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Unformatted text preview: 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 = 37 . Under static equilibrium summationdisplay vector F = 0 and summationdisplay vector = 0 . The sum of the torques about the pivot is T L 2 sin  mg L 2 2 M 1 g L 1 M 2 g L 2 = 0 T = mg L 2 2 + M 1 g L 1 + M 2 g L 2 L 2 sin = mg 2 sin + M 1 g L 1 L 2 sin + M 2 g sin = (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 = . 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 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 = 0 4. R x F x sin = 0 5. F x cos  R x sin = 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 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 + F y sin  W cos = 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  W 2 sin  F x sin = 0 2. F y cos  W 2 cos  F x sin = 0 correct 3. 2 F y sin  W cos  2 F x sin = 0 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 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.
 Spring '08
 Turner
 Acceleration, Gravity, Mass, Work

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