9-17-08LectPHY2048

9-17-08LectPHY2048 - Last time: v F2 F1 Note that Fnet = F...

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xnet x x FF m a = = net m a == GG G Because is a vector equation it can be resolved into 3 equations, one for each coordinate axis. ynet y y m a = = znet z z m a = = x: y: z: 1 F G 2 F G v G Note that says nothing about any constant velocity the object might have. net m a G G G net F0 = G says that the acceleration of the object is zero, but implies nothing about its velocity. Each is true independent of what happens along the other coordinate directions. Last time:
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Free body diagram Arlo Chad Bubba Example - Three way tug of war (stalemate ) over the tire. AA x A y ˆˆ FF i F j =+ G BB x B y i F j G CC x C y i F j G Net A B C Ax Bx Cx Ay By Cy F F ( F F F ) i ( F F F ) j ma m(0) 0 += + + + + + = = = GG G G G a0 = G Which means that, Ax Bx Cx FFF0 ++= Ay By Cy + & Given the magnitudes F A and F C find φ , and F B . = G since
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Ax Cx FF0 += Ay By Cy FFF0 + For the direction of given, so, B F G Bx F0 = & Now, o Ax A FF c o s 4 7 =− The magnitudes F A and F C are known and we’re asked to find φ , and F B . while Cx C c o s = φ So the x equation gives o1 o A AC C F F cos47 F cos 0 cos cos47 F ⎡⎤ −+ φ = φ = ⎢⎥ ⎣⎦ Also, o Ay A s i n 4 7 = Cy C s i n = φ , and By B = − So from the y equation, oo AB C B F sin47 F F sin 0 F F sin φ = = + φ Done.
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rope rope The elevator and the hanging mass. m Fm a = mg F T F T mg ma =+ =− = Tm a m gm ( a g ) = += + So: ( a g ) = + 2 , g 9.8 m /s = m elevator = 1 kg Draw the free body diagram m T = 1 kg ˆ j g g = − What’s the tension, T, in the rope?
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Tm ( a g ) = + 2 , g 9.8 m /s = Suppose the elevator is stationary. Then a = 0 , so, 2 2 kg m T (1 kg)(9.8 m /s ) 9.8 9.8 N s == = 2 g Fm g ( 1 k g ) ( 9 . 8 m / s ) 9 . 8 N = −= = m T F g = 1 kg ˆ j The tension must supply a force equal but opposite in direction (up) to the force due to gravity (down). Otherwise the mass would accelerate. ( a g )m ( 0 g g =+ = recall (positives so up)
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Tm ( a g ) =+ m T F g = 1 kg ˆ j The solution Is the general case where a need not be zero. Suppose that the elevator has a constant velocity downward, (Recall our earlier statement that F net = 0 says nothing about any constant velocity the object may have) v = – 2 m/s Since a is still zero, T remains mg.
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Suppose now that the elevator has an initial velocity v = – 2 m/s downward but then decelerates uniformly to a stop in 2 seconds. v G m T F g elevator rope = 1 kg ˆ j What is the deceleration and the tension during that time? 22 T( 1 k g)(1.0 m /s 9.8 m /s ) 10.8 N =+ = The tension must be larger now since in addition to balancing the force of gravity, it must also supply the force to decelerate the mass.
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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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9-17-08LectPHY2048 - Last time: v F2 F1 Note that Fnet = F...

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