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 <title>resultant force example</title>
 <name>ResultantForceExample</name>
 <created>2025-05-30 06:36:32</created>
 <modified>2026-02-14 18:44:52</modified>
 <type>Example</type>
<parent id="989">resultant force</parent>
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 <content>\section*{Resultant Forces with Vectors}

In physics, the \textbf{resultant force} is the single force that represents the combined effect of multiple forces acting on an object. Forces are vector quantities, characterized by both magnitude and direction. To find the resultant force, we add the individual force vectors using vector addition.

\subsection*{Example: Two Forces Acting on an Object}

Consider an object subjected to two forces:
\begin{itemize}
\item \(\vec{F_1} = 5 \, \text{N}\) at \(0^\circ\) (along the positive \(x\)-axis).
\item \(\vec{F_2} = 3 \, \text{N}\) at \(60^\circ\) from the positive \(x\)-axis.
\end{itemize}

We aim to find the resultant force \(\vec{F_R} = \vec{F_1} + \vec{F_2}\).

\subsubsection*{Step 1: Resolve Forces into Components}

Express each force in terms of its \(x\)- and \(y\)-components:
\begin{itemize}
\item For \(\vec{F_1}\):
\[
F_{1x} = 5 \cos(0^\circ) = 5 \, \text{N}, \quad F_{1y} = 5 \sin(0^\circ) = 0 \, \text{N}.
\]
So, \(\vec{F_1} = (5, 0) \, \text{N}\).
\item For \(\vec{F_2}\):
\[
F_{2x} = 3 \cos(60^\circ) = 3 \cdot 0.5 = 1.5 \, \text{N}, \quad F_{2y} = 3 \sin(60^\circ) = 3 \cdot \frac{\sqrt{3}}{2} \approx 2.598 \, \text{N}.
\]
So, \(\vec{F_2} \approx (1.5, 2.598) \, \text{N}\).
\end{itemize}

\subsubsection*{Step 2: Sum the Components}

The components of the resultant force \(\vec{F_R} = (F_{Rx}, F_{Ry})\) are:
\[
F_{Rx} = F_{1x} + F_{2x} = 5 + 1.5 = 6.5 \, \text{N},
\]
\[
F_{Ry} = F_{1y} + F_{2y} = 0 + 2.598 \approx 2.598 \, \text{N}.
\]
Thus, \(\vec{F_R} \approx (6.5, 2.598) \, \text{N}\).

\subsubsection*{Step 3: Calculate Magnitude and Direction}

The magnitude of \(\vec{F_R}\) is:
\[
|\vec{F_R}| = \sqrt{F_{Rx}^2 + F_{Ry}^2} = \sqrt{6.5^2 + 2.598^2} = \sqrt{42.25 + 6.75} \approx \sqrt{49} \approx 7 \, \text{N}.
\]

The direction (angle \(\theta\) relative to the positive \(x\)-axis) is:
\[
\theta = \tan^{-1}\left(\frac{F_{Ry}}{F_{Rx}}\right) = \tan^{-1}\left(\frac{2.598}{6.5}\right) \approx \tan^{-1}(0.4) \approx 21.8^\circ.
\]

Thus, the resultant force is approximately \(7 \, \text{N}\) at \(21.8^\circ\) from the positive \(x\)-axis.

\subsection*{Diagram of Forces}

The following diagram illustrates the two forces and their resultant:

\begin{figure}[h]
\centering
\resizebox{\textwidth}{!}{\includegraphics{forces.png}}
\end{figure}

This example was generated by Grok, an AI developed by xAI, on June 6, 2025.</content>
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