Hibbeler Statics 14E P2.1 — Solving for the Magnitude and Direction of the Resultant of Two Coplanar-Concurrent Forces

_{Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-1}

Solution:

The parallelogram law and the triangulation rule are shown in the figures below.

Considering figure (b), we can solve for the magnitude of $\textbf{F}_R$ using the cosine law.

\begin{align*} \textbf{F}_R & = \sqrt{700^2+450^2-2\left( 700 \right)\left( 450 \right)\cos45^{\circ}}\\ & = 497.01 \ \text{N}\\ & = 497 \ \text{N} \end{align*}

Then we use the sine law to solve for the interior angle $\theta$ .

\begin{align*} \frac{\sin \theta}{700} & = \frac{\sin 45^{\circ}}{497.01}\\ \sin \theta & =\frac{700\ \sin 45^{\circ }}{497.01}\\ \theta & = \sin^{-1} \left( \frac{700\ \sin 45^{\circ }}{497.01} \right)\\ & \text{This is an ambiguous case }\\ \theta & = 84.81^\circ \ or \ \theta =95.19^\circ \\ \end{align*}

In here, the correct angle measurement is $\theta = 95.19^{\circ}$ .

Thus, the direction angle $\phi$ of $\textbf{F}_R$ measured counterclockwise from the positive x-axis, is

\begin{align*} \phi & = \theta +60^\circ \\ & = 95.19^\circ +60^\circ \\ & = 155^\circ \end{align*}

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