Question

problem 10 <hw 03 - attempt 1> determine the force in cables ab and ac necessary to support the 15 - kg traffic light in (figure 1). < 1 of 1 > part a determine f_ab. express your answer to three significant figures and include the appropriate units. f_ab = 298 n previous answers all attempts used; correct answer displayed part b determine f_ac. express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Calculate the weight of the traffic - light

The weight of the traffic - light is given by $W = mg$, where $m = 15$ kg and $g=9.81$ m/s². So $W=15\times9.81 = 147.15$ N.

Step2: Set up equilibrium equations

Let's assume the forces in cables $AB$ and $AC$ are $F_{AB}$ and $F_{AC}$ respectively. In the x - direction: $\sum F_x=F_{AB}\cos12^{\circ}-F_{AC}\cos25^{\circ}=0$. In the y - direction: $\sum F_y=F_{AB}\sin12^{\circ}+F_{AC}\sin25^{\circ}-W = 0$.
From $F_{AB}\cos12^{\circ}-F_{AC}\cos25^{\circ}=0$, we have $F_{AC}=\frac{F_{AB}\cos12^{\circ}}{\cos25^{\circ}}$.
Substitute $F_{AC}$ into the y - direction equilibrium equation:
\[

$$\begin{align*} F_{AB}\sin12^{\circ}+\frac{F_{AB}\cos12^{\circ}}{\cos25^{\circ}}\sin25^{\circ}-147.15&=0\\ F_{AB}(\sin12^{\circ}+\frac{\cos12^{\circ}\sin25^{\circ}}{\cos25^{\circ}})&=147.15\\ F_{AB}(\sin12^{\circ}+\tan25^{\circ}\cos12^{\circ})&=147.15\\ F_{AB}(0.2079 + 0.4663\times0.9781)&=147.15\\ F_{AB}(0.2079+0.4562)&=147.15\\ F_{AB}\times0.6641&=147.15\\ F_{AB}&=\frac{147.15}{0.6641}\approx222\text{ N} \end{align*}$$

\]

Step3: Calculate $F_{AC}$

Substitute $F_{AB}$ into $F_{AC}=\frac{F_{AB}\cos12^{\circ}}{\cos25^{\circ}}$.
$F_{AC}=\frac{222\times\cos12^{\circ}}{\cos25^{\circ}}=\frac{222\times0.9781}{0.9063}\approx240$ N

Answer:

Part A: $F_{AB}=222$ N
Part B: $F_{AC}=240$ N