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notes5a - nd law gives F y = N − mg = 0 or N = mg ∑ = W...

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1D Dynamics The Normal Force Consider the textbook on the table F ? F G =mg Consider Newton’s 2 nd law in y-direction: F ? mg = = ? y y ma mg F F but book is at rest. So, a y =0, gives = = = 0 ? ? mg F mg F F y New force has same magnitude as the weight, but opposite direction m
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New force is a result of the contact between the book and the table New force is called the Normal Force, n , N or F N In general it is not equal to mg , - we must usually solve for N ``Normal’’ means ``perpendicular’’ (to the surface of contact) Now, apply an additional force, F A to the book N mg F A N mg F A F y = N mg F A = 0 N = mg + F A The normal force is not mg!
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Normal Force (Revisited) Put textbook on a scale in an elevator N mg a If elevator is at rest or moving with a constant velocity up or down, a=0. Then Newton’s 2
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Unformatted text preview: nd law gives: F y = N − mg = 0 or N = mg ∑ = W N mg Demos 4 and 7 If elevator is accelerating? F y = N − mg = ma or N = mg ∑ + ma N = m ( g + a ) If a > 0, N > mg If a < 0, N < mg If a = -g, N = 0 (“weightless”) Book on an Incline (Frictionless) θ m g N x y N m g θ θ y x Using Newton’s 2 nd Law, find the normal force and the acceleration of the book As we did for 2D kinematics, break problem into x- and y-components ∑ ∑ = = y y x x ma F ma F FBD mg sin θ = ma x N − mg cos = ma y = a x = g sin N = mg cos If θ → 0°, a x = 0 and N = mg • If θ → 90°, a x = g, N = 0...
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This note was uploaded on 12/10/2011 for the course PHY 1111 taught by Professor Stencil during the Fall '11 term at UGA.

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notes5a - nd law gives F y = N − mg = 0 or N = mg ∑ = W...

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