hw14 - Muraj, Hamza Homework 14 Due: Feb 24 2006, 4:00 am...

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Muraj, Hamza – Homework 14 – Due: Feb 24 2006, 4:00 am – Inst: Florin 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 5) 10 points Given: The mass on the le±t is equal to the sum o± the other two masses; i.e. , m 1 = m 2 + m 3 , and the center mass is greater than the mass on the right hand side; i.e. , m 2 > m 3 . A massless pulley is attached to the ceiling, in a uni±orm gravitational feld g , and rotates with no ±riction about its pivot. Another massless pulley is attached to the end o± the cord on the right and rotates with no ±riction about its pivot. m 1 T 1 m 2 m 3 T 2 T 3 T 4 Select the correct expression. 1. Cannot be determined. 2. T 2 = m 2 g 3. T 2 > m 2 g 4. T 2 < m 2 g correct Explanation: One can immediately see that the mass m 2 is accelerating downward, so T 1 < m 2 g , which can be seen in Eq. 2 below. However, the dynamic expressions ±or the three masses are derived below ±or your inter- est. Consider the ±ree body diagram ±or the right hand pulley. T 23 T 23 T 1 There±ore : T 23 = T 2 = T 3 , T 1 = T 4 , and T 1 = T 23 + T 23 . Consider the ±ree body diagrams ±or mass m 1 and the right hand pulley where m 23 = m 2 + m 3 . m 1 m 23 T 1 m g 23 a Consider the ±ree body diagrams ±or masses m 2 and m 3 . Note: The right hand pulley system has an acceleration a 1 , which is downward. m 2 m 3 2 3 - + The equations o± motion using these ±ree body diagram are T 1 - m 1 g = - m 1 a 1 (1) T 23 - m 2 g = - m 2 ( a 23 - a 1 ) (2) T 23 - m 3 g = + m 3 ( a 23 + a 1 ) . (3) Since T 1 = 2 T 23 and m 1 = m 2 + m 3 , Eq. 4 is Eq. 1 and Eq. 5 is Eqs. 2 plus 3. 2 T 23 - ( m 2 + m 3 ) g = - ( m 2 + m 3 ) a 1 (4) 2 T 23 - ( m 2 + m 3 ) g = +( m 2 + m 3 ) a 1 + ( m 2 - m 3 ) a 23 . (5)
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Muraj, Hamza – Homework 14 – Due: Feb 24 2006, 4:00 am – Inst: Florin 2 Subtracting Eq. 4 from Eq. 5 and solving for a 1 , we have 0 = 2 ( m 2 + m 3 ) a 1 - ( m 2 - m 3 ) a 23 2 ( m 2 + m 3 ) a 1 = ( m 2 - m 3 ) a 23 a 1 = ( m 2 - m 3 ) 2 ( m 2 + m 3 ) a 23 . (6) Rearranging Eqs. 2 and 3, we have T 23 = m 2 g - m 2 ( a 23 - a 1 ) (7) T 23 = m 3 g + m 3 ( a 23 + a 1 ) . (8) Eliminating T 23 , we have ( m 2 - m 3 ) g = m 2 ( a 23 - a 1 ) + m 3 ( a 23 + a 1 ) ( m 2 - m 3 ) g = ( m 2 + m 3 ) a 23 - ( m 2 - m 3 ) a 1 . (9) Substituting a 1 from Eq. 6, we have ( m 2 - m 3 ) g = ( m 2 + m 3 ) a 23 - ( m 2 - m 3 ) 2 2 ( m 2 + m 3 ) a 23 (10) = 2 ( m 2 + m 3 ) 2 - ( m 2 - m 3 ) 2 2 ( m 2 + m 3 ) a 23 . Solving for a 23 , we have a 23 = 2 ( m 2 + m 3 )( m 2 - m 3 ) 2 ( m 2 + m 3 ) 2 - ( m 2 - m 3 ) 2 g . (11) Use a 23 from Eq. 11 to substitute into Eq. 6 to solve for a 1 a 1 = ( m 2 - m 3 ) 2 ( m 2 + m 3 ) a 23 (6) = ( m 2 - m 3 ) 2 ( m 2 + m 3 ) × 2 ( m 2 + m 3 ) ( m 2 - m 3 ) 2 ( m 2 + m 3 ) 2 - ( m 2 - m 3 ) 2 g = ( m 2 - m 3 ) 2 2 ( m 2 + m 3
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.

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hw14 - Muraj, Hamza Homework 14 Due: Feb 24 2006, 4:00 am...

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