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Unformatted text preview: homework 05 – ANDERSON, ZACH – Due: Sep 29 2007, 4:00 am 1 Question 1, chap 5, sect 6. part 1 of 1 10 points A block of mass m accelerates with acceler ation g up a frictionless plane that is inclined at an angle α above the horizontal. The force F o that pushes the block is at an angle β above the horizontal. Find the force F o . m α β F o 1. mg 1 − sin β cos( β − α ) 2. mg 1 cos( α + β ) 3. mg 1 + sin α cos( α + β ) correct 4. mg 1 + sin α cos( β − α ) 5. mg 1 + sin α cos( α − β ) 6. mg 1 − sin β cos( α + β ) 7. mg sin β cos( α + β ) 8. mg 1 + sin β cos( β − α ) 9. mg 1 + sin β cos( α + β ) 10. mg 1 + sin β cos( α − β ) Explanation: The acceleration up the plane is given, a = g . The angle that F o makes with respect to the inclined plane is α + β . So the component of F o which is parallel to the inclined plane is F o cos( α + β ) Therefore F o cos( α + β ) − mg sin α = ma = mg F o = mg 1 + sin α cos( α + β ) Question 2, chap 6, sect 1. part 1 of 3 10 points The suspended 2 . 7 kg mass on the right is moving up, the 2 . 3 kg mass slides down the ramp, and the suspended 8 . 5 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 12 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 3 k g μ = . 1 2 32 ◦ 8 . 5 kg 2 . 7 kg What is the acceleration of the three block system? Correct answer: 4 . 92523 m / s 2 (tolerance ± 1 %). Explanation: Let : m 1 = 2 . 7 kg , m 2 = 2 . 3 kg , m 3 = 8 . 5 kg , and θ = 32 ◦ . Basic Concept: F net = ma negationslash = 0 homework 05 – ANDERSON, ZACH – Due: Sep 29 2007, 4:00 am 2 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di rected upward F net 1 = m 1 a = T 1 − m 1 g (1) For the mass on the table, the parallel compo nent of its weight is mg sin θ and the perpen dicular component of its weight is mg cos θ . ( N = mg cos θ from equilibrium). The accel eration a is directed down the table, T 3 and the parallel weight component m 2 g sin θ act down the table, and T 1 and the frictional force μ N = μm 2 g cos θ act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin θ − T 1 − μm 2 g cos θ . For the mass m 3 , T 3 acts up and the weight m 3 g acts down, with the acceleration a di rected downward F net 3 = m 3 a = m 3 g − T 3 . (3) Adding Eqs. (1), (2), & (3) yields ( m 1 + m 2 + m 3 ) a = m 3 g + m 2 g sin θ − μm 2 g cos θ − m 1 g ....
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This note was uploaded on 04/10/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Acceleration, Force, Friction, Mass, Work

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