# Determine the force in cables AB and AC necessary 1

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

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### 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)

Show me the free body diagram again

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