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HW6-7Sol - HW PROBLEM 1 The block(m = 30 kg is on a level...

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Oregon State University PH 201 Fall 2009 HW6-7 Solutions Page 1 x-analysis—situation (a) Σ F x = ma x F T2 x F T1 x F max s = ma x F T2 cos θ 2 F T1 cos θ 1 μ s F N = 0 y-analysis—situation (a) Σ F y = ma y F N + F T1y + F T2 y mg = ma y F N + F T1 sin θ 1 + F T2 sin θ 2 mg = 0 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 coefficients μ s = 0.750 and μ k = 0.500. If F T1 = 100 N, θ 1 = 40.0°, and θ 2 = 25.0°, find F T2 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 x F T1 x F k = ma x F T2 cos θ 2 F T1 cos θ 1 μ k F N = 0 y-analysis—situation (b) Σ F y = ma y F N + F T1y + F T2 y mg = ma y F N + F T1 sin θ 1 + F T2 sin θ 2 mg = 0 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 )—solve simultaneously: Solve for F N : F N = mg F T1 sin θ 1 F T2 sin θ 2 Substitute: F T2 cos θ 2 F T1 cos θ 1 μ s ( mg F T1 sin θ 1 F T2 sin θ 2 ) = 0 Collect terms: F T2 cos θ 2 + μ s F T2 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 For (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
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Oregon State University PH 201 Fall 2009 HW6-7 Solutions Page 2 x-analysis Σ F x = ma x 0 = ma x 0 = 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 floor. From the moment the man releases the coin, how long does it take to hit the floor 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 floor, 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 floorʼs 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 floor as the both accelerate downward—at different rates? Object A (coin): Δ y A = v A0 ( Δ t ) + ( 1 / 2 )(– g )( Δ t ) 2 Object B (elevator floor): Δ y B = v B0 ( Δ t ) + ( 1 / 2 )( a y )( Δ t ) 2 But Δ y A = Δ y B – 1 (Itʼs –1 because Δ y downward is negative; A travels downward 1 meter more than B.) And v A0 = v B0 (When the coin is released, its velocity is the same as the elevator floorʼs.) Substituting: v A0 ( Δ t ) + ( 1 / 2 )(– g )( Δ t ) 2 = v A0 ( Δ t ) + ( 1 / 2 )( a y )( Δ t ) 2 – 1 Solve for Δ t : –( 1 / 2 )( a y
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