Section12 - Physics 204A Class Notes Section 12 Equilibrium...

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Physics 204A Class Notes 11-1 Section 12 – Equilibrium Now we have a fairly complete answer to our question, “what do objects do and why do they do it?” We will spend the rest of the semester applying our knowledge to a variety of new situations. In this chapter, we will focus our attention on objects that are at rest and remain at rest. Such objects are said to be in “static equilibrium.” This is highly desired for buildings, bridges and many other structures. An object in static equilibrium has no acceleration and no angular acceleration. According to the Second Laws for Translation and Rotation, ! r F = m r a " ! r F = 0 and ! r " = m r # $ ! r " = 0 . That is, the net force on the object and the net torque on the object must be zero. The rest of this section consists of example problems on increasing complexity. Example 12.1: A 5.00kg piñata is hung from the middle of a 4.00m long rope. Find the tension in the rope if it sags 20.0cm. Given: m = 5.00kg, l = 4.00m, and s = 20.0cm Find: F t1 = ? and F t2 = ? The angle that the tension makes with the horizontal can be found using the definition of the sine, sin " = s l 2 = 2 s l # = arcsin 2 s l $ % & ( ) = arcsin 2 0.2 ( ) 4 $ % & ( ) # = 5.74˚ Applying the Second Law to the point where the rope connects to the piñata, " F x = ma x # F t 2 cos $ % F t 1 cos = 0 # F t 2 = F t 1 = F t Since the piñata is in the middle of the rope, both angles are equal and the tension in both sides must be the same. Looking at the y-direction, ! F y = ma y " F t sin # + F t sin # $ mg = 0 " 2F t sin # = mg " F t = mg 2sin # Putting in the numbers, F t = (5.00)(9.80) 2sin 5.74˚ ! F t = 245N . This problem did not require the use of torques. Now, let’s look at a problem that involves some torques. x y F g F t1 F t2 s θ θ
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Physics 204A Class Notes 11-2 Example 12.2: A 20.0kg board 4.00m long is supported at each end. A 70.0kg person stands 1.00m from the left side. Find the force exerted by each support. Given: m = 20.0kg, M = 70.0kg, x 1 = 1.00m, x 2 = 2.00m, and x 3 = 4.00m Find: F L = ? and F R = ? Note that choosing the origin where one of the unknown forces acts simplifies the algebra of the solution because the torque about the origin caused by this unknown force will be zero. So this unknown won’t appear in the equation produced by the Second Law of Rotation. Applying the Second Law for Rotation to the board about the origin using the convention that counterclockwise torques are positive, !" o = I # $ (0)F L % x 1 Mg % x 2 mg + x 3 F R = 0 .
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This note was uploaded on 03/11/2012 for the course PHYSICS 204A taught by Professor Kagan during the Fall '09 term at CSU Chico.

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Section12 - Physics 204A Class Notes Section 12 Equilibrium...

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