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      <text>9.1.3.5 The force vectors on an aeroplane</text>
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      <text><![CDATA[<h4>The force vectors on an aeroplane in steady flight</h4><p>The forces acting on an aeroplane are shown in the figure below.<br></p>
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<img src="@@PLUGINFILE@@/9.1.3.4AeroplaneExample.svg" alt="Aeroplane flying at an angle beta to the horizon and alpha to its own horizontal.  The forces of Thrust and drag are marked as arrows pointing along the aeroplanes horizontal forward and backward respectively. The lift force is marked perpendicular to these and the aeroplane's weight acts vertically down." class="img-responsive atto_image_button_text-bottom" width="496" height="156">
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<p>The <span><strong>magnitude</strong></span> (strength) of the forces are indicated by</p> <p>\(T\): the thrust provided by the engines,</p>
<p>\(W\): the weight,</p>
<p>\(D\): the drag (acting against the direction of flight path) and </p>
<p>\(L\): the lift (taken perpendicular to the path.)</p>
<p>In a more realistic situation force vectors in <span><em>three</em></span> dimensions would need to be considered. As the plane is in <span><em>steady</em></span> flight the sum of the forces in any direction is zero. (If this were not the case, then, by Newton’s second law, the non-zero resultant force would cause the aeroplane to accelerate.)</p>
<p>So, resolving forces in the direction of the path:
 \[T\cos\alpha - D − W\sin\beta = 0\] 
Then, resolving forces perpendicular to the path:
\[T\sin\alpha + L − W\cos\beta = 0\].</p>
<hr>
<h4 class="HELM_example">Example<br></h4>
<p>If the plane has mass \(72\, 000\) tonnes, the drag is \(130\ \mathrm{kN}\), the lift is \(690\ \mathrm{kN}\) and \(\beta = 6^\circ\) find the magnitude of the thrust and the value of \(\alpha\) to maintain steady flight.</p>
<h4 class="HELM_solution">Solution</h4>
<p> From these two equations we see:\[T\cos\alpha = D + W\sin\beta = 130000 + (72000)(9.81)\sin6^\circ = 203830.54\] and \[T\sin\alpha = W\cos\beta − L = (72000)(9.81)\cos 6^\circ − 690000 = 12450.71\] hence \[\tan\alpha=\frac{T\sin\alpha}{T\cos\alpha}=\frac{12450.71}{203830.54}=0.061084\quad\rightarrow\quad \alpha=3.50^\circ\] and consequently, for the thrust:\[T = 204210\ \mathrm{N}\].</p>]]></text>
<file name="9.1.3.4AeroplaneExample.svg" path="/" encoding="base64">[encoded data, 41756 characters]</file>
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