# Boom AB is held in the position shown by three cables. Knowing that the tensions in calbe AC and AC are 900lb and 1200lb,respectively, determne (a) the tension in cable AE if the resultant of thettensions exerted at point A of the boom must be directed along AB, (b) the corresponding magnitude o the resultant. 02510500651.png

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Boom AB is held in the position shown by three cables. Knowing that the tensions in calbe AC and AC are 900lb and 1200lb,respectively, determne (a) the tension in cable AE if the resultant of thettensions exerted at point A of the boom must be directed along AB, (b) the corresponding magnitude o the resultant.

2021-02-25
$$\displaystyle\sum{F}_{{x}}$$ Tension=R=AE
$$\displaystyle{A}{C}{x}=-{900}{l}{b}{s}$$
$$\displaystyle{A}{D}{x}=-{12000}{\cos{{\left({30}\right)}}}=-{1039.23}{l}{b}{s}$$
$$\displaystyle{A}{E}{x}={R}{\cos{{\left({50}\right)}}}$$
$$\displaystyle\sum{F}_{{y}}$$
$$\displaystyle{A}{C}{y}={0}$$
$$\displaystyle{A}{D}{y}=-{600}{l}{b}{s}$$
$$\displaystyle{A}{E}{y}=-{R}{\sin{{\left({50}\right)}}}$$
$$\displaystyle{A}{C}{x}+{A}{D}{x}=-{1939.23}{l}{b}{s}$$
$$\displaystyle{A}{C}{y}+{A}{D}{y}=-{600}{l}{b}{s}$$
Just solve the following equations for R and P...
$$\displaystyle-{19.39}{.23}-{R}{\cos{{\left({50}\right)}}}=-{P}{\cos{{\left({65}\right)}}}$$
$$\displaystyle-{600}-{R}{\sin{{\left({50}\right)}}}=-{P}{\sin{{\left({65}\right)}}}$$
You should get...
$$\displaystyle{R}={1659.45}{l}{b}{s}$$
$$\displaystyle{P}={2064.65}{l}{b}{s}$$

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