<section>
<h2 class="title">Components of Vectors</h2>
<p>In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called <strong>components</strong> of the original vector. The process of breaking a vector into its components is called <strong>resolving into components</strong>.</p>
<p>In practise it is most useful to resolve a vector into components which are at right angles to one another, usually horizontal and vertical. Think about all the problems we've solved so far. If we have vectors parallel to the \(x\)- and \(y\)-axes problems are straightforward to solve.</p>
<p>Any vector can be resolved into a horizontal and a vertical component. If \(\stackrel{\to }{R}\) is a vector, then the horizontal component of \(\stackrel{\to }{R}\) is \({\stackrel{\to }{R}}_{x}\) and the vertical component is \({\stackrel{\to }{R}}_{y}\).</p>
<img alt="6f53fd6df983bd86925e844f2c9ef291.png" src="https://data.ulearngo.com/assets/3d4bfe3c-a21e-4ee8-9de9-7ecb3c098101"/>
<p>When resolving into components that are parallel to the \(x\)- and \(y\)-axes we are always dealing with a right-angled triangle. This means that we can use trigonometric identities to determine the magnitudes of the components (we know the directions because they are aligned with the axes).</p>
<p>From the triangle in the diagram above we know that \begin{align*} \cos(\theta) &amp;= \frac{R_x}{R}\\ \frac{R_x}{R} &amp;= \cos(\theta)\\ R_x &amp;=R\cos(\theta) \end{align*} and \begin{align*} \sin\theta &amp;= \frac{R_y}{R}\\ \frac{R_y}{R} &amp;= \sin(\theta)\\ R_y &amp;=R\sin(\theta) \end{align*}</p>
<p>\[\boxed{R_x=R\cos(\theta)}\]</p>
<p>\[\boxed{R_y=R\sin(\theta)}\]</p>
<p><strong>Note</strong> that the angle is measured counter-clockwise from the positive \(x\)-axis.</p>
<div class="ns-school-emp">
<h2>Example 1: Resolving a Vector Into Components </h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>A force of \(\text{250}\) \(\text{N}\) acts at an angle of \(\text{30}\)\(\text{°}\) to the positive \(x\)-axis. Resolve this force into components parallel to the \(x\)- and \(y\)-axes.</p>
</div>
<div>
<h3>Step 1: Draw a rough sketch of the original vector</h3>
<img alt="b9cf2abd659a6c45ab37aa8d7ee4784a.png" src="https://data.ulearngo.com/assets/9fbf7902-6a0c-4fef-a537-4f461b414c48"/></div>
<div>
<h3>Step 2: Determine the vector components</h3>
<p>Next we resolve the force into components parallel to the axes. Since these directions are perpendicular to one another, the components form a right-angled triangle with the original force as its hypotenuse.</p>
<img alt="5703c0ebd69449e8da137a9c5b2cc827.png" src="https://data.ulearngo.com/assets/d9d92d4f-23f3-4530-a616-c5a2f6fe90e8"/>
<p>Notice how the two components acting together give the original vector astheir resultant.</p>
</div>
<div>
<h3>Step 3: Determine the magnitudes of the component vectors</h3>
<p>Now we can use trigonometry to calculate the magnitudes of thecomponents of the original displacement:</p>
<p>\begin{align*} F_y &amp;= \text{250} \sin(30°)\\ &amp;=\text{125}\text{ N} \end{align*}</p>
<p>and</p>
<p>\begin{align*} F_x &amp;= \text{250} \cos(30°)\\ &amp;=\text{216,5}\text{ N} \end{align*}</p>
<p>Remember \({F}_{x}\) and \({F}_{y}\) are the magnitudes of the components. \(\vec{F}_x\) is in the positive \(x\)-direction and \(\vec{F}_y\) is in the positive \(y\)-direction.</p>
</div>
</div>
</div>
<div class="ns-school-emp">
<h2>Example 2: Resolving a Vector Into Components</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>A force of \(\text{12.5}\) \(\text{N}\) acts at an angle of \(\text{230}\)\(\text{°}\) to the positive \(x\)-axis. Resolve this force into components parallel to the \(x\)- and \(y\)-axes.</p>
</div>
<div>
<h3>Step 1: Draw a rough sketch of the original vector</h3>
<img alt="237558d0d2f3372af9e58f9a39d3e84e.png" src="https://data.ulearngo.com/assets/3284d5a2-8520-4e38-b260-f60cd8b1820c"/></div>
<div>
<h3>Step 2: Determine the vector components</h3>
<p>Next we resolve the force into components parallel to the axes. Since these directions are perpendicular to one another, the components form a right-angled triangle with the original force as its hypotenuse.</p>
<p>Now we can use trigonometry to calculate the magnitudes of thecomponents of the original force:</p>
<p>\begin{align*} F_y &amp;= \text{12,5} \sin(\text{230}\text{°})\\ &amp;=-\text{9,58}\text{ N} \end{align*}</p>
<p>and</p>
<p>\begin{align*} F_x &amp;= \text{12,5} \cos(\text{230}\text{°})\\ &amp;=-\text{8,03}\text{ N} \end{align*}</p>
<p><strong>Notice</strong> that by using the full angle we actually get the correct signs for the components if we use the standard Cartesian coordinates. \(\vec{F}_x\) is in the negative \(x\)-direction and \(\vec{F}_y\) is in the negative \(y\)-direction.</p>
</div>
</div>
</div>
</section>
<section></section>

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Components of Vectors

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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.


Components of Vectors

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