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318 lines
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HTML
318 lines
16 KiB
HTML
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<!-- #BeginEditable "Body" -->
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<h1 class="center">Force Calculations</h1>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/force-push-pull.jpg" width="240" height="229" alt="force push pull wrestlers" /></p>
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<p> </p>
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<p class="large"><a href="force.html">Force</a> is push or pull.</p>
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<p> </p>
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<p>Forces on an object are usually <b>balanced</b> (if unbalanced the object accelerates):</p>
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<div style="clear:both"></div>
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<div class="simple">
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<table style="border: 0; margin:auto;">
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<tr>
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<td><span class="center"><img src="images/forces-balanced.gif" width="135" height="206" alt="forces balanced" /><br>
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</span></td>
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<td> </td>
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<td valign="bottom"><span class="center"><img src="images/forces-unbalanced.gif" width="132" height="187" alt="forces unbalanced" /><br>
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</span></td>
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</tr>
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<tr align="center">
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<th><span class="center">Balanced </span></th>
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<td> </td>
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<th valign="bottom"><span class="center">Unbalanced </span></th>
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</tr>
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<tr align="center">
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<td><span class="center"><i>No Acceleration</i></span></td>
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<td> </td>
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<td valign="bottom"><span class="center"><i>Acceleration</i></span></td>
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</tr>
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</table>
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</div>
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<div style="clear:both"></div>
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<div class="example">
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<h3>Example: The forces at the top of this bridge tower are <b>in balance</b> (it is not accelerating):</h3>
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<p class="center"><img src="images/suspension-bridge.jpg" width="240" height="142" alt="suspension bridge" /></p>
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<p>The cables pull <b>downwards</b> equally to the left and right, and that is balanced by the tower's <b>upwards</b> push. (Does the tower push? Yes! Imagine you stand there instead of the tower.)</p>
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<p>We can <a href="../algebra/mathematical-models.html">model</a> the forces like this:</p>
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<p class="center"><img src="images/suspension-bridge-forces1.svg" alt="suspension bridge forces" /></p>
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<p>And when we put them <b>head-to-tail </b>we see they <b>close back on themselves</b>, meaning the net effect is zero:</p>
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<p class="center"><img src="images/suspension-bridge-forces2.svg" alt="suspension bridge forces" /><br>
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The forces are in balance.</p>
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</div>
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<p class="words">Forces in balance are said to be <b>in equilibrium</b>: there is also no change in motion.</p>
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<h2>Free Body Diagrams</h2>
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<p>The first step is to draw a Free Body Diagram (also called a Force Diagram)</p>
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<p class="words"><b>Free Body Diagram</b>: A sketch where a body is cut free from the world except for the forces acting on it.</p>
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<p>In the bridge example the free body diagram for the top of the tower is:</p>
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<p class="center"><img src="images/suspension-bridge-free-body.svg" alt="suspension bridge free body" /><br>
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<b>Free Body Diagram</b></p>
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<p>It helps us to think clearly about the <b>forces acting on the body</b>.</p>
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<div class="example">
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<h3>Example: Car on a Highway</h3>
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<p>What are the forces on a car cruising down the highway?</p>
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<p class="center"><img src="images/force-ex3a.gif" width="300" height="103" alt="car moving" /></p>
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<p>The engine is working hard, so why doesn't the car continue to accelerate?</p>
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<p>Because the driving force is balanced by:</p>
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<ul>
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<li>Air resistance (put simply: the air resists being pushed around),</li>
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<li>Rolling resistance, also called rolling friction (the tires resist having their shape changed)</li>
