Lect06 830 - Physics 211: Lecture 6 Todays Agenda q q Recap

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Physics 211: Lecture 6, Pg 1 Physics 211: Lecture 6 Physics 211: Lecture 6 Today’s Agenda Today’s Agenda Recap Recap Problems. ..problems. ..problems!! Problems. ..problems. ..problems!! Accelerometer Inclined plane Motion in a circle
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Physics 211: Lecture 6, Pg 2 Review Review Discussion of dynamics. Review Newton’s 3 Laws The Free Body Diagram Free Body Diagram The tools we have for making & solving problems: » Ropes & Pulleys (tension) » Hooke’s Law (springs)
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Physics 211: Lecture 6, Pg 3 Review: Pegs & Pulleys Review: Pegs & Pulleys Used to change the direction of forces An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude: The tension is the same on both sides The tension is the same on both sides for a massless rope! for a massless rope! F 1 = -T i ideal peg or pulley F 2 = T j | F 1 | = | F 2 | massless rope
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Physics 211: Lecture 6, Pg 4 Review: Springs Review: Springs Hooke’s Law: Hooke’s Law: The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position. F X = -kx Where x is the displacement from the equilibrium and k is the constant of proportionality. relaxed position F X = 0 x Spring demo
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Physics 211: Lecture 6, Pg 5 Lecture 6, Lecture 6, Act 1 Act 1 Springs A spring with spring constant 40 N/m has a relaxed length of 1 m . When the spring is stretched so that it is 1.5 m long, what force is exerted on a block attached to the end of the spring? x = 0 M k k M x = 0 x = 1.5 (a) -20 N (b) 60 N (c) - 60 N x = 1
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Physics 211: Lecture 6, Pg 6 Lecture 6, Lecture 6, Act 1 Act 1 Solution Recall Hooke’s law: F X = -kx Where x is the displacement from equilibrium . F X = - (40) ( .5) F X = - 20 N (a) -20 N (b) 60 N (c) - 60 N
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Physics 211: Lecture 6, Pg 7 Problem: Problem: Pendulum Accelerometer Pendulum Accelerometer A weight of mass m is hung from the ceiling of a car with a massless string. The car travels on a horizontal road (duh, it’s Illinois), and has an acceleration a in the x direction. The string makes an angle θ with respect to the vertical ( y ) axis. Solve for θ in terms of a and g .
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Lect06 830 - Physics 211: Lecture 6 Todays Agenda q q Recap

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