Question

The 500-N weight is supported by two cables, the cable forces being F1 and F2. Knowing that the resultant of F1 and F2 is a force of magnitude 500 N acting in the y-direction, determine F1 and F2.

Answer 1

Gives

$F=500N$

Assume diagram

Force horizontal components

$\Sigma F_H=0$

$-T_1cos 35°+T_2cos50°=0$

$T_2=\frac{T_1cos35}{cos50}$

Vertical components

$\Sigma F_v=0$

$T_1sin35°+T_2sin50°-500=0$

Put T₂ value

$$T_1sin35°+\frac{T_1cos35}{cos50}\times sin50°-500°=0$$

$T_1=322.631477N$

$T_2=T_1\frac{cos35°}{cos50°}$

$T_2=411.14N$