7 - roofner(bar784 Homework 7 Weathers(17104 This print-out...

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roofner (bar784) – Homework 7 – Weathers – (17104) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A force of 34 . 3 N pushes and pulls to blocks as shown in the figure below. The vertical contact surfaces between the two blocks are frictionless. The contact between the blocks and the horizontal surface has a coefficient of friction of 0 . 17. The acceleration of gravity is 9 . 8 m / s 2 . 4 . 9 kg 9 . 8 kg F F 50 μ k = 0 . 17 a What is the magnitude a of the acceleration of the blocks? Correct answer: 2 . 47103 m / s 2 . Explanation: Given : M = 4 . 9 kg , 2 M = 9 . 8 kg , θ = 50 , and μ k = 0 . 17 . Basic Concept: vector F net = mvectora Solution: The frictional forces are f M = μ k N M = μ k M g. f 2 M = μ k N 2 M = μ k (2 M g F sin θ ) . F net = F + F cos θ f M f 2 M = 3 M a Therefore a = F + F cos θ f M f 2 M 3 M = F (1 + cos θ + μ k sin θ ) 3 M 3 μ k M g 3 M = (34 . 3 N)(1 + cos 50 + 0 . 17 sin 50 ) 3 (4 . 9 kg) 3 (0 . 17) (4 . 9 kg) (9 . 8 m / s 2 ) 3 (4 . 9 kg) = 2 . 47103 m / s 2 . 002 10.0 points Three blocks in contact with each other are pushed across a rough horizontal surface by a 72 N force as shown. The acceleration of gravity is 9 . 8 m / s 2 . 2 kg 5 . 2 kg 2 . 5 kg F μ = 0 . 19 If the coefficient of kinetic friction between each of the blocks and the surface is 0 . 19, find the magnitude of the force exerted on the 5 . 2 kg block by the 2 . 5 kg block. Correct answer: 18 . 5567 N. Explanation: M 1 M 2 M 3 F μ = 0 . 19 Given : F = 72 N , μ = 0 . 19 , M 1 = 2 kg , M 2 = 5 . 2 kg , M 3 = 2 . 5 kg , and g = 9 . 8 m / s 2 . Note: F is acting on the combined mass of the three blocks, resulting in a common acceleration after accounting for friction. m 1 + m 2 + m 3 N g F f μ ( m 1 + m 2 + m 3 ) g If we treat the blocks as one object of mass M 1 + M 2 + M 3 , the frictional force opposing the applied force will be f μ = μ ( M 1 + M 2 + M 3 ) g

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roofner (bar784) – Homework 7 – Weathers – (17104) 2 and the net force equals M total a is F f μ = ( M 1 + M 2 + M 3 ) a . Since M 1 + M 2 + M 3 = 2 kg + 5 . 2 kg + 2 . 5 kg = 9 . 7 kg , a = F μ ( M 1 + M 2 + M 3 ) g M 1 + M 2 + M 3 = 72 N 9 . 7 kg (0 . 19) (9 . 7 kg) (9 . 8 m / s 2 ) 9 . 7 kg = 5 . 56068 m / s 2 , Now that we know the acceleration, the force exerted on M 3 by M 2 is F 3 = M 3 a + μ M 3 g = (2 . 5 kg) (5 . 56068 m / s 2 ) + (0 . 19) (2 . 5 kg) (9 . 8 m / s 2 ) = 18 . 5567 N . According to Newton’s third law, this is also the force exerted on M 2 by M 3 . Using the free-body diagram for the third block, we have F 32 μ M 3 g = M 3 a F 32 = M 3 bracketleftbigg F M 1 + M 2 + M 3 μ g + μ g bracketrightbigg = bracketleftbigg M 3 M 1 + M 2 + M 3 bracketrightbigg F (1) = bracketleftbigg (2 . 5 kg) (2 kg) + (5 . 2 kg) + (2 . 5 kg) bracketrightbigg (72 N) = bracketleftbigg 2 . 5 kg 9 . 7 kg bracketrightbigg (72 N) = 18 . 5567 N . Using the free-body diagram for the first and second blocks, we have F F 23 μ [ M 1 + M 2 ] g = [ M 1 + M 2 ] a F 23 = [ M 1 + M 2 ] bracketleftbigg F M 1 + M 2 + M 3 μ g + μ g bracketrightbigg + F = bracketleftbigg M 1 + M 2 M 1 + M 2 + M 3 + 1 bracketrightbigg F = bracketleftbigg M 3 M 1 + M 2 + M 3 bracketrightbigg F (1 ) which is the same as Eq. 1 above, since
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