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Homework 11 - Solutions

Homework 11 - Solutions - homework 11 FRENNEA KYLE Due 4:00...

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homework 11 – FRENNEA, KYLE – Due: Apr 13 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A weight (with a mass of 66 kg) is sus- pended from a point near the right-hand end of a uniform boom with a mass of 73 kg . To support the uniform boom a cable runs from this same point to a wall (the left-hand verti- cal coordinate in the figure) and by a pivot on the same wall at an elevation of 5 m . The acceleration of gravity is 9 . 8 m / s 2 . 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 Vertical Height (m) Horizontal Distance (m) Boom and Weight 73 kg T 66 kg Figure: Drawn to scale. Calculate the tension in the cable. Correct answer: 2729 . 23 N (tolerance ± 1 %). Explanation: Let : M b = 73 kg , M w = 66 kg , = 10 m boom length , x b = 2 = 5 m , x = 8 m , y = 5 m , and h = 8 . 5 m . The static equilibrium conditions are summationdisplay T x = 0 , (1) summationdisplay T y = 0 , and (2) summationdisplay τ = 0 , or (3) W b x b + W w x = T y x . (4) From the figure, θ = arctan parenleftbigg h - y x parenrightbigg (5) = arctan parenleftbigg 8 . 5 m - 5 m 8 m parenrightbigg = 23 . 6294 . 0 1 2 3 4 5 6 7 8 9 10 -2 -1 0 1 2 3 4 5 6 7 8 9 10 Vertical Height (m) Horizontal Distance (m) Force Scale is 312 . 55 N / m (0 , h ) ( x, y ) (0 , y ) W w T x T T y W b θ Let the cable make an angle θ with the horizontal boom. Then using Eq. 5 and the figure, we have sin θ = T y T = h - y radicalbig ( h - y ) 2 + x 2 , T y = T ( h - y ) radicalbig ( h - y ) 2 + x 2 , (6) Solving Eq. 4 for T y , we have T y = ( W b x b + W w x ) x , and substituting T y from Eq. 6, we have T ( h - y ) radicalbig ( h - y ) 2 + x 2 = W b x b + W w x x

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homework 11 – FRENNEA, KYLE – Due: Apr 13 2007, 4:00 am 2 T = ( W b x b + W w x ) radicalbig ( h - y ) 2 + x 2 x ( h - y ) = [(73 kg) (5 m) + (66 kg) (8 m)] × (9 . 8 m / s 2 ) × radicalbig (8 . 5 m - 5 m) 2 + (8 m) 2 (8 m) (8 . 5 m - 5 m) = 2729 . 23 N . Question 2 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 ) g cos θ 2. T = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g sin θ 3. T = 0 4. T = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g cos θ 5. T = ( m 1 + m 2 ) parenleftbigg L x cord parenrightbigg parenleftBig g 2 parenrightBig 6. T = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g cos θ 7. T = ( m 1 + m 2 ) g sin θ 8. T = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g correct 9. T = parenleftbigg m 1 + 1 2 m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g 10. T = parenleftbigg 1 2 m 1 + m 2 parenrightbiggparenleftbigg L x cord parenrightbigg g sin θ Explanation: Under static equilibrium, summationdisplay F = 0 and summationdisplay τ = 0 .
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Homework 11 - Solutions - homework 11 FRENNEA KYLE Due 4:00...

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