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# oldHWK28 - oldhomewk 28 VALENCIA DANIEL Due Apr 6 2008 4:00...

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oldhomewk 28 – VALENCIA, DANIEL – Due: Apr 6 2008, 4:00 am 1 Question 1, chap 14, sect 3. part 1 of 2 10 points A solid bar of length L has a mass m 1 . The bar is fastened by a pivot at one end to a wall which is at an angle θ with respect to the horizontal. The bar is held horizontally by a vertical cord that is fastened to the bar at a distance x cord from the wall. A mass m 2 is suspended from the free end of the bar. T m 2 m 1 θ x cord L Find the tension T in the cord. 1. T = 0 2. T = ( m 1 + m 2 ) g sin θ 3. T = parenleftbigg 1 2 m 1 + m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g correct 4. T = parenleftbigg 1 2 m 1 + m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g cos θ 5. T = parenleftbigg m 1 + 1 2 m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g sin θ 6. T = ( m 1 + m 2 ) g cos θ 7. T = ( m 1 + m 2 ) parenleftbigg L x cord parenrightbigg parenleftBig g 2 parenrightBig 8. T = parenleftbigg m 1 + 1 2 m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g 9. T = parenleftbigg m 1 + 1 2 m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g cos θ 10. T = parenleftbigg 1 2 m 1 + m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g sin θ Explanation: Under static equilibrium, summationdisplay F = 0 and summationdisplay τ = 0 . Since we know little about the reaction force between the bar and the wall, it is easi- est to begin by examining the torques around the connection between the bar and the wall, because the reaction forces will produce no torques around that point. summationdisplay τ = - L m 2 g + x cord T - L 2 m 1 g = 0 . T = parenleftbigg 1 2 m 1 + m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g . Question 2, chap 14, sect 3. part 2 of 2 10 points Find the the horizontal component of the force exerted on the bar by the wall. (Take right to be the positive direction.) 1. F x = parenleftbigg 1 2 m 1 + m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g cos θ 2. F x = parenleftbigg m 1 + 1 2 m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g cos θ 3. F x = 0 correct 4. F x = parenleftbigg 1 2 m 1 + m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g sin θ 5. F x = ( m 1 + m 2 ) g sin θ 6. F x = ( m 1 + m 2 ) parenleftbigg L x cord parenrightbigg parenleftBig g 2 parenrightBig 7. F x = parenleftbigg m 1 + 1 2 m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g sin θ 8. F x = parenleftbigg m 1 + 1 2 m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g 9. F x = parenleftbigg 1 2 m 1 + m 2 parenrightbigg parenleftbigg L x cord parenrightbigg g 10. F x = ( m 1 + m 2 ) g cos θ Explanation: In general, there will be a horizontal reac- tion force R x at the connection between an object and a wall. However, none of the other forces in this situation act in the horizontal direction, so summationdisplay F x = R x = 0 .

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oldhomewk 28 – VALENCIA, DANIEL – Due: Apr 6 2008, 4:00 am 2 Question 3, chap 14, sect 1.
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oldHWK28 - oldhomewk 28 VALENCIA DANIEL Due Apr 6 2008 4:00...

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