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</ul>
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<p>Like this:</p>
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<p class="center"><img src="images/force-ex3b.svg" alt="car forces: driving, weight, air, tires" /><br>
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<b>Free Body Diagram</b></p>
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<p><b>W</b> is the car's weight,</p>
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<p><b>R<sub>1</sub></b> and <b>R<sub>2</sub></b> are the rolling resistance of the tires,</p>
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<p><b>N<sub>1</sub></b> and <b>N<sub>2</sub></b> are the reaction forces (balancing out the car's weight).</p>
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<p><i>Note: steel wheels (like on trains) have less rolling resistance, but are way too slippery on the road!</i></p>
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</div>
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<h2>Calculations</h2>
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<p>Force is a <a href="../algebra/vectors.html">vector</a>. A vector has <b>magnitude</b> (size) and <b>direction</b>:</p>
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<p class="center"><img src="../algebra/images/vector-mag-dir.svg" alt="vector magnitude and direction" /></p>
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<p>We can model the forces by drawing arrows of the correct size and direction. Like this:</p>
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<div class="example">
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<h3>Example: Admiring the View</h3>
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<p>Brady stands on the edge of a balcony supported by a horizontal beam and a strut:</p>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/force-ex2a.svg" alt="force man on beam with strut at 60 degrees" /></p>
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<p> </p>
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<p>He weighs 80kg.</p>
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<p>What are the forces?</p>
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<p> </p>
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<p>Let's take the spot he is standing on and think about the forces just there:</p>
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<p class="center"><img src="images/force-ex2b.svg" alt="free body diagram man on beam with strut at 60 degrees" /></p>
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<h3>His Weight</h3>
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<p>His 80 kg mass creates a downward force due to Gravity.</p>
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<p>Force is mass times acceleration: <span class="large"><b>F</b> = m<b>a</b></span></p>
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<p>The acceleration due to gravity on Earth is 9.81 m/s<sup>2</sup>, so <span class="large"> <b>a</b> = 9.81 m/s<sup>2</sup></span></p>
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<p class="center"><b>F</b> = 80 kg × 9.81 m/s<sup>2</sup></p>
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<p class="center"><b>F</b> = 785 N</p>
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<h3>The Other Forces</h3>
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<p>The forces are balanced, so they should close back on themselves like this:</p>
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<p class="center"><img src="images/force-ex2c.svg" alt="force beam strut" /></p>
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<p class="center">We can use <a href="../algebra/trig-solving-triangles.html">trigonometry</a> to solve it.<br>
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Because it is a <b>right-angled triangle</b>, <a href="../algebra/sohcahtoa.html">SOHCAHTOA</a> will help.</p>
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<p>For the <b>Beam</b>, we know the Adjacent, we want to know the Opposite, and "TOA" tells us to use Tangent:</p>
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<p class="center larger">tan(60°) = Beam/785 N</p>
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<p class="center larger">Beam/785 N = tan(60°)</p>
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<p class="center larger">Beam = tan(60°) × 785 N</p>
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<p class="center larger">Beam = 1.732... × 785 N = 1360 N</p>
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<p>For the <b>Strut</b>, we know the Adjacent, we want to know the Hypotenuse, and "CAH" tells us to use Cosine:</p>
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<p class="center larger">cos(60°) = 785 N / Strut</p>
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<p class="center larger">Strut × cos(60°) = 785 N</p>
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<p class="center larger">Strut = 785 N / cos(60°)</p>
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<p class="center larger">Strut = 785 N / 0.5 = 1570 N</p>
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<p>Solved:</p>
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<p class="center"><img src="images/force-ex2d.svg" alt="force beam 1360 strut 1570" /></p>
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<p>Interesting how much force is on the beam and strut compared to the weight being supported!</p>
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</div>
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<h2>Torque (or Moment)</h2>
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<p>What if the beam is just stuck into the wall (called a cantilever)?</p>
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<p><img src="images/force-ex5a.svg" alt="force man cantilever" /></p>
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<p>There is no supporting strut, so what happens to the forces?</p>
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<p>The Free Body Diagram looks like this:</p>
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<p class="center"><img src="images/force-ex5b.svg" alt="force cantilever free body diagram" /></p>
