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Homework 9

# Homework 9 - sanchez(ds28677 homework 09 Turner(58220 This...

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sanchez (ds28677) – homework 09 – Turner – (58220) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A block is pushed upward along a frictionless inclined plane. m F θ Determine the horizontal force F on the block which causes it to move upward with a constant velocity. 1. F = m g cos θ 2. F = m g cot θ 3. F = m g cos θ 4. F = m g tan θ correct 5. F = m g sin θ 6. F = m g sin θ Explanation: The block is subject to three forces: its weight vector W , the normal force vector N of the surface, and the external force vector F which pushes it to the right. The external force acts in the hor- izontal direction, the weight acts vertically, and the normal force acts in the direction nor- mal (perpendicular) to the surface — leftward from vertically up by angle θ . Let the x axis point horizontally to the left while the y axix points vertically up. In these coordinates, the net force on the block has components F net x = F N sin θ, F net y = N cos θ W. Since the block does not accelerate, both com- ponents must vanish, hence N = W cos θ = mg cos θ and F = N sin θ = mg sin θ cos θ = mg tan θ. 002 (part 1 of 2) 10.0 points The 6 . 9 N weight is in equilibrium under the influence of the three forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 6 N force acts from above on the right at an angle of 57 with the horizontal. The force 6 . 9 N acts straight down. F 6 N 6 . 9 N 57 α What is the magnitude of the force F ? Correct answer: 3 . 76405 N. Explanation: Standard angular measurements are from the positive x -axis in a counter-clockwise di- rection. Let : F 1 = F , F 2 = 6 N , α 2 = 57 , and F 3 = 6 . 9 N . Consider the free body diagram. The green vectors are the components of the slanted forces. F 1 F 2 F 3 α 2 α

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sanchez (ds28677) – homework 09 – Turner – (58220) 2 The weight is is equilibrium, so summationdisplay F x = F 1 x + F 2 x + F 3 x = 0 F 1 x = F 2 cos α 2 0 = (6 N) cos 57 = 3 . 26783 N and summationdisplay F y = F 1 y + F 2 y + F 3 y = 0 F 1 y = F 2 sin α 2 F 3 = (6 N) sin 57 ( 6 . 9 N) = 1 . 86798 N , and F 1 = radicalBig F 2 1 x + F 2 1 y = radicalBig ( 3 . 26783 N) 2 + (1 . 86798 N)
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