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	<title>Static Equilibrium EX 3 - Revision history</title>
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		<title>imported&gt;Patrick: Created page with &#039;We will consider the board and the diver as one object. Then the forces exerted on this system are the gravitational force on the diver (the magnitude is 60 N), &lt;math&gt;F_G&lt;/math&gt; …&#039;</title>
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		<updated>2011-07-20T20:06:13Z</updated>

		<summary type="html">&lt;p&gt;Created page with &amp;#039;We will consider the board and the diver as one object. Then the forces exerted on this system are the gravitational force on the diver (the magnitude is 60 N), &amp;lt;math&amp;gt;F_G&amp;lt;/math&amp;gt; …&amp;#039;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;We will consider the board and the diver as one object. Then the forces exerted on this system are the gravitational force on the diver (the magnitude is 60 N), &amp;lt;math&amp;gt;F_G&amp;lt;/math&amp;gt; and the forces exerted by the two supports, &amp;lt;math&amp;gt;F_1&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;F_2&amp;lt;/math&amp;gt;. The force &amp;lt;math&amp;gt;F_1&amp;lt;/math&amp;gt; acts on the system at the pivot, the force &amp;lt;math&amp;gt;F_2&amp;lt;/math&amp;gt; acts at a distance 0.5 m from the pivot and the gravitational force acts at the distance of 3 m from the pivot. The diving board is in equilibrium and therefore&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\Sigma F_y = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;math&amp;gt;\Sigma \tau = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The diagram below shows the free body diagram and it also shows the force vectors tail-to-tail with the lever arm vectors.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
[[image:Helena_Stat_Equil_Ex_3_SolnA.png|TOP]]&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We will compute the torques. Note that the torque exerted by the force &amp;lt;math&amp;gt;F_1&amp;lt;/math&amp;gt; is zero:&lt;br /&gt;
&lt;br /&gt;
{| border=&amp;quot;1&amp;quot; cellpadding=&amp;quot;2&amp;quot; style=&amp;quot;text-align:center&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
! Forces&lt;br /&gt;
! Direction&lt;br /&gt;
! Torque&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;F_G&amp;lt;/math&amp;gt;&lt;br /&gt;
| CW&lt;br /&gt;
| &amp;lt;math&amp;gt;\tau = -(3)(600)\sin{90^{\circ}} = -1800 Nm&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;F_2&amp;lt;/math&amp;gt;&lt;br /&gt;
| CCW&lt;br /&gt;
| &amp;lt;math&amp;gt;\tau = (0.5) F_2 \sin{90^{\circ}} = 0.5 F_2&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;\Sigma \tau&amp;lt;/math&amp;gt;&lt;br /&gt;
| &lt;br /&gt;
| 0&lt;br /&gt;
|}&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
Solving the equation&lt;br /&gt;
&lt;br /&gt;
 &amp;lt;math&amp;gt;F_2 = {1800 \over 0.5} = 3600 N&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
The net force must be zero:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;F_1 + F_2 - 600 = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;F_1 + 3600 - 600 = 0&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
 &amp;lt;math&amp;gt;F_1 = -3000 N&amp;lt;/math&amp;gt;&lt;br /&gt;
&amp;lt;br&amp;gt;&lt;br /&gt;
The negative sign indicates that the force &amp;lt;math&amp;gt;F_1&amp;lt;/math&amp;gt; points downward. It means that the diving board has to be &amp;quot;clamped&amp;quot; to the supporting pilon. The clamps that exert the downward force on the diving board.&lt;/div&gt;</summary>
		<author><name>imported&gt;Patrick</name></author>
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