Homework_7_solutions

Homework_7_solutions - Homework 7 1) (2m)(10N) + (1m)(30N)...

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Homework 7 1) (2m)(10N) + (1m)(30N) = (1m)(5N) + (d4)(15N), d4 = [(20+30-5)/15]m = (45/15)m = 3m 2) Draw pulley systems with mechanical advantages of 1, 2, 3, 4, 5, 6, 7, 8, and 9. 3) Repeat the upper tension problem on slide 4 of lecture series 2, with blocks 1, 2, &3 having a mass of 5, 6, & 9kg respectively. T 1 = (6kg +9kg)(1ms -2 ) = 15 N, for T 2 = (9kg)(1ms -2 ) = 9N 4) Repeat the lower tension problem on slide 4 of lecture series 2, with a 5kg mass and a horizontal tension of 7N. Calculate the original angle A as well as the new angle with the above mass and force. T = F M-Wt /CosA = F/SinA = [(50N) 2 +(7N) 2 ] = 50.5N, A = Cos -1 (50/50.5) = Sin -1 (7/50.5) = 8.0 deg. 2 1 4 5 6 7 9 3 8
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5) Repeat the upper tension problem on slide 6, of lecture series 2, with m 1 = 3kg and m 2 = 6kg. This is the dynamic case in which the system is moving up the incline. Therefore, this is similar to the lower example on the same slide where: T = [2(M 1 X M 2 )g/(M 1 +M 2 )], however the global equation must be changed to account for the mass 1 being on an incline: Global Eqn.: m 1 a + m 2 a = m 2 g – (Sin30 o )m 1 g a = [m 2 g - ½ m 1 g]/[m 1 + m 2 ] Tension Eqn.: T - (Sin30 o )m 1 g = m 1 a T = ½ m 1 g + m 1 a T = ½ m 1 g + m 1 [m 2 g - ½ m 1 g]/[m 1 + m 2 ] = m 1 g + [m 2 - ½ m 1 ]/[m 1 + m 2 ]) = m 1 g [m 1 + m 2 ]/[m 1 + m 2 ] + [m 2 - ½ m 1 ]/[m 1 + m 2 ]) = m 1 g m 1 + ½ m 2 + m 2 - ½ m 1 )/[m 1 + m 2 ]) = (1.5) m 1 m 2 g/[m 1 + m 2 ]) = (1.5)(3kg)(6kg)(10m/s 2 )/(3kg+6kg)] = (270/9)N = 30N 6) Repeat the lower tension problem on slide 6 of lecture series 2, with m 1 = 3kg and m 2 = 6kg.
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Homework_7_solutions - Homework 7 1) (2m)(10N) + (1m)(30N)...

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