This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hiarker (srh959) 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 Determine the horizontal force F on the block which causes it to move upward with a constant velocity. 1. F = m g cot 2. F = m g sin 3. F = m g sin 4. F = m g tan correct 5. F = m g cos 6. F = m g cos 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 . 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 , F net y = N cos W. Since the block does not accelerate, both com ponents must vanish, hence N = W cos = mg cos and F = N sin = mg sin cos = mg tan . 002 (part 1 of 2) 10.0 points The 7 . 1 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 with the horizontal. The 5 . 9 N force acts from above on the right at an angle of 51 with the horizontal. The force 7 . 1 N acts straight down. F 5 . 9 N 7 . 1 N 5 1 What is the magnitude of the force F ? Correct answer: 4 . 4845 N. Explanation: Standard angular measurements are from the positive xaxis in a counterclockwise di rection. Let : F 1 = F , F 2 = 5 . 9 N , 2 = 51 , and F 3 = 7 . 1 N . Consider the free body diagram. The green vectors are the components of the slanted forces. F 1 F 2 F 3 2 hiarker (srh959) 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 2 = (5 . 9 N) cos 51 = 3 . 71299 N and summationdisplay F y = F 1 y + F 2 y + F 3 y = 0 F 1 y = F 2 sin 2 F 3 = (5 . 9 N) sin51 ( 7 . 1 N) = 2 . 51484 N...
View
Full
Document
 Spring '09
 NUI
 Physics, Friction, Work

Click to edit the document details