oldhw28_s - oldhomewk 28 – YOO, HEE – Due: Apr 6 2008,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: oldhomewk 28 – YOO, HEE – 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 = ( m 1 + m 2 ) parenleftbigg L x cord parenrightbigg parenleftBig g 2 parenrightBig 2. T = 0 3. T = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g cos θ 4. T = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g 5. T = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g sin θ 6. T = ( m 1 + m 2 ) g sin θ 7. T = ( m 1 + m 2 ) g cos θ 8. T = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g cos θ 9. T = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g sin θ 10. T = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g correct 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 =- Lm 2 g + x cord T- L 2 m 1 g = 0 . T = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg 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 = ( m 1 + m 2 ) g sin θ 2. F x = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g 3. F x = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g sin θ 4. F x = ( m 1 + m 2 ) g cos θ 5. F x = 0 correct 6. F x = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g 7. F x = ( m 1 + m 2 ) parenleftbigg L x cord parenrightbigg parenleftBig g 2 parenrightBig 8. F x = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g cos θ 9. F x = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g cos θ 10. F x = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g sin 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 . oldhomewk 28 – YOO, HEE – Due: Apr 6 2008, 4:00 am 2 Question 3, chap 14, sect 1....
View Full Document

This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 8

oldhw28_s - oldhomewk 28 – YOO, HEE – Due: Apr 6 2008,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online