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	<title>Torque EX 2 - Revision history</title>
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	<updated>2026-04-20T19:35:05Z</updated>
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	<entry>
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		<title>imported&gt;Patrick at 17:06, 31 July 2011</title>
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		<updated>2011-07-31T17:06:00Z</updated>

		<summary type="html">&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;The diagram shows the gravitational force acting on the pendulum, the pivot (black dot) and the lever arm.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
[[image:Helena_Torque_Ex_2_Soln_A.png|TOP]]&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
To solve this problem we first move the two vectors &amp;lt;math&amp;gt;\vec{r}&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\vec{F}_G&amp;lt;/math&amp;gt; so that they are placed &amp;quot;tail to tail&amp;quot;, then we draw the x-axis parallel to the lever arm and we determine the angle between them:&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
[[image:Helena_Torque_Ex_2_Soln_B.png|TOP]]&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Since we drew the vector of the force in the &amp;quot;fourth quadrant&amp;quot; we conclude that the direction of the torque is clockwise and therefore the torque is negative:&lt;br /&gt;
&lt;br /&gt;
 &amp;lt;math&amp;gt;\tau = -F_G r \sin{\theta} = -(2.5)(0.6)\sin{37^{\circ}} = 0.9 Nm&amp;lt;/math&amp;gt;&lt;/div&gt;</summary>
		<author><name>imported&gt;Patrick</name></author>
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