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Homework 28 - homework 28 BAUTISTA ALDO Due 4:00 am Version...

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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 θ = 44 . Basic Concepts: X ~ F = 0 X ~ τ = 0 . Solution: The sum of the torques about the pivot is T L 2 sin θ - 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 θ = (19 kg)(9 . 8 m / s 2 ) 6 . 6 m 2 (6 . 6 m) sin θ + (25 kg)(9 . 8 m / s 2 )(3 . 2 m) (6 . 6 m) sin θ + (13 kg)(9 . 8 m / s 2 )(6 . 6 m) (6 . 6 m) sin θ = 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 θ 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 θ = 0 3. R x - F x cos θ = 0 4. - R x + F x = 0 correct 5. F x cos θ - R x sin θ = 0
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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 θ + F y sin θ - W cos θ = 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 θ - W ‘ cos θ - 2 F x sin θ = 0 2. F y sin θ - W 2 sin θ - F x sin θ = 0 3. F y cos θ - W 2 cos θ - F x sin θ = 0 correct 4. W 2 - F x cos θ - F y cos θ = 0 5. F y cos θ - W ‘ sin θ + F x sin θ = 0 Explanation: For rotational equilibrium about point O, the net torque on the system about that point must vanish. The angle θ appears as follows x y F x F y W W cos θ F y cos θ F x sin θ θ θ θ θ 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 θ - 2 W cos θ - ‘ F x sin θ = 0 .
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