solutions 7

# Solutions 7 - Patel Kinal – Homework 7 – Due 7:00 pm – Inst D Weathers 1 This print-out should have 10 questions Multiple-choice questions

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Unformatted text preview: Patel, Kinal – Homework 7 – Due: Sep 28 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 10 questions. Multiple-choice 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 . 5 kg is pushed forward by a horizontal force F . A block of mass m B = 1 . 3 kg is in turn pushed forward by the cart as show on the picture: 1 . 3 kg F 3 . 5 kg μ =0 . 54 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 . 54 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: 87 . 1111 N. Explanation: Consider the free-body 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 = F-N , (1) m A a Ay = X cart F y = N floor- m A g-F , (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 = 87 . 1111 N . (10) keywords: 002 (part 1 of 1) 10 points A force of 13 . 44 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 . 2. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 6 kg 3 . 2 kg F F 6 ◦ μ k = 0 . 2 a Patel, Kinal – 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: 2 . 72497 m / s 2 . Explanation: Given : M = 1 . 6 kg , 2 M = 3 . 2 kg , θ = 60 ◦ , and μ k = 0 . 2 ....
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## This note was uploaded on 10/22/2010 for the course PHYS PHYS 1710 taught by Professor Weathers during the Fall '07 term at North Texas.

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Solutions 7 - Patel Kinal – Homework 7 – Due 7:00 pm – Inst D Weathers 1 This print-out should have 10 questions Multiple-choice questions

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