PHY 303K - Homework 8

PHY 303K - Homework 8 - homework 08 – FIERRO JEFFREY –...

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Unformatted text preview: homework 08 – FIERRO, JEFFREY – Due: Feb 17 2008, 11:00 pm 1 Question 1, chap 5, sect 5. part 1 of 1 10 points The pulley system is in equilibrium and the pulleys are weightless and frictionless. The weights are 184 N, 28 N, and 37 N. The acceleration of gravity is 9 . 8 m / s 2 . 184 N 28 N 37 N T Find the tension T . Correct answer: 406 . 5 N (tolerance ± 1 %). Explanation: Let : W 1 = 184 N , W 2 = 28 N , W 3 = 37 N , and g = 9 . 8 m / s 2 . W 1 W 2 W 3 4 7 5 3 6 1 2 8 T 7 T 1 T 7 T 7 T 8 T 1 T 1 T 6 T 6 T 2 T 5 T 6 T 5 T 5 T 4 T 3 T T T We will start at the loose end with ten- sion T and work toward the weight W 1 by first moving to the left and down, then by moving up. The equation will come from the equilibrium at the suspended weight W 1 . 1) At the weight W 2 , T 1 = W 2 + T . At pulley 1, T 2 = 2 T 1 = 2 W 2 + 2 T . 2 ) Starting up from pulley 2, T 3 = 2 T . At the weight W 3 , T 4 = W 3 + T 3 = W 3 + 2 T . At pulley 3, 2 T 5 = T 4 T 5 = T 4 2 = W 3 2 + T . At pulley 4, T 5 = 2 T 6 homework 08 – FIERRO, JEFFREY – Due: Feb 17 2008, 11:00 pm 2 T 6 = T 5 2 = W 3 4 + T 2 . At pulley 5, T 7 = 2 T 6 = W 3 2 + T . At pulley 6, T 8 = 2 T 7 = W 3 + 2 T . 3) Finally, for equilibrium at W 1 , T 7 + T 8 + T 6 = W 1 + T 1 + T 2 7 4 W 3 + 7 2 T = W 1 + 3 W 2 + 3 T 7 W 3 + 14 T = 4 W 1 + 12 W 2 + 12 T T = 2 W 1 + 6 W 2- 7 2 W 3 = 2 (184 N) + 6 (28 N)- 7 2 (37 N) = 406 . 5 N . Question 2, chap 5, sect 5. part 1 of 1 10 points Three strings (labeled A , B , and C ) at- tached to the sides of a square box are tied together by a knot as shown in the figure. The tension in the string labeled C is 64 N. 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 y Distance(m) x Distance (m) Three Strings Knotted Together A T A B T B C 64 N Figure: Drawn to scale. Calculate the magnitude of the tension in the string marked A . Correct answer: 33 . 435 N (tolerance ± 3 %). Explanation: Basic Concepts: summationdisplay F x = ma x , (1) summationdisplay F y = ma y , (2) summationdisplay F z = ma z , and summationdisplay τ = I α. There is no acceleration, so a x = a y = 0 . Solution: Using the knot (4 m , 4 . 5 m) as the coordinate origin (see figure below), we have x a = 0 m- 4 m =- 4 m , y a = 1 m- 4 . 5 m =- 3 . 5 m , r a = radicalBig x 2 a + y 2 a = 5 . 31507 m , x b = 0 m- 4 m =- 4 m , y b = 10 m- 4 . 5 m = 5 . 5 m , r b = radicalBig x 2 b + y 2 b = 6 . 80074 m , x c = 10 m- 4 m = 6 m , y c = 2 m- 4 . 5 m =- 2 . 5 m , r c = radicalBig x 2 c + y 2 c = 6 . 5 m ,-4-3-2-1 1 2 3 4 5 6-4-3-2-1 1 2 3 4 5 y Distance(m) x Distance (m) Three Strings Knotted Together A T A B T B C 64 N homework 08 – FIERRO, JEFFREY – Due: Feb 17 2008, 11:00 pm 3 Figure: Drawn to scale. Using Eqs. 1 and 2, we have T A x a r a + T B x b r b + T C x c r c = 0 (1) T A y a r a + T B y b r b + T C y c r c = 0 . (2) Isolating T B T A r b x b x a r a + T B + T C r b x b x c r c = 0 (1) T A r b y b y a r a + T B + T C r b y b y c r c = 0 . (2) Subtracting Eq . 2 from 1 T A parenleftbigg r b x a x...
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PHY 303K - Homework 8 - homework 08 – FIERRO JEFFREY –...

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