HW14.sol - homework 14 RAHMAN, TARIQUE Due: Mar 26 2008,...

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Unformatted text preview: homework 14 RAHMAN, TARIQUE Due: Mar 26 2008, 3:00 am 1 Question 1, chap 10, sect 2. part 1 of 2 10 points Three spherical masses are located in a plane at the positions shown in the figure be- low. A has mass 15 . 6 kg, B has mass 50 . 7 kg, and C has mass 46 . 7 kg. 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 A B C y Distance(m) x Distance (m) Three Masses in a Plane Figure: Drawn to scale. Calculate the x-coordinate of the center of mass. Correct answer: 6 . 0323 m (tolerance 3 %). Explanation: Basic Concepts: x cm = i m i x i i m i (1) y cm = i m i y i i m i (2) There is no acceleration, so a x = a y = 0 . Solution: Let: x a = 2 m , y a = 8 . 5 m , m a = 15 . 6 kg , x b = 5 m , y b = 4 m , m b = 50 . 7 kg , x c = 8 . 5 m , y c = 6 m , m c = 46 . 7 kg . x cm = i m i x i i m i (1) = m a x a + m b x b + m c x c m a + m b + m c = (15 . 6 kg) (2 m) + (50 . 7 kg) (5 m) (15 . 6 kg) + (50 . 7 kg) + (46 . 7 kg) + (46 . 7 kg) (8 . 5 m) (15 . 6 kg) + (50 . 7 kg) + (46 . 7 kg) = (681 . 65 m kg) (113 kg) = 6 . 0323 m . Question 2, chap 10, sect 2. part 2 of 2 10 points Calculate the y-coordinate of the center of mass. Correct answer: 5 . 44779 m (tolerance 3 %). Explanation: y cm = i m i y i i m i (1) = m a y a + m b y b + m c y c m a + m b + m c = (15 . 6 kg) (8 . 5 m) + (50 . 7 kg) (4 m) (15 . 6 kg) + (50 . 7 kg) + (46 . 7 kg) + (46 . 7 kg) (6 m) (15 . 6 kg) + (50 . 7 kg) + (46 . 7 kg) = (615 . 6 m kg) (113 kg) = 5 . 44779 m . Question 3, chap 10, sect 2. part 1 of 2 10 points A square plate is produced by welding to- gether four smaller square plates, each of side a . The weight of each of the four plates is shown in the figure. x y 60 N 20 N 60 N 30 N (0 , 0) (2 a, 0) (0 , 2 a ) (2 a, 2 a ) homework 14 RAHMAN, TARIQUE Due: Mar 26 2008, 3:00 am 2 Find the x-coordinate of the center of grav- ity (as a multiple of a ). Correct answer: 1 . 02941 a (tolerance 1 %). Explanation: Let : x 1 = x 2 = 1 2 a, x 3 = x 4 = 3 2 a, W 1 = 60 N , W 2 = 20 N , W 3 = 60 N , and W 4 = 30 N . y x W 1 W 2 W 3 W 4 a 2 3 a 2 a 2 3 a 2 The total weight is W = W 1 + W 2 + W 3 + W 4 = 170 N . Applying the definition of center of gravity, x cg = i W i x i W 1 + W 2 + W 3 + W 4 = ( W 1 + W 2 ) a 2 + ( W 3 + W 4 ) 3 a 2 W = (60 N + 20 N) + 3 (60 N + 30 N) 2 (170 N) a = 1 . 02941 a . Question 4, chap 10, sect 2. part 2 of 2 10 points Find the y-coordinate of the center of grav- ity (as a multiple of a ). Correct answer: 0 . 970588 a (tolerance 1 %). Explanation: Let : y 1 = y 4 = 1 2 a and y 2 = y 3 = 3 2 a. y cg = i W i y i W 1 + W 2 + W 3 + W 4 = ( W 1 + W 4 ) a 2 + ( W 2 + W 3 ) 3 a 2 W = (60 N + 30 N) + 3 (20 N + 60 N) 2 (170 N) a = . 970588 a ....
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This note was uploaded on 02/03/2010 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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HW14.sol - homework 14 RAHMAN, TARIQUE Due: Mar 26 2008,...

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