Answer
$F_3=776N$
$F_1=466N$
$F_2=879N$
Work Step by Step
We can find the magnitude of the required forces as follows:
The sum of the forces in the x-direction
$\Sigma F_x=0$
$\implies (\frac{3}{5})F_3(\frac{3}{5})+600-F_2=0$.eq(1)
The sum of the forces in the y-direction
$\Sigma F_y=0$
$\implies (\frac{4}{5})F_1-(\frac{3}{5})F_3(\frac{4}{5})=0$.eq(2)
and the sum of the forces in the z-direction
$F_z=0$
$\implies (\frac{4}{5})F_3+(\frac{3}{5})F_1-900=0$..eq(3)
From eq(2), we obtain:
$5F_1-3F_3=0$
$\implies F_1=\frac{3}{5}F_3$..eq(4)
From eq(3), we obtain:
$4F_3+3F_1=4500$eq(5)
We plug in the value of $F_1$ from eq(4) into eq(5) to obtain:
$4F_3+3(\frac{3}{5}F_3)=4500$
This simplifies to:
$F_3=776N$
We plug in the value of $F_3$ in eq(4) to obtain:
$F_1=\frac{3}{5}(776)=466N$
From eq(1), we have
$(\frac{9}{25})F_3+600-F_2=0$
$\implies (\frac{9}{25})(776)+600-F_2=0$
This simplifies to:
$F_2=879N$
