Lecture_6

Lecture_6 - P123Lecture68October2010...

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P123 Lecture 6 8 October 2010 Applying Newton’s Laws of Motion REVIEW OF LAST WEEKS LECTURE F R Forces: Add vectorially R F F 2 1 x i x F R 2 F 1 Equilibrium: i i F F F R ... 2 1 i y i y i F R i F 0 3 F F Newton’s 1 st Law: i 0 Then 0 If a F i 2 F 1 Newton’s 2 nd Law: i ) constant ( v a m F z y x i ma F y , , Newton’s 3 rd Law: A on B B on A F F i i i z x , , 2 (action-reaction pair)
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NEWTON’S 1 ST LAW: const. v o F i Important Case: = Static Equilibrium i o v EXAMPLE : A light fixture that weights 20 N is suspended from the kitchen ceiling by a single chain. (massless) • What is the tension in the chain? T - Draw a free body w = mg diagram for fixture: Equilibrium => ii i y x F F i 0 ; 0 0 i x i F N 20 0 mg T mg T F i y i 3 Must have a chain that can support a tension of 20 N.
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EXAMPLE : Suppose I wish to hang the (20 N) light fixture with two cables at 45 o from vertical. What is minimum tension each cable must withstand before T (45 o breaking? Free body diagram: 45 o 45 o T 1 T 2 T 2 cos(45 o ) 1 cos(45 ) 45 o 45 o mg T 2 sin(45 o ) T 1 sin(45 o ) i x T T T T T F i 1 2 1 2 0 ) 45 sin( ) 45 sin( ) 45
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This note was uploaded on 03/14/2011 for the course P 123 taught by Professor Stevens during the Spring '11 term at Rust.

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Lecture_6 - P123Lecture68October2010...

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