbf{k} \\
\mathbf{M}_{C 2} & =-P(0.6) \mathbf{k} \\
\mathbf{M}_{C 3} & =-F(0.6) \mathbf{j}
\end{aligned}
$$
Require
$$
\begin{aligned}
& \mathbf{M}_{C 1}+\mathbf{M}_{C 2}+\mathbf{M}_{C 3}=\mathbf{0} \\
& 45 \mathbf{j}+60 \mathbf{k}-P(0.6) \mathbf{k}-F(0.6) \mathbf{j}=\mathbf{0}
\end{aligned}
$$
Equate the $\mathbf{j}$ and $\mathbf{k}$ components
$$
\begin{aligned}
& 45-F(0.6)=0 \\
& F=75 \mathrm{~N} \\
& 60-P(0.6)=0 \\
& P=100 \mathrm{~N}
\end{aligned}
$$
