Chapter 2, Problem 2.127RP

Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.

Fig. P2.127

To determine

The direction and magnitude of the resultant of two forces.

Answer to Problem 2.127RP

The magnitude of the resultant of two forces is $\underset{\_}{R=104.4\text{N}}$, and the direction is $\underset{\_}{\varphi \text{=86}\text{.7}°}$.

Explanation of Solution

The sketch of the triangle formed by the two forces $P$, $Q$, and resultant $R$ is shown in figure $1$.

Write the expression for the total angle of a triangle

$\alpha +\varphi +\text{γ}=180°$ (I)

Here, $\alpha$, $\varphi$, and $\text{γ}$ are the three angles of a triangle.

Write the expression for the angle $\varphi$ formed at the base

$\varphi =180°-\alpha -80°$ (II)

The sum of the three angles of a triangle is equal to $180°$.

Write the expression for cosine law for finding $R$ in figure

$R^2=A^2+B^2-2\left(AB\cos \text{γ}\right)$ (III)

Here, $R$ is the hypotenuse which is representing resultant of two forces $P$ and $Q$

Write the expression for law of Sines

$\frac{A}{\sin \alpha }=\frac{B}{\sin \varphi }=\frac{R}{\sin \text{γ}}$ (IV)

Here, $A$, $B$, and $C$ are the side lengths of triangle.

Conclusion:

Use equation (I) to obtain the angle $\text{γ}$ as shown in figure

$\begin{array}{c}\text{γ=}180°-\left(20°+10°\right)\\ =150°\end{array}$

Substitute $48\text{N}$ for $A$, $60\text{N}$ for $B$, and $150°$ for $\text{γ}$ to obtain resultant force $R$

$\begin{array}{c}{R}^2={\left(48\text{N}\right)}^2+{\left(60\text{N}\right)}^2-2\left(48\text{N}\times 60\text{N}\times \cos 150°\right)\\ =10892.3\\ R=104.366\text{N}\end{array}$

Use the law of sines to find the angle $\alpha$.

Substitute $48\text{N}$ for $A$, $104.366\text{N}$ for $R$, and $150°$ for $\text{γ}$ in equation (III) to find $\alpha$

$\begin{array}{c}\frac{48\text{N}}{\sin \alpha }=\frac{104.366\text{N}}{\sin \text{150}°\text{N}}\\ \sin \alpha =\frac{48\text{N}\times \sin \text{150}°}{104.366\text{N}}\\ =0.22996\\ \alpha =13.2947°\end{array}$

Substitute $13.2947°$ for $\alpha$ in equation (II)

$\begin{array}{c}\varphi =180°-13.2947°-80°\\ =86.705°\end{array}$

Therefore the magnitude of resultant force is $\underset{\_}{R=104.4\text{N}}$, and the direction is $\underset{\_}{\varphi =86.7°}$.