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<p>The upwards force <b>R</b> balances the downwards <b>Weight</b>.</p>
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<p>With only those two forces the beam will spin like a propeller! But there is also a "turning effect" <b>M</b> called <b>Moment</b> (or <b>Torque</b>) that balances it out:</p>
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<p class="words"><b>Moment</b>: Force times the Distance at right angles.</p>
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<p>We know the <b>Weight</b> is 785 N, and we also need to know the <b>distance at right angles</b>, which in this case is 3.2 m.</p>
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<p class="center larger"><b>M</b> = 785 N x 3.2 m = <b>2512 Nm</b></p>
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<p>And that moment is what stops the beam from rotating.</p>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/moment-fishing-rod.jpg" width="180" height="235" alt="moment on fishing rod" /></p>
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<p> </p>
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<p>You can feel moment when holding onto a fishing rod.</p>
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<p>As well as holding up its weight you have to stop it from rotating downwards.</p>
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<div style="clear:both"></div>
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<h2>Friction</h2>
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<div style="clear:both"></div>
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<div class="example">
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<h3>Box on a Ramp</h3>
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<p style="float:left; margin: 0 10px 5px 0;"><img src="images/force-ex4a.gif" width="200" height="255" alt="forces box on 20 degress incline: W, f, R" /></p>
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<p>The box weighs 100 kg.</p>
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<p>The friction force is enough to keep it where it is.</p>
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<p>The reaction force R is at right angles to the ramp.</p>
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<p>The box is not accelerating, so the forces are in balance:</p>
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<p class="center"><img src="images/force-ex4b.gif" width="94" height="167" alt="force diagram: W, f, R" /></p>
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<p>The 100 kg mass creates a downward force due to Gravity:</p>
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<p class="so"><b>W</b> = 100 kg × 9.81 m/s<sup>2</sup> = 981 N</p>
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<p> </p>
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<p>We can use <span class="center"><a href="../algebra/sohcahtoa.html">SOHCAHTOA</a> to solve the triangle.</span></p>
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<p>Friction <b>f</b>:</p>
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<p class="so">sin(20°) = <b>f</b>/981 N</p>
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<p class="so"><b>f</b> = sin(20°) × 981 N = 336 N</p>
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<p>Reaction <b>N</b>:</p>
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<p class="so">cos(20°) = R/981 N</p>
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<p class="so"><b>R</b> = cos(20°) × 981 N = 922 N</p>
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<p>And we get:</p>
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<p class="center"><img src="images/force-ex4c.gif" width="193" height="162" alt="force diagram: W=981N, f=336N, R=922N" /></p>
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</div>
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<h2>Tips for Drawing Free Body Diagrams</h2>
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<ul>
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<li>Draw as simply as possible. A box is often good enough.</li>
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<li>Forces point in the <b>direction they act on the body</b></li>
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<li>straight arrows for <b>forces</b></li>
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<li>curved arrows for <b>moments</b></li>
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</ul>
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<h2>Sam and Alex Pull a Box</h2>
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<p>The calculations can sometimes be easier when we turn <b>magnitude and direction</b> into <b>x and y</b>:</p>
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<table style="border: 0; margin:auto;">
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<tr>
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<td align="center"><img src="../algebra/images/vector-polar.svg" alt="vector polar" /></td>
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<td align="center"><=></td>
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<td align="center"><img src="../algebra/images/vector-cartesian.svg" alt="vector cartesian" /></td>
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</tr>
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<tr>
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<td align="center">Vector <b>a</b> in Polar<br>
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Coordinates</td>
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<td align="center"> </td>
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<td align="center">Vector <b>a</b> in Cartesian<br>
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Coordinates</td>
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</tr>
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</table>
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<p>You can read how to convert them at <a href="../polar-cartesian-coordinates.html">Polar and Cartesian Coordinates</a>, but here is a quick summary:</p>
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<div class="simple">
