Tag Archives: Force Vectors

Hibbeler Statics 14E P2.2 — Solving for an Unknown Force Given the Magnitude and Direction of a Resultant and Another Force

If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction $\theta$ .

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

Solution:

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

Engineering Mechanics by RC Hibbeler Problem 2.2 Parallelogram Law — Parallelogram Law

Engineering Mechanics by RC Hibbeler Problem 2.2 Triangulation Rule — Triangulation Rule

Considering the figure of the triangulation rule, we can solve for the magnitude of $\textbf{F}$ using the cosine law.

\begin{align*} \textbf{F} & = \sqrt{700^2+500^2-2\left( 700 \right)\left( 500 \right)\cos105^{\circ}}\\ & = 959.78 \ \text{N}\\ & = 960 \ \text{N}\\ \end{align*}

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

\begin{align*} \frac{\sin \left(90^{\circ}-\theta \right)}{700} & = \frac{\sin 105^{\circ}}{959.78}\\ \sin \left(90^{\circ}-\theta \right) & =\frac{700 \sin 105^{\circ }}{959.78}\\ 90^{\circ}-\theta & = \sin^{-1} \left( \frac{700 \sin 105^{\circ }}{959.78} \right)\\ \theta & = 90^\circ-\sin^{-1} \left( \frac{700 \sin 105^{\circ }}{959.78} \right) \\ \theta & = 90^\circ-44.79^\circ\\ \theta & = 45.2^\circ\\ \end{align*}

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

If $\theta = 60 \degree$ and $\textbf{F} = 450 \ \text{N}$ , determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

_{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*}

Chapter 2: Force Vectors

Vector Addition of Forces

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Addition of a System of Coplanar Forces

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Cartesian Vectors | Addition of Cartesian Vectors

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Force Vector Directed Along a Line

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Dot Product

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Problem 139

Hibbeler Statics 14E P2.2 — Resultant of a System of Two Forces

Determine the magnitude of the resultant force $\textbf{F}_{\text{R}} = \textbf{F}_1 + \textbf{F}_2$ and its direction, measured counterclockwise from the positive x axis.

Engineering Mechanics: Statics figure for Problem 2-3

_{Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 2-1
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-3}

SOLUTION:

The parallelogram law of the force system is shown.

Consider the triangle AOB.

Using cosine law to solve for the resultant force $\textbf{F}_{\text{R}}$

\begin{align*} \textbf{F}_\text{R} & =\sqrt{\left(250\right)^2+\left(375\right)^2-2\left(250\right)\left(375\right) \cos\:75^{\circ} }\\ & =393.2 \ \text{lb}\\ & =393\:\text{lb}\\ \end{align*}

The value of angle θ can be solved using sine law.

\begin{align*} \frac{393.2}{\sin\:\left(75^{\circ} \right)} & = \frac{250}{\sin\:\theta } \\ \sin \theta & = \frac{250 \ \sin75 \degree}{393.2}\\ \theta & =\sin^{-1} \left(\frac{250 \ \sin75 \degree}{393.2}\right)\\ \theta & = 37.89^{\circ}\\ \end{align*}

Solve for the unknown angle $\phi$ .

\phi =360^{\circ} -45^{\circ} +37.89^{\circ} =353^{\circ}

The resultant force has a magnitude of 393 lb and is located 353º measured counterclockwise from the positive x-axis.

If the magnitude of the resultant force is to be 500 N, directed along the positive y-axis, determine the magnitude of force F and its direction \theta .

If \theta = 60 \degree and \textbf{F} = 450 \ \text{N}, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.