11-3-08LectPHY2048

11-3-08LectPHY2048 - Things you knew & should remember...

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Things you knew & should remember (in the onslaught of new material) going into Exam 2: The fundamental definition of work is f i r r WF d r =⋅ G G G G Hence to get the work done on an object by , first consider the direction of the displacement and the direction of the force during the trajectory. If for the entire trajectory Where Fd r Fd rco s G G dr G F G φ G F G G F G 0 φ= Work done is positive G F G o 180 φ = Work done is negative r r = −Δ G F G o 90 φ = Work done is zero W0 = F G
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And a final example. A ball of mass m is swung on the end of a rope in a vertical circle of radius R . At the top of the trajectory it has just enough speed to stay on the circular path . What is it’s speed at the bottom of the trajectory? T v B v At the top, T Fm a = 2 T v Tm g m R −− = R Since the speed is just enough to stay on the trajectory T = 0 (top only), mg C a 2 T v m R −=
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T v B v T R mg C a 2 T vg R = Solving for 2 T v Now since the tension in the rope is always z to the displacement along the trajectory it does no work as the ball goes from the top to the bottom. Then by conservation of mechanical energy, mech E0 Δ= UK 0 Δ+ The ball “falls” through height 2R making Δ U = –mg(2R), so, 22 BT 11 2mgR mv 0 −+ = 2R
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T v B v T R mg C a 22 BT 11 mv 2mgR =+ v4 g R v 2 B v 4gR gR 2 B v5 g R = B g R =
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T v B v T R mg C a What is the tension at the bottom? T C a At the bottom, Fm a = 2 B v Tm g m R −= 2 B v m g R =+ Above we found, 2 B v5 g R = So, 5gR m g6 m g R =
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Structures like bridges, buildings, signposts and ladders all need to withstand the loads (i.e. forces and torques) imposed on them by their own weight as well as the stresses placed on them in their intended use. Statics Consideration of the forces and torques on structures, which forms the starting point for ensuring that the structures do their job, is called Statics . The principal requirement of statics is that the body or structure be in equilibrium . As the name statics suggests, for the most part this means that the structure is static or stationary despite the varied forces that act on its various parts.
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The requirements for such equilibrium of a body are that net F0 = G net 0 τ= G The vector sum of all external forces acting on the body equals zero. The vector sum of all external torques, measured about any point , equals zero. These are vector equations that each yield 3 separate equations, one for each coordinate direction. Balance of forces Balance of torques net,x = net,y = net,z = 0 τ = 0 τ = 0 τ =
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We will only consider examples in 2-dimensions which reduces this set to the three equations: net,x F0 = net,y = net,z 0 τ = & The prevalent external force is the force of gravity. As we’ll see in the next chapter the force of gravity acts on all points of an extended object.
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This note was uploaded on 08/25/2011 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.

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11-3-08LectPHY2048 - Things you knew & should remember...

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