Problem 10.25

10.25.

If we sum the forces in the $\hat{x}$ and the $\hat{y}$ directions we find that
$\sum F_{x} = f_{x} - T\cos 25{}^{\circ}= 0$
where the term f_x is the force in the $\hat{x}$ direction at the pivot point.
$\sum F_{y} = f_{y} + t\sin 25{}^{\circ}- 3200 N = 0$
again, here f_y is the force at the pivot in the $\hat{y}$ direction.
Now, if we take the perpendicular components of the forces and sum the torques, we find
$\sum \vec{\tau} = (T)(\frac{3\ell}{4}) - (1200 N)(\sin 25{}^{\circ})(\frac{\ell}{2}) - (2000 N)(\sin 25{}^{\circ})(\ell) ~=~ 0$
Solving this equation for the tension, T, we find
T = 1465 N
Now, plugging this into the x equation, we find
f_x = 1.33 kN
and substituting T into the y equation, we see that
f_y = 2.58 kN.