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hw 8 Solutions

# hw 8 Solutions - Bell Evan Homework 8 Due 11:00 pm Inst...

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Bell, Evan – Homework 8 – Due: Mar 20 2005, 11:00 pm – Inst: Furlan 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Two objects, of masses 30 kg and 54 kg, are hung from the ends of a stick that is 70 cm long and has marks every 10 cm, as shown. 30 kg 54 kg A B C D E F G 10 20 30 40 50 60 If the mass of the stick is negligible, at which of the points indicated should a cord be attached if the stick is to remain horizontal when suspended from the cord? 1. E 2. D 3. F correct 4. C 5. G 6. A 7. B Explanation: Let : = 70 cm , m 1 = 30 kg , m 2 = 54 kg . For static equilibrium, τ net = 0. Denote x the distance from the left end point of the stick to the point where the cord is attached. m 1 g x - m 2 g [ - x ] = τ = 0 ( m 1 - m 2 ) x = m 2 x = m 2 m 1 - m 2 = (54 kg) (70 cm) 30 kg - 54 kg = 45 cm . Therefore the point should be point F . 002 (part 1 of 2) 10 points A uniform rod pivoted at one end “point O is free to swing in a vertical plane in a gravita- tional field. However, it is held in equilibrium by a force F at its other end. The rod makes an angle 36 with the hori- zontal. The length of the rod is 7 m. The force makes and angle 58 with the horizontal. The weight of the rod is 7 . 4 N. x y 7 m F 58 36 O What is the magnitude of the force F ? Correct answer: 7 . 99068 N. Explanation: Let : = 7 m , W = 7 . 4 N , θ = 36 , and φ = 58 .

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Bell, Evan – Homework 8 – Due: Mar 20 2005, 11:00 pm – Inst: Furlan 2 x y ψ W F φ θ O R Basic Concepts: For static equilibrium, we have X i F x i = 0 , (1) X i F y i = 0 , and (2) X i τ i = 0 . (3) Solution: Using the torque equation, about point O , (Eq. 3 with counter clock- wise positive), we have 2 W cos θ - ‘ F cos θ sin φ + ‘ F sin θ cos φ = 0 W 2 - F sin φ + F tan θ cos φ = 0 , so F = W 2 [sin φ - cos φ · tan θ ] (4) = (7 . 4 N) 2 [sin 58 - cos 58 · tan 36 ] = 7 . 99068 N . Alternate Solution: The angle between the force F and the rod is β = φ - θ = (58 ) - (36 ) = 22 . Using the torque equation, about point O , (Eq. 3 with counter clockwise positive), we have 2 W cos θ = ‘ F sin β , so F = W 2 cos θ sin β (5) = (7 . 4 N) 2 cos 36 sin 22 = 7 . 99068 N . 003 (part 2 of 2) 10 points What is the magnitude of the force the pivot exerts on the rod at point O ? Correct answer: 4 . 28008 N. Explanation: Using the translational equations (Eqs. 1 and 2), where R x and R y are the horizontal and vertical components of the force R at the pivot, we have F x + R x = 0 , so (1) R x = - F x = - F cos φ = - (7 . 99068 N) cos 58 = - 4 . 23442 N , and F y + R y = W , so (2) R y = W - F y = W - F sin φ = (7 . 4 N) - (7 . 99068 N) sin 58 = 0 . 623516 N , and k ~ R k = q R 2 x + R 2 y = q ( - 4 . 23442 N) 2 + (0 . 623516 N) 2 = 4 . 28008 N . The angle ψ from the positive x axis to the vector R is ψ = arctan R y R x = arctan 0 . 623516 N - 4 . 23442 N = 171 . 623 .
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hw 8 Solutions - Bell Evan Homework 8 Due 11:00 pm Inst...

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