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HW #4-solutions

# HW #4-solutions - rathbun(rar2954 HW#4 Antoniewicz(57380...

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rathbun (rar2954) – HW #4 – Antoniewicz – (57380) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9 . 8 m / s 2 . 2 kg 9 kg 6 kg F F r If F = 27 N and F r = 6 N, what is the magnitude of the force exerted on the block with mass 9 kg by the block with mass 6 kg? Correct answer: 13 . 4118 N. Explanation: m 1 m 2 m 3 F F r Given : F = +27 N ˆ ı , F r = - 6 N ˆ ı , m 1 = 2 kg , m 2 = 9 kg , m 3 = 6 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 F 21 F m 2 F 32 F 12 m 3 F r F 23 Let F , F r , F 32 represent the force exerted on the system from the right, from the left, and the force exerted on m 2 by m 3 , respec- tively. Note: F 21 = - F 12 and F 32 = - F 23 , and F 21 = F 12 and F 32 = F 23 , where F F is the magnitude of F . The sum of the forces acting on each block separately are m 1 a = + F - F 21 = + F - F 12 (1) m 2 a = + F 12 - F 32 = + F 12 - F 23 (2) m 3 a = + F 23 - F r (3) To find the acceleration we can treat the three masses as a single object or add the forces acting on each component of the sys- tem, Eqs. 1, 2, and 3. F - F r = ( m 1 + m 2 + m 3 ) a Solving for a, we have a = F - F r m 1 + m 2 + m 3 (4) = 27 N - 6 N 2 kg + 9 kg + 6 kg = 1 . 23529 m / s 2 . We can solve for F 23 using Eq. 3 and sub- stituting a from Eq. 4. The result is F 23 = m 3 a + F r (3) = m 3 ( F - F r ) m 1 + m 2 + m 3 + F r ( m 1 + m 2 + m 3 ) m 1 + m 2 + m 3 = m 3 F - ( m 1 + m 2 ) F r m 1 + m 2 + m 3 (5) = (6 kg) (27 N) 2 kg + 9 kg + 6 kg + (2 kg + 9 kg) (6 N) 2 kg + 9 kg + 6 kg = 13 . 4118 N . Alternative Solution: Using Eq. 3 and the numerical result of Eq. 4, we have F 23 = m 3 a + F r (3) = (6 kg) (1 . 23529 m / s 2 ) + 6 N = 13 . 4118 N . 002 (part 1 of 3) 4.0 points

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rathbun (rar2954) – HW #4 – Antoniewicz – (57380) 2 Two forces, 440 N at 8 and 378 N at 25 are applied to a car in an e ff ort to accelerate it. 3543 kg 440 N 8 378 N 25 What is the magnitude of the resultant of these two forces? Correct answer: 784 . 512 N. Explanation: Given : m = 3543 kg , F 1 = 440 N , θ 1 = 8 , F 2 = 378 N , and θ 2 = - 25 . Consider the side forces: F 1 ,y = F 1 sin θ 1 F 2 ,y = F 2 sin θ 2 F y,net = (440 N) sin 8 + (378 N) sin( - 25 ) = - 98 . 5135 N Consider the forward forces: F 1 ,x = F 1 cos θ 1 F 2 ,x = F 2 cos θ 2 F x,net = (440 N) cos 8 + (378 N) cos( - 25 ) = 778 . 302 N Thus the net force is F net = (778 . 302 N) 2 + ( - 98 . 5135 N) 2 = 784 . 512 N . 003 (part 2 of 3) 3.0 points Find the direction of the resultant force (in re- lation to forward, with counterclockwise con- sidered positive). Answer in degrees from the positive x -axis, with counter-clockwise positive, within the limits of - 180 to 180 . Correct answer: - 7 . 21385 . Explanation: tan θ = F y,net F y,net θ = tan - 1 F y,net F x,net = tan - 1 - 98 . 5135 N 778 . 302 N = - 7 . 21385 , which is 7 . 21385 to the right of forward. 004 (part 3 of 3) 3.0 points If the car has a mass of 3543 kg, what magni- tude of acceleration does it have?
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HW #4-solutions - rathbun(rar2954 HW#4 Antoniewicz(57380...

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