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Practice Homework 11 Solutions

# Practice Homework 11 Solutions - practice 11 ALIBHAI ZAHID...

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practice 11 – ALIBHAI, ZAHID – Due: Apr 8 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Two weights attached to a uniform beam of mass 38 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 . 23 kg 20 kg 2 . 9 m 8 m 37 38 kg What is the tension in the cable which sup- ports the beam? Correct answer: 0 . 770847 kN (tolerance ± 1 %). Explanation: Let : m = 38 kg , M 1 = 23 kg , M 2 = 20 kg , L 1 = 2 . 9 m , L 2 = 8 m , and θ = 37 . Under static equilibrium summationdisplay vector F = 0 and summationdisplay vector τ = 0 . The sum of the torques about the pivot is T L 2 sin θ - m g L 2 2 - M 1 g L 1 - M 2 g L 2 = 0 T = m g L 2 2 + M 1 g L 1 + M 2 g L 2 L 2 sin θ = m g 2 sin θ + M 1 g L 1 L 2 sin θ + M 2 g sin θ = (38 kg)(9 . 8 m / s 2 ) 2 sin 37 + (23 kg) (9 . 8 m / s 2 ) (2 . 9 m) (8 m) sin 37 + (20 kg) (9 . 8 m / s 2 ) sin 37 = 770 . 847 N = 0 . 770847 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 = 0 correct 3. R x - F x cos θ = 0 4. R x - F x sin θ = 0 5. F x cos θ - R x sin θ = 0 Explanation: Using the distances, angles and forces de- picted in the figure, the condition summationdisplay F x = 0 for translational equilibrium in the x direction gives - R x + F x = 0 . Question 3

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practice 11 – ALIBHAI, ZAHID – Due: Apr 8 2007, 4:00 am 2 part 2 of 3 10 points What is the condition for translational equilibrium along the vertical y direction? 1. R y + F y - W = 0 correct 2. R y - F y + W = 0 3. R y + F y = 0 4. R y sin θ + F y sin θ - W cos θ = 0 5. W - R y + F y = 0 Explanation: For the equilibrium in the y direction, summationdisplay F y = 0 gives 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. F y sin θ - W 2 sin θ - F x sin θ = 0 2. F y cos θ - W 2 cos θ - F x sin θ = 0 correct 3. 2 F y sin θ - W ℓ cos θ - 2 F x sin θ = 0 4. W 2 - F x cos θ - F y cos θ = 0 5. F y cos θ - W ℓ 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 ℓ F y cos θ - 2 W cos θ - ℓ F x sin θ = 0 . Question 5 part 1 of 1 10 points A light string has its ends tied to two walls separated by a distance equal to four-fifth the length L of the string as shown in the figure. A 25 kg mass is suspended from the center of the string, applying a tension in the string.
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Practice Homework 11 Solutions - practice 11 ALIBHAI ZAHID...

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