Homework 28 - homework 28 BAUTISTA, ALDO Due: Apr 10 2006,...

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Unformatted text preview: homework 28 BAUTISTA, ALDO Due: Apr 10 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. Two weights attached to a uniform beam of mass 19 kg are supported in a horizontal position by a pin and cable as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 25 kg 13 kg 3 . 2 m 6 . 6 m 44 19 kg What is the tension in the cable which sup- ports the beam? Correct answer: 0 . 488424 kN (tolerance 1 %). Explanation: Let : m = 19 kg , M 1 = 25 kg , M 2 = 13 kg , L 1 = 3 . 2 m , L 2 = 6 . 6 m , and = 44 . Basic Concepts: X ~ F = 0 X ~ = 0 . Solution: The sum of the torques about the pivot is T L 2 sin - mg L 2 2- M 1 gL 1- M 2 gL 2 = 0 Solving for the tension T T = m g L 2 2 + M 1 g L 1 + M 2 g L 2 L 2 sin = (19 kg)(9 . 8 m / s 2 ) 6 . 6 m 2 (6 . 6 m) sin + (25 kg)(9 . 8 m / s 2 )(3 . 2 m) (6 . 6 m) sin + (13 kg)(9 . 8 m / s 2 )(6 . 6 m) (6 . 6 m) sin = 0 . 488424 kN . Question 2 Part 1 of 3. 10 points. A uniform rod pivoted at one end point O is free to swing in a vertical plane in a gravitational field. However, it is held in equilibrium by a force F at its other end. x y F F x F y W R x R R y O Force vectors are drawn to scale. What is the condition for translational equilibrium along the horizontal x direction? 1. F x = 0 2. R x- F x sin = 0 3. R x- F x cos = 0 4.- R x + F x = 0 correct 5. F x cos - R x sin = 0 homework 28 BAUTISTA, ALDO Due: Apr 10 2006, 4:00 am 2 Explanation: Using the distances, angles and forces de- picted in the figure, the condition X F x = 0 for translational equilibrium in the x direction is- R x + F x = 0 . Question 3 Part 2 of 3. 10 points. What is the condition for translational equilibrium along the vertical y direction? 1. W- R y + F y = 0 2. R y sin + F y sin - W cos = 0 3. R y + F y = 0 4. R y- F y + W = 0 5. R y + F y- W = 0 correct Explanation: For the equilibrium in the y direction, X F y = 0 turns out to be R y + F y- W = 0 . Question 4 Part 3 of 3. 10 points. Taking the origin (point O ) as the pivot point, what is the condition for rotational equilibrium? 1. 2 F y sin - W cos - 2 F x sin = 0 2. F y sin - W 2 sin - F x sin = 0 3. F y cos - W 2 cos - F x sin = 0 correct 4. W 2- F x cos - F y cos = 0 5. F y cos - W sin + F x sin = 0 Explanation: For rotational equilibrium about point O, the net torque on the system about that point must vanish. The angle appears as follows x y F x F y W W cos F y cos F x sin O and we see that the forces R x and R y exert no torque on the point O. From the figure we have, counterclockwise F y cos - 2 W cos - F x sin = 0 ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Homework 28 - homework 28 BAUTISTA, ALDO Due: Apr 10 2006,...

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