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Unformatted text preview: Nguyen, Thanh – Homework 7 – Due: Sep 28 2007, 7:00 pm – Inst: D Weathers 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A cart of mass m A = 3 kg is pushed forward by a horizontal force F . A block of mass m B = 0 . 35 kg is in turn pushed forward by the cart as show on the picture: . 35 kg F 3 kg μ =0 . 76 9 . 8m / s 2 If the cart and the block accelerate forward fast enough, the friction force between the block and the cart would keep the block sus pended above the floor without falling down. Given g = 9 . 8 m / s 2 and the static friction coefficient μ s = 0 . 76 between the block and the cart; the floor is horizontal and there is no friction between the cart and the floor. Calculate the minimal force F on the cart that would keep the block from falling down. Correct answer: 43 . 1974 N. Explanation: Consider the freebody diagrams for the cart and the block: m A m B = F N N m A g N floor m B g F F where N is the normal force between the cart and the block and F is the friction force be tween them. According to the above diagrams, Newton’s Second Law says m A a Ax = X cart F x = FN , (1) m A a Ay = X cart F y = N floor m A gF , (2) m B a Bx = X block F x = + N , (3) m B a By = X block F y = + F  m B g. (4) We want both the cart and the block to move horizontally with the same acceleration a Ax = a Bx = a while the block does not fall down i.e. a By = 0; of course, the cart does not fall down either, thus a Ay = 0 as well. Sub stituting these acceleration into eqs. (1–4) we easily solve for all the forces and find N = m B × a, (5) F = m B × g, (6) F = ( m A + m B ) × a, (7) N floor = ( m A + m B ) × g. (8) Finally, me apply the static friction law F ≤ μ s N . According to eqs. (5) and (6) this requires m B g ≤ μ s × m B a = ⇒ a ≥ g μ s (9) and therefore — thanks to eq. (7) — F ≥ ( m A + m B ) g μ s = 43 . 1974 N . (10) keywords: 002 (part 1 of 1) 10 points A force of 5 . 3 N pushes and pulls to blocks as shown in the figure below. The vertical contact surfaces between the two blocks are frictionless. The contact between the blocks and the horizontal surface has a coefficient of friction of 0 . 27. The acceleration of gravity is 9 . 8 m / s 2 . 1 kg 2 kg F F 5 ◦ μ k = 0 . 27 a Nguyen, Thanh – Homework 7 – Due: Sep 28 2007, 7:00 pm – Inst: D Weathers 2 What is the magnitude a of the acceleration of the blocks? Correct answer: 0 . 621661 m / s 2 . Explanation: Given : M = 1 kg , 2 M = 2 kg , θ = 50 ◦ , and μ k = 0 . 27 . Basic Concept: ~ F net = m~a Solution: The frictional forces are f M = μ k N M = μ k M g....
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This note was uploaded on 10/24/2011 for the course PHYS 1710 taught by Professor Weathers during the Winter '08 term at North Texas.
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