{"id":647,"date":"2020-04-22T01:23:35","date_gmt":"2020-04-22T01:23:35","guid":{"rendered":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/?p=647"},"modified":"2020-06-01T19:59:36","modified_gmt":"2020-06-01T19:59:36","slug":"section-11-4-11-5","status":"publish","type":"post","link":"https:\/\/books.compclassnotes.com\/rothphys110-2e\/2020\/04\/22\/section-11-4-11-5\/","title":{"rendered":"Chapter 11: Torque"},"content":{"rendered":"\n

11.4 Problem solving<\/h2>\n\n\n\n

The process for solving problems with torque is nearly identical to the process for solving problems with force (see section 9.4<\/a>). However, instead of a free body diagram\u2014which represents the system as a single point\u2014we\u2019ll use an extended<\/em> free body diagram (E-FBD); with torque, the location where the force is applied is important. You\u2019ll see an example in the next section.<\/p>\n\n\n\n

11.5 Equilibrium revisited<\/h2>\n\n\n\n

An object is in static equilibrium<\/em> when the following conditions are true:<\/p>\n\n\n\n

  1. The net torque acting on the object is zero

    \u03c4net<\/sub> = 0<\/em><\/p><\/li>

  2. The net force acting on the object is zero

    F<\/b>net<\/sub> = 0<\/em><\/p>

    Remember that force is a vector, so both both force components (Fx<\/sub><\/em> and Fy<\/sub><\/em>) are zero.<\/p><\/li><\/ol>\n\n\n\n

    When an object is in static equilibrium, and not rotating at all, we are free to choose a convenient reference for calculating torques. This of course raises the question, “what makes a convenient<\/em> choice?”<\/p>\n\n\n\n

    Often, you won’t immediately know the magnitude of a particular force exerted on an object. If the object is in static equilibrium, you can use the point where this force acts as the reference point for torque calculations, totally bypassing the problem of not knowing anything about that force! How convenient!<\/p>\n\n\n\n

    example<\/h4>\n\n\n\n\n