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Unformatted text preview: sanchez (ds28677) – homework 09 – Turner – (58220) 1 This printout should have 13 questions. Multiplechoice 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 xaxis in a counterclockwise 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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 Spring '08
 Turner
 Physics, Force, Friction, Work, kg, Sanchez

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