<h2>Algebraic Addition and Subtraction of Vectors</h2>
<section>
<p>In a previous lesson, you learnt about addition and subtraction of vectors in one dimension. The following worked example provides a refresher of the concepts.</p>
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<h2>Example 1: Adding Vectors Algebraically</h2>
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<h3>Question</h3>
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<p>A force of \(\text{5}\) \(\text{N}\) to the right is applied to a crate. A second force of \(\text{2}\) \(\text{N}\) to the left is also applied to the crate. Calculate algebraically the resultant of the forces applied to the crate.</p>
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<div>
<h3>Step 1: Draw a sketch</h3>
<p>A simple sketch will help us understand the problem.</p>
<img alt="e96dfa70c9d2649e71fa12efc9d013ad.png" src="https://data.ulearngo.com/assets/8d15e067-3813-44d2-9810-68166a9d4ea4"/></div>
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<h3>Step 2: Decide which method to use to calculate the resultant</h3>
<p>Remember that force is a vector. Since the forces act along a straight line (i.e. the \(x\)-direction), wecan use the algebraic technique of vector addition.</p>
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<h3>Step 3: Choose a positive direction</h3>
<p>Choose the <strong>positive</strong> direction to be to the right. This means that the <strong>negative</strong> direction is to the left.</p>
<p>Rewriting the problem using the choice of a positive direction gives us a force of \(\text{5}\) \(\text{N}\) in the positive \(x\)-direction and force of \(\text{2}\) \(\text{N}\) in the negative \(x\)-direction being applied to the crate.</p>
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<h3>Step 4: Now define our vectors algebraically</h3>

\begin{align*} \vec{F}_{1}&amp; = \text{5}\text{ N} \\ \vec{F}_{2}&amp; = -\text{2}\text{ N} \end{align*}</div>
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<h3>Step 5: Add the vectors</h3>
<p>Thus, the resultant force is:</p>

\begin{align*} \vec{F}_{1}+\vec{F}_{2}&amp; = (5)+(-2) \\ &amp; = \text{3}\text{ N} \end{align*}</div>
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<h3>Step 6: Quote the resultant</h3>
<p>Remember that in this case a positive force means to the right: \(\text{3}\) \(\text{N}\) to the right.</p>
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<p>We can now expand on this work to include vectors in two dimensions.</p>
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<h2>Example 2: Algebraic Solution in Two Dimensions</h2>
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<h3>Question</h3>
<div>
<p>A force of \(\text{40}\) \(\text{N}\) in the positive \(x\)-direction acts simultaneously (at the same time) to a force of \(\text{30}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the magnitude of the resultant force.</p>
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<div>
<h3>Step 1: Draw a rough sketch</h3>
<p>As before, the rough sketch looks as follows:</p>
<img alt="657c31a3a9b72e75fbfdc3dfaed7464b.png" src="https://data.ulearngo.com/assets/ddef807d-a4f9-47c8-af84-2b48fbb90c38"/></div>
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<h3>Step 2: Determine the length of the resultant</h3>
<p>Note that the triangle formed by the two force vectors and the resultant vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let \(R\) represent the length of the resultant vector. Then: \begin{align*} F_x^{2} + F_y^{2} &amp;= R^{2}\ \text{Pythagoras' theorem}\\ (40)^{2} + (30)^{2} &amp;= R^{2}\\ R &amp;= \text{50}\text{ N} \end{align*}</p>
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<h3>Step 3: Quote the resultant</h3>
<p>The magnitude of the resultant force is then \(\text{50}\) \(\text{N}\).</p>
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</section>
<section>
<h1 class="title">Direction</h1>
<p>For two dimensional vectors we have only covered finding the magnitude of vectors algebraically. We also need to know the direction. For vectors in one dimension this was simple. We chose a positive direction and then the resultant was either in the positive or in the negative direction. In grade 10 you learnt about the different ways to specify direction. We will now look at using trigonometry to determine the direction of the resultant vector.</p>
<p>We can use simple trigonometric identities to calculate the direction. We can calculate the direction of the resultant in the previous worked example.</p>
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<h2>Example 1: Direction of the Resultant</h2>
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<h3>Question</h3>
<div>
<p>A force of \(\text{40}\) \(\text{N}\) in the positive \(x\)-direction acts simultaneously (at the same time) to a force of \(\text{30}\) \(\text{N}\) in the positive \(y\)-direction. Calculate the magnitude of the resultant force.</p>
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<div>
<h3>Step 1: Magnitude</h3>
<p>We determined the magnitude of the resultant vector in the previous worked example to be \(\text{50}\) \(\text{N}\). The sketch of the situation is:</p>
<img alt="9f576aef35b9271f3509a098d88e8486.png" src="https://data.ulearngo.com/assets/766681ee-9090-4dc4-a2d4-6ed1cc012879"/></div>
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<h3>Step 2: Determine the direction of the resultant</h3>
<p>To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive \(x\)-axis, by using simple trigonometry: \begin{align*} \tan\alpha &amp;= \frac{\text{opposite side}}{\text{adjacent side}} \\ \tan\alpha &amp;= \frac{\text{30}}{\text{40}} \\ \alpha &amp;= \tan^{-1}(\text{0,75}) \\ \alpha &amp;= \text{36,87}\text{°} \end{align*}</p>
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<h3>Step 3: Quote the resultant</h3>
<p>The resultant force is then \(\text{50}\) \(\text{N}\) at \(\text{36.9}\)\(\text{°}\) to the positive \(x\)-axis.</p>
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</section>

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Algebraic Methods Continued

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Discover the principles and operations of scalars and vectors such as graphical and algebraic techniques of vector addition, finding magnitude with Pythagoras theorem, and resultant force on a submarine.


Algebraic Methods Continued

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