determine the force in cables ab and ac necessary to support the 13 - kg traffic light in (figure 1). part a…

Explanation:

Step1: Calculate the weight of the traffic - light

The weight of the traffic - light $W =mg$, where $m = 13\ kg$ and $g=9.81\ m/s^{2}$. So $W=13\times9.81 = 127.53\ N$.

Step2: Set up force equilibrium equations

Let's assume the angles and directions of forces. In the x - direction, $\sum F_x=F_{AB}\cos12^{\circ}-F_{AC}\times\frac{7}{25}=0$. In the y - direction, $\sum F_y=F_{AB}\sin12^{\circ}+F_{AC}\times\frac{24}{25}-W = 0$. From the x - direction equation, we have $F_{AB}\cos12^{\circ}=F_{AC}\times\frac{7}{25}$, so $F_{AB}=\frac{7F_{AC}}{25\cos12^{\circ}}$. Substitute $F_{AB}$ into the y - direction equation: [ \begin{align*} \frac{7F_{AC}}{25\cos12^{\circ}}\sin12^{\circ}+F_{AC}\times\frac{24}{25}-127.53&=0\ F_{AC}\left(\frac{7\tan12^{\circ}}{25}+\frac{24}{25}\right)&=127.53\ F_{AC}\left(\frac{7\times0.21256}{25}+\frac{24}{25}\right)&=127.53\ F_{AC}\left(\frac{1.48792 + 24}{25}\right)&=127.53\ F_{AC}\times\frac{25.48792}{25}&=127.53\ F_{AC}&=\frac{127.53\times25}{25.48792}\approx125\ N \end{align*} ]

Step3: Calculate $F_{AB}$

Substitute $F_{AC}$ into the x - direction relationship. $F_{AB}=\frac{7\times125}{25\cos12^{\circ}}=\frac{875}{25\times0.9781}\approx36.2\ N$

Answer:

Part A: $F_{AB}=36.2\ N$ Part B: $F_{AC}=125\ N$