HW6-7Sol - Oregon State University PH 201 Fall 2009 HW6-7...

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Unformatted text preview: Oregon State University PH 201 Fall 2009 HW6-7 Solutions Page 1 x-analysissituation (a) F x = ma x F T2 T x F T1 T x F max s = ma x F T2 T cos 2 F T1 cos 1 s F N = y-analysissituation (a) F y = ma y F N + F T1y + F T2 T2 T y mg = ma y F N + F T1 sin 1 + F T2 sin 2 mg = FBD of msituation (a) coord. axes m F T1 1 HW PROBLEM 1: The block ( m = 30 kg) is on a level surface with friction coefF cients s = 0.750 and k = 0.500. k If F T1 = 100 N, 1 = 40.0, and 2 = 25.0, F nd F T2 T so that: (a) the block is still at rest but just about to start sliding to the right. (b) the block slides to the right at constant velocity. F T2 2 x-analysissituation (b) F x = ma x F T2 T x F T1 T x F k = ma x F T2 T cos 2 F T1 cos 1 k F N = y-analysissituation (b) F y = ma y F N + F T1y + F T2 T2 T y mg = ma y F N + F T1 sin 1 + F T2 sin 2 mg = y x mg F N F max s 2 1 F T2 F T1 F max s becomes F k in situation (b). These are two equations in two unknowns ( F N and F T2 T )solve simultaneously: Solve for F N : F N = mg F T1 sin 1 F T2 T sin 2 Substitute: F T2 cos 2 F T1 cos 1 s ( mg F T1 sin 1 F T2 T sin 2 ) = 0 Collect terms: F T2 cos 2 + s F T2 T sin 2 = F T1 cos 1 + s mg s F T1 sin 1 Solve: F T2 = ( F T1 cos 1 + s mg s F T1 sin 1 )/(cos 2 + s sin 2 ) = [(100)(cos40) + (.75)(30)(9.80) (.75)(100)(sin40)]/[cos25 + (.75)(sin25)] = 203 N or (b): F T2 = ( F T1 cos 1 + k mg k F T1 sin 1 )/(cos 2 + k sin 2 ) = [(100)(cos40) + (.50)(30)(9.80) (.50)(100)(sin40)]/[cos25 + (.50)(sin25)] = 171 N Oregon State University PH 201 Fall 2009 HW6-7 Solutions Page 2 x-analysis F x = ma x 0 = ma x 0 = y-analysis F y = ma y F N W = ma y F N mg = ma y HW PROBLEM 2: A 90-kg man boards an elevator and stands on an accurate bathroom scale (which just happens to be there). During his elevator ride, he observes that the scale reads a steady 700 N. Also during this ride, he happens to drop a coin from a height of 1 m above the elevator oor. From the moment the man releases the coin, how long does it take to hit the oor of the elevator? FBD of m coord. axes W F N y x (scale) elevator m (person) The x-direction has no forces and thus zero acceleration. In the y-direction, a y is easily solved for: a y = ( F N mg )/ m This is the acceleration of the elevator oor, the man, and the coin in his handthey all move as oneuntil the moment when he releases the coin. After tha t moment, the coins acceleration is g . But the elevator oors acceleration continues to be a y . Essentially then, the question becomes this: How long does it take the coin to catch up to the elevator oor as the both accelerate downwardat different rates?...
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This note was uploaded on 01/13/2010 for the course PH 201 taught by Professor Staff during the Fall '08 term at Oregon State.

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HW6-7Sol - Oregon State University PH 201 Fall 2009 HW6-7...

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