Homework 5 - Solutions

# Homework 5 - Solutions - homework 05 – FRENNEA KYLE –...

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Unformatted text preview: homework 05 – FRENNEA, KYLE – Due: Feb 22 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A force F acts to the right on a 4 . 18 kg block. A 2 . 24 kg block is stacked on top of the 4 . 18 kg block and can slide on it with a coefficient of friction of 0 . 1between the blocks. The table has a coefficient of friction of 0 . 19. The acceleration of gravity is 9 . 8 m / s 2 . The system is in equilibrium. 4 . 18 kg 2 . 24 kg μ 2 μ 1 F Find the force F required to accelerate the 4 . 18 kg block at 1 . 8 m / s 2 . Correct answer: 23 . 7788 N (tolerance ± 1 %). Explanation: Given : m 1 = 2 . 24 kg , m 2 = 4 . 18 kg , μ 1 = 0 . 1 , and μ 2 = 0 . 19 . Consider the free body diagrams below for each mass m 2 m 1 F T μ 2 ( m 1 + m 2 ) g 2 T μ 1 m 1 g μ 1 m 1 g Basic Concepts : The acceleration of the two masses will be different because of the pulley system between them. The tensions in the strings will be different. Solution Let T be the tension in the string attached to m 2 on its left. Its acceleration will be a and directed to the right. Because of the pulley system between m 1 and m 2 , 2 T will be the tension in the string attached to m 1 on its left, and the acceleration of that string, and of m 1 , will be a 2 and will be directed to the left. For the mass m 1 , 2 T acts to the left and the motion to the left defines the frictional force μ 1 m 1 g acting to the right, with the acceleration a 2 directed to the left F net 1 = m 1 parenleftBig a 2 parenrightBig = 2 T − μ 1 m 1 g . Multiplying by 2, m 1 a = 4 T − 2 μ 1 m 1 g . (1) For the mass m 2 , notice first that the normal force will be N = ( m 1 + m 2 ) g , and since the frictional force μ 1 m 1 g acted to the right on m 1 , it acts to the left on m 2 . For the block m 2 , the force F acts to the right, and the tension T and the two frictional forces μ 1 m 1 g and μ 2 ( m 1 + m 2 ) g act to the left, with the acceleration a directed to the right F net 2 = m 2 a = F − T − μ 1 m 1 g − μ 2 ( m 1 + m 2 ) g . Multiplying by 4 4 m 2 a = 4 F − 4 T − 4 μ 1 m 1 g − 4 μ 2 ( m 1 + m 2 ) g . (2) Adding equations (1) and (2) yields ( m 1 + 4 m 2 ) a = 4 F − 6 μ 1 m 1 g − 4 μ 2 ( m 1 + m 2 ) g . Therefore F = ( m 1 + 4 m 2 ) a 4 + 3 μ 1 m 1 g 2 + μ 2 ( m 1 + m 2 ) g . = [2 . 24 kg + 4 (4 . 18 kg)] (1 . 8 m / s 2 ) 4 + 3 (0 . 1) (2 . 24 kg) (9 . 8 m / s 2 ) 2 + (0 . 19) (2 . 24 kg + 4 . 18 kg) (9 . 8 m / s 2 ) . = 23 . 7788 N . Question 2 part 1 of 1 10 points homework 05 – FRENNEA, KYLE – Due: Feb 22 2007, 4:00 am 2 A block is at rest on an inclined plane. 1 4 k g μ s = . 3 9 θ c Find the critical angle, θ c , at which the block just begins to slide. Correct answer: 21 . 3058 ◦ (tolerance ± 1 %)....
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## This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Homework 5 - Solutions - homework 05 – FRENNEA KYLE –...

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