91Ó°ÊÓ

weight =$1000$pounds

and figure showing the arrangement

When two parallel lines are cut by a transverse, the alternate angles are equal.
Here, the angles A and B form the alternate angles.

Draw the force diagram using vectors and mark the corresponding angles using the alternate angle principle which is stated above. Let the tensions in the two cables are F₁ and F₂ and the tension in the weight be F₃.

The vector having angle $\mathrm{Î\pm }$and magnitude$\left|\right|v\left|\right|$is,

$v=\left|\right|v\left|\right|(\cos \mathrm{Î\pm }i+\sin \mathrm{Î\pm }j),$where$\mathrm{Î\pm }$is the angle between the vector and positive x- axis.

From the figure the force $F_1$is expressed as,

$$\begin{gathered} F_1=\left|\right|F_1\left|\right|(\cos 40°i+\sin 40°j) \\ =\left|\right|F_1\left|\right|(0.7660444431i+0.6427876096j) \\ =0.7660444431\left|\right|F_1\left|\right|i+0.6427876096\left|\right|F_1\left|\right|j \end{gathered}$$

Again from the figure the force F₂is expressed as,

$$\begin{gathered} F_2=\left|\right|F_2\left|\right|(\cos 155°i+\sin 155°j) \\ =\left|\right|F_2\left|\right|(-0.9063077870i+0.4226182617j) \\ =-0.9063077870\left|\right|F_2\left|\right|i+0.4226182617\left|\right|F_2\left|\right|j \end{gathered}$$

From figure F₃ equals to 1000 pounds the weight of the suspended weight

Thus,the vector F₃is suspended as

$$\begin{gathered} F_3=1000(\cos 270°i+\sin 270°j) \\ =1000(0i+(-1\left)j\right) \\ =-1000j \end{gathered}$$

$F_1+F_2+F_3=0$

$0.7660444431\left|\right|F_1\left|\right|i+0.6427876096\left|\right|F_1\left|\right|j-0.9063077870\left|\right|F_2\left|\right|i+0.4226182617\left|\right|F_2\left|\right|j+(-1000j)=0$

From equation 1 we will get the value of $\left|\right|F_1\left|\right|$as

role="math" localid="1649678254737" $$\begin{gathered} 0.7660444431\left|\right|F_1\left|\right|-0.9063077870\left|\right|F_2\left|\right|=0 \\ 0.7660444431\left|\right|F_1|=0.9063077870|\left|F_2\right| \\ \left|\right|F_1\left|\right|=\frac{0.9063077870}{0.7660444431}\left|\right|F_2| \\ |\left|F_1\right||=1.1831007915|\left|F_2\right|| \end{gathered}$$

Substitute $1.1831007915$$\left|\right|F_2\left|\right|$for $\left|\right|F_1\left|\right|$in equation (2)

role="math" localid="1649678541554" $$\begin{gathered} 0.6427876096(1.1831007915|\left|F_2\right|\left|\right)+0.4226182617\left|\right|F_2\left|\right|=1000 \\ 0.7606825296\left|\right|F_2\left|\right|+0.4226182617\left|\right|F_2\left|\right|=1000 \\ 1.1831007913\left|\right|F_2\left|\right|=1000 \\ \left|\right|F_2\left|\right|=\frac{1000}{1.1831007913} \\ \left|\right|F_2\left|\right|=845.2365236787 \\ ≈845.2lb \end{gathered}$$

$$\begin{gathered} 0.7660444431\left|\right|F_1\left|\right|-0.9063077870(845.2)=0 \\ 0.7660444431\left|\right|F_1\left|\right|-766.0113415724=0 \\ 0.766\left|\right|F_1\left|\right|=766.0113415724 \\ \left|\right|F_1\left|\right|=\frac{766.0113415724}{0.7660444431} \\ \left|\right|F_1\left|\right|=999.9567890245 \\ ≈1000lb \end{gathered}$$

Therefore,the tensions in the right and left cables are$1000lband$ $845.2lb$respectively