Chapter 2, Problem 2.127RP

To determine

The magnitude (R) and direction $\left(\varphi \right)$ of the resultant of the force applied to the bracket by member A and B by using trigonometry.

Answer to Problem 2.127RP

The magnitude (R) and direction $\left(\varphi \right)$ of the resultant of the force applied to the bracket by member A and B are $\underset{\_}{21.8\text{kN}}$ and $\underset{\_}{73.4°(\searrow )}$ respectively.

Explanation of Solution

Given information:

The force $\left(F_A\right)$ in member A is 15 kN at an angle $\left(\phi \right)$ is $40°$ with the vertical axis.

The force $\left(F_B\right)$ in member B is 10 kN at an angle $\left(\beta \right)$ is $20°$ with the vertical axis.

Calculation:

Draw the force triangle as in Figure (1).

Calculate the angle $\left(\gamma \right)$ using the relation:

$\gamma =180°-\left(\phi +\beta \right)$

Substitute $40°$ for $\phi$ and $20°$ for $\beta$.

$\begin{array}{c}\gamma =180°-\left(40°+20°\right)\\ =120°\end{array}$

Calculate the magnitude of the resultant force by applying law of cosine in Figure 1.

$R^2=F_A^2+F_B^2-2F_A{F}_B\cos \gamma$

Substitute 15 kN for $F_A$, 10 kN for $F_B$, and $120°$ for $\gamma$.

$\begin{array}{l}{R}^2={15}^2+{10}^2-2\left(15\times 10\right)\cos \left(120°\right)\\ =225+100-\left(-150\right)\\ R=\sqrt{475}\\ R=21.\text{8}\text{kN}\end{array}$

Calculate the angle $\left(\alpha \right)$ using the law of sines:

$\frac{F_B}{\sin \alpha }=\frac{R}{\sin \gamma }$

Substitute 21.8 kN for R and $120°$ for $\gamma$.

$\begin{array}{l}\frac{10}{\sin \alpha }=\frac{21.8}{\sin \left(120°\right)}\\ \frac{10}{\sin \alpha }=25.1725\\ \alpha ={\sin }^{-1}\left(\frac{10}{25.1725}\right)\\ \alpha =23.414°\end{array}$

Calculate the angle $\left(\varphi \right)$ using the relation:

$\varphi =\alpha +\left(90°-\phi \right)$

Substitute $23.414°$ for $\alpha$ and $40°$ for $\phi$.

$\begin{array}{c}\varphi =23.414°+\left(90°-40°\right)\\ =73.4°(\searrow )\end{array}$

Thus, the magnitude (R) and direction $\left(\varphi \right)$ of the resultant of the force applied to the bracket by member A and B are $\underset{\_}{21.8\text{kN}}$ and $\underset{\_}{73.4°(\searrow )}$ respectively.