HW6-7Sol

# HW6-7Sol - Oregon State University PH 201 Fall 2008 HW6-7 Solutions Page 1 x-analysis—situation(a Σ F x = ma x F T2 T x F T1 T x – F max s =

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Unformatted text preview: Oregon State University PH 201 Fall 2008 HW6-7 Solutions Page 1 x-analysis—situation (a) Σ F x = ma x F T2 T x F T1 T x – F max s = ma x F T2 T cos θ 2 F T1 cos θ 1 – μ s F N = y-analysis—situation (a) Σ F y = ma y F N + F T1y + F T2 T2 T y – mg = ma y F N + F T1 sin θ 1 + F T2 sin θ 2 – mg = FBD of m—situation (a) coord. axes m F T1 θ 1 HW PROBLEM 1: The block ( m = 30 kg) is on a level surface with friction coefF cients μ s = 0.750 and μ k = 0.500. k If F T1 = 100 N, θ 1 = 40.0°, and θ 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 θ 2 x-analysis—situation (b) Σ F x = ma x F T2 T x F T1 T x – F k = ma x F T2 T cos θ 2 F T1 cos θ 1 – μ k F N = y-analysis—situation (b) Σ F y = ma y F N + F T1y + F T2 T2 T y – mg = ma y F N + F T1 sin θ 1 + F T2 sin θ 2 – mg = y x mg F N F max s θ 2 θ 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 θ 1 – F T2 T sin θ 2 Substitute: F T2 cos θ 2 F T1 cos θ 1 – μ s ( mg – F T1 sin θ 1 – F T2 T sin θ 2 ) = 0 Collect terms: F T2 cos θ 2 + μ s F T2 T sin θ 2 = F T1 cos θ 1 + μ s mg – μ s F T1 sin θ 1 Solve: F T2 = ( F T1 cos θ 1 + μ s mg – μ s F T1 sin θ 1 )/(cos θ 2 + μ s sin θ 2 ) = [(100)(cos40) + (.75)(30)(9.80) – (.75)(100)(sin40)]/[cos25 + (.75)(sin25)] = 203 N ¡or (b): F T2 = ( F T1 cos θ 1 + μ k mg – μ k F T1 sin θ 1 )/(cos θ 2 + μ k sin θ 2 ) = [(100)(cos40) + (.50)(30)(9.80) – (.50)(100)(sin40)]/[cos25 + (.50)(sin25)] = 171 N 2 pts. 1 pt. 1 pt. 1 pt. 1 pt. 1 pt. 1 pt. 1 pt. 1 pt. Oregon State University PH 201 Fall 2008 HW6-7 Solutions Page 2 x-analysis Σ F x = ma x 0 = ma x 0 = y-analysis Σ 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 ﬂ oor. From the moment the man releases the coin, how long does it take to hit the ﬂ 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 ﬂ oor, the man, and the coin in his hand—they all move as one—until the moment when he releases the coin. After tha t moment, the coinʼs acceleration is – g . But the elevator ﬂ oorʼs 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 ﬂ oor as the both accelerate downward—at 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 2008 HW6-7 Solutions Page 1 x-analysis—situation(a Σ F x = ma x F T2 T x F T1 T x – F max s =

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