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<table style="border: 0; margin:auto;">
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<tr>
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<th>From Polar Coordinates (r,<i>θ</i>)<br>
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to Cartesian Coordinates (x,y)</th>
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<td> </td>
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<th><span class="Larger">From Cartesian Coordinates (x,y)<br>
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to Polar Coordinates (r,θ)</span></th>
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</tr>
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<tr>
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<td><ul>
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<li><b>x = r</b> × <b>cos( <i>θ</i> )</b></li>
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<li><b>y = r</b> × <b>sin(<i> θ</i> )</b></li>
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</ul></td>
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<td> </td>
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<td><ul>
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<li><b>r = √ ( x<sup>2</sup> + y<sup>2 </sup>)</b></li>
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<li><b><i>θ</i> = tan<sup>-1 </sup>( y / x )</b></li>
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</ul></td>
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</tr>
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</table>
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</div>
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<p>Let's use them!</p>
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<div class="example">
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<p style="float:right; margin: 0 0 5px 10px;"><img src="../algebra/images/vector-ex1c.svg" alt="vector example: 2 people pulling box" /></p>
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<h3>Example: Pulling a Box</h3>
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<p>Sam and Alex are pulling a box <i>(viewed from above)</i>:</p>
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<ul>
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<li>Sam pulls with 200 Newtons of force at 60°</li>
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<li>Alex pulls with 120 Newtons of force at 45° as shown</li>
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</ul>
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<p>What is the combined <a href="force.html">force</a>, and its direction?</p>
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<p>Let us add the two vectors head to tail:</p>
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<p class="center"><img src="../algebra/images/vector-ex1a.gif" width="176" height="146" alt="vector example: 200 at 60, 120 at 45" /></p>
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<p>First convert from polar to Cartesian (to 2 decimals):</p>
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<p>Sam's Vector:</p>
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<ul>
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<li><b>x = r × cos( <i>θ</i> ) = 200 × cos(60°) = 200 × 0.5 = 100</b></li>
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<li><b>y = r × sin(<i> θ</i> ) = 200 × sin(60°) = 200 × 0.8660 = 173.21</b></li>
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</ul>
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<p>Alex's Vector:</p>
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<ul>
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|
<li><b>x = r × cos( <i>θ</i> ) = 120 × cos(−45°) = 120 × 0.7071 = 84.85</b></li>
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|
<li><b>y = r × sin(<i> θ</i> ) = 120 × sin(−45°) = 120 × -0.7071 = −84.85</b></li>
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</ul>
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<p>Now we have:</p>
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<p class="center"><img src="../algebra/images/vector-ex1b.gif" width="190" height="137" alt="vector example: force components" /></p>
|
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<p>Add them:</p>
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|
<p class="center larger">(100, 173.21) + (84.85, −84.85) = (184.85, 88.36)</p>
|
|
<p>That answer is valid, but let's convert back to polar as the question was in polar:</p>
|
|
<ul>
|
|
<li><b>r = √ ( x<sup>2</sup> + y<sup>2 </sup>) = √ ( 184.85<sup>2</sup> + 88.36<sup>2 </sup>)</b> = <b> 204.88</b></li>
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|
<li><b><i>θ</i> = tan<sup>-1 </sup>( y / x ) = tan<sup>-1 </sup>( 88.36 / 184.85 ) = 25.5°</b></li>
|
|
</ul>
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|
<p class="center"><span class="large">And we have this (rounded) result:</span><br>
|
|
<img src="../algebra/images/vector-ex1d.gif" width="138" height="140" alt="vector example: forces in trianle" /></p>
|
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<p class="center"><span class="large">And it looks like this for Sam and Alex:</span><br>
|
|
<img src="../algebra/images/vector-ex1e.svg" alt="vector example: combined forces" /></p>
|
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<p>They might get a better result if they were shoulder-to-shoulder!</p>
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</div>
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<p> </p>
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<div class="questions">
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<script>
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getQ(11933, 11934, 11935, 11936, 11937, 11938, 9313, 9314, 9315, 9316);
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</script> </div>
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<div class="related">
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<a href="force.html">Force</a>
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<a href="../measure/metric-acceleration.html">Acceleration</a>
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<a href="index.html">Physics Index</a>
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