Determine the force in cables AB and AC necessary to support the 12-kg traffic light.

#### Solution:

We will first draw the free body diagram representing the forces at point A (the traffic light).

We will assume forces going \rightarrow^+ to be positive and \uparrow+ to be positive and write our equations of equilibrium.

\sum \text{F}_\text{x}=0

F_{AB}\text{cos}\,(12^0)\,-\,\frac{24}{25}F_{AC}=0 ——————(eq.1)

\sum \text{F}_\text{y}=0

F_{AB}\text{sin}\,(12^0)\,+\,\frac{7}{25}F_{AC}\,-\,117.72=0 ——————(eq.2)

To solve for F_{AB} and F_{AC}, we will isolate for F_{AB} in eq.1.

F_{AB}=\frac{24\,F_{AC}}{25\,\text{cos}\,(12^0)}

(Simplify)

F_{AB}\,=\,0.9814F_{AC} ——————(eq.3)

We will now substitute this value into eq.2.

0.9814F_{AC}\text{sin}\,(12^0)\,+\,\frac{7}{25}F_{AC}\,-\,117.72=0

(solve for F_{AC})

F_{AC}=243.2 N

We can now substitute the value of F_{AC} into eq.3 to figure out F_{AB}.

F_{AB}\,=\,(0.9814)(243.2)

F_{AB}\,=\,238.7 N

good