solution_hwk9 - homework 09 – GUPTA, SHIVANI – Due: Feb...

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Unformatted text preview: homework 09 – GUPTA, SHIVANI – Due: Feb 18 2008, 11:00 pm 1 Question 1, chap 5, sect 6. part 1 of 1 10 points The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9 . 8 m / s 2 . 1 kg 7 kg 5 kg F ℓ F r μ = 0 If F ℓ = 21 N and F r = 5 N, what is the magnitude of the force exerted on the block with mass 7 kg by the block with mass 5 kg? Correct answer: 11 . 1538 N (tolerance ± 1 %). Explanation: m 1 m 2 m 3 F ℓ F r μ = 0 Given : vector F ℓ = +21 N ˆ ı, vector F r = − 5 N ˆ ı , μ = 0 , m 1 = 1 kg , m 2 = 7 kg , m 3 = 5 kg , and g = 9 . 8 m / s 2 . Note: F is acting on the combined mass of the three blocks, resulting in a common acceleration after accounting for friction. m 1 F 21 F ℓ m 2 F 32 F 12 m 3 F r F 23 Let F ℓ , F r , F 32 represent the force exerted on the system from the right, from the left, and the force exerted on m 2 by m 3 , respec- tively. Note: vector F 21 = − vector F 12 and vector F 32 = − vector F 23 , and bardbl vector F 21 bardbl = bardbl vector F 12 bardbl and bardbl vector F 32 bardbl = bardbl vector F 23 bardbl , where bardbl vector F bardbl ≡ F is the magnitude of vector F . The sum of the forces acting on each block separately are m 1 a = + F ℓ − F 21 = + F ℓ − F 12 (1) m 2 a = + F 12 − F 32 = + F 12 − F 23 (2) m 3 a = + F 23 − F r (3) To find the acceleration we can treat the three masses as a single object or add the forces acting on each component of the sys- tem, Eqs. 1, 2, and 3. F ℓ − F r = ( m 1 + m 2 + m 3 ) a Solving for a, we have a = F ℓ − F r m 1 + m 2 + m 3 (4) = 21 N − 5 N 1 kg + 7 kg + 5 kg = 1 . 23077 m / s 2 . We can solve for F 23 using Eq. 3 and sub- stituting a from Eq. 4. The result is F 23 = m 3 a + F r (3) = m 3 ( F ℓ − F r ) m 1 + m 2 + m 3 + F r ( m 1 + m 2 + m 3 ) m 1 + m 2 + m 3 = m 3 F ℓ − ( m 1 + m 2 ) F r m 1 + m 2 + m 3 (5) = (5 kg) (21 N) 1 kg + 7 kg + 5 kg + (1 kg + 7 kg) (5 N) 1 kg + 7 kg + 5 kg = 11 . 1538 N . Alternative Solution: Using Eq. 3 and the numerical result of Eq. 4, we have F 23 = m 3 a + F r (3) = (5 kg) (1 . 23077 m / s 2 ) + 5 N = 11 . 1538 N . Question 2, chap 5, sect 6. part 1 of 1 10 points homework 09 – GUPTA, SHIVANI – Due: Feb 18 2008, 11:00 pm 2 The surfaces between a 5 kg block and the 15 kg wedge and between the 15 kg wedge and the horizontal plane is smooth (without friction). The acceleration of gravity is 9 . 8 m / s 2 . A block is released on the inclined plane (top side of the wedge). 5 k g μ = 21 ◦ 15 kg μ = 0 F 9 . 8m / s 2 What is the force F which must be exerted on the 15 kg block in order that the 5 kg block does not move up or down the plane? Correct answer: 75 . 2374 N (tolerance ± 1 %)....
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This note was uploaded on 10/13/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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solution_hwk9 - homework 09 – GUPTA, SHIVANI – Due: Feb...

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