Chapter 2, Problem 104RP

Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 15 kN in member A and 10 kN in member B, determine by trigonometry the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.

To determine

The magnitude and direction of the of the resultant of the forces applied to the bracket by members $A$ and $B$.

Answer to Problem 104RP

The magnitude of the resultant force applied to the bracket by members $A$ and $B$ is $21.7945\text{kN}$

The resultant force vector will be at an angle of $73.4145°$ from the horizontal.

Explanation of Solution

Given information:

The force in member $A$ is $15\text{kN}$ and force in member $B$ is $10\text{kN}$.

The given figure below shows the force diagram.

Figure-(1)

Write angle between the given forces.

$\begin{array}{c}\theta =180°-20°-40°\\ =120°\end{array}$

Write the expression for law of cosines.

$R^2=A^2+B^2-2AB\cos \theta$ ...... (I)

Write the expression for sine law.

$\frac{\sin \left(20°+\alpha \right)}{A}=\frac{\sin \theta }{R}$ ...... (II)

Here, angle made by the resultant with vertical is $\alpha$ ,and angle made by the forces $A$ and $B$ is $\theta$.

Write the expression for the angle made by the resultant with the horizontal.

$\phi =90°-\alpha$ ...... (III)

Calculation:

Substitute the value $15\text{kN}$ for $A$, $10\text{kN}$ for $B$ and $120°$ for $\theta$ in Equation (I)

$\begin{array}{l}{R}^2={\left(15\text{kN}\right)}^2+{\left(10\text{kN}\right)}^2-2\left(15\text{kN}\right)\left(10\text{kN}\right)\cos 120°\\ R^2=\left(225{\text{kN}}^2+100{\text{kN}}^2-\left(300\right)\left(-0.5\right){\text{kN}}^2\right)\\ R^2=475{\text{kN}}^2\\ R=21.7945\text{kN}\end{array}$

Substitute the value $21.7945\text{kN}$ for $R$, $120°$ for $\theta$ and $15\text{kN}$ for $A$ in Equation (II)

$\begin{array}{l}\frac{\sin \left(20+\alpha \right)}{15\text{kN}}=\frac{\sin 120°}{21.7945\text{kN}}\\ \sin \left(20+\alpha \right)=\frac{\left(0.866\right)\left(15\text{kN}\right)}{21.7945\text{kN}}\\ 20+\alpha =0.596\\ \alpha =16.5855°\end{array}$

Substitute the value $16.5855°$ for $\alpha$ in Equation (III)

$\begin{array}{c}\phi =90°-16.5855°\\ =73.4145°\end{array}$

Conclusion:

The magnitude of the resultant of the forces applied to the brackets is $21.7945\text{kN}$.

The resultant force vector will be at an angle of $73.4145°$ from the horizontal.