The eyebolt is fastened to a thick base plate, and it supports three steel cables with tensions of 150 lb, 350 lb, and 800 lb. Determine the resultant force that acts on the eyebolt by using the vector algebra approach. The unit vectors I and j are oriented with the x-y coordinates as shown.

Solution:

Approach-

Given those, the magnitude and angle of action of R fallow from Equation,

$ \theta=\tan ^{-1}\left(\frac{R_y}{R_x}\right) \\ $

Solution:-

The components of the $8001 \mathrm{~b}$ force are,

$ \begin{aligned} \\ F_{x, 1} &=(800 \mathrm{lb}) \cos 45^{\circ} \\\\ &=56.7 \mathrm{lb} \\\\ F_{y, 1} &=(800 \mathrm{~J}) \sin 45^{\circ} \\\\ &=565.7 \mathrm{lb}\\ \end{aligned}\\ $

and $F_1$ is written in vector from as,

$ F_1=565.7 j+565 l 7 j 13\\ $

By Using the same procedure for the other two forces,

$ \begin{aligned}\\ F_2 &=-\left(350 \sin 20^{\circ}\right) j+\left(350 \cos 20^{\circ}\right) j \mathrm{db} \\\\ &=-119.7 j+328.9 j \mathrm{Jb} \\\\ F_3 &=-150 i \mathrm{Jb}\\ \end{aligned}\\ $

To calculate the components of the resultant, the horizontal and,

$ \begin{aligned} \\ R_x &=565.7-119.7-150 \mathrm{lb} \\\\ &=296.0 \mathrm{lb} \\\\ R_y &=565.7+328.9 \mathrm{lb} \\\\ &=894.9 \mathrm{lb}\\ \end{aligned}\\ $

The magnitude of the resultant force is

$ \begin{aligned}\\ R &=\sqrt{(296.0 \mathrm{Jb})^2+(894.6 \mathrm{Jb})^2} \\\\ &=942.3 \mathrm{lb}\\ \end{aligned}\\ $

and it acts at the angle

$ \begin{aligned}\\ a &=\tan ^{-1}\left(\frac{894.6 \lambda_b}{296.0 \lambda_b}\right) \\\\ &=\tan ^{-1}(3.022) \\\\ \Delta &=71.69^{\circ}\\ \end{aligned}\\ $

Discussion :-

The assistant force is larger than any one force, since $R_x$ and $R_y$ are positive,

The tip of the resultant vector lies in the first quadrant of the $x-y$ Plame.

$ \begin{aligned}\\ &R=942.3 \mathrm{lb} \\\\ &\partial=71.69^{\circ}\\ \end{aligned}\\ $