Homework 9 - sanchez(ds28677 – homework 09 – Turner...

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Unformatted text preview: 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 5 7 ◦ 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 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 − = − (6 N) cos57 ◦ = − 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) sin57 ◦ − ( − 6 . 9 N) = 1 . 86798 N , and F 1...
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Homework 9 - sanchez(ds28677 – homework 09 – Turner...

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