Lecture_8

Lecture_8 - 4.12 Non-equilibrium Conditions If a system is...

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4.12 Non-equilibrium Conditions If a system is in non-equilibrium, then it must be accelerating. v v By Newton’s 2 nd Law then: = a m F 2 dimensions then: a In 2 dimensions then: x x ma F = y y ma F All the problems we will do can be classified as either an equilibrium or a non-equilibrium problem. If it’s an equilibrium problem, then the sum of the forces in both the x - and y - direction will be zero. If it’s a non-equilibrium problem, then the sum of the forces in the x -direction will be ma x and the sum of the forces in the y -direction will be ma y .
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Example : The steel I-beam in the drawing has a weight of 8.0 kN and is being lifted upward at a constant velocity. What is the tension in each cable attached to its ends. Is this an equilibrium or non-equilibrium problem? It is an equilibrium problem, since the beam is tl t i T h f not accelerating. Therefore: = 0 x F = 0 y F Choose a coordinate system. y x Draw the FBD. Now break the vectors down into their components. W is along y , so it’s done, but T needs to be broken T T down. T y T y = = 0 x x x T T F T x T x 0 = + = W T T F y y y W T y = 2 From the gure: o y T T 70 sin = 70 o 70 o W figure: W T o = 70 sin 2 o W T 70 sin 2 = N 4260 70 sin 2 N 8000 = = o
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Example : Ignore all frictional effects and assume the rope and pulley are massless. What is the acceleration of the two blocks and the tension in the rope? hen the 185 block is hung on the rope this system W 1 When the 185-N block is hung on the rope, this system will begin to accelerate, since there is no friction. Therefore, this is a non-equilibrium problem. + y Thus, the non-equilibrium conditions apply: = x x ma F a W 2 = y y ma F Choose coordinate axes. + x Draw the FBD for each mass. ccelerates in the irection and W 1 W 2 W 1 accelerates in the x -direction, and W 2 accelerates in the - y -direction. Thus, the following must be true: F N1 T T W 1 = x x ma F = 0 x F W 2 W 1 W 2 = 0 y F = y y ma F
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W 1 = x x a m F 1 = = x x a m T F 1 = 0 y F = = = 1 1 1 1 0 W F W F F N N y = 0 x F W 2 Must include a minus sign here, since W 2 is accelerating in – y -direction!
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Lecture_8 - 4.12 Non-equilibrium Conditions If a system is...

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