Question

f the magnitude of the resultant force is to be 9kN directed along the

positive x axis, determine the magnitude of force T acting on the eyebolt

and its angle delta.

Answer 1

$T=\sqrt{{F_1}^2+{F_2}^2-2F_1F_2\cos45°},$

$T=6.57~kN,$

$\delta= \arctan \frac {F_1\sin 45°}{T},$

$\delta= 30.6°.$