# hw27 - Overton Mays Homework 27 Due 4:00 am Inst Turner...

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Overton, Mays – Homework 27 – Due: Apr 11 2005, 4:00 am – Inst: Turner 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (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 gravita- tional feld. However, it is held in equilibrium by a Force F at its other end. The rod makes an angle 40 with the hori- zontal. The length oF the rod is 5 m. The Force makes and angle 68 with the horizontal. The weight oF the rod is 9 . 8 N. x y 5 m F 68 40 O What is the magnitude oF the Force F ? Correct answer: 7 . 99541 N. Explanation: Let : = 5 m , W = 9 . 8 N , θ = 40 , and φ = 68 . x y ψ W φ θ O R Basic Concepts: ±or static equilibrium, we have X i F xi = 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) = (9 . 8 N) 2[sin68 - cos68 · tan40 ] = 7 . 99541 N . Alternate Solution: The angle between the Force F and the rod is β = φ - θ = (68 ) - (40 ) = 28 . 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)

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Overton, Mays – Homework 27 – Due: Apr 11 2005, 4:00 am – Inst: Turner 2 = (9 . 8 N) 2 cos40 sin28 = 7 . 99541 N . 002 (part 2 of 3) 10 points What is the magnitude of the force the pivot exerts on the rod at point O ? Correct answer: 3 . 82983 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 . 99541 N)cos68 = - 2 . 99513 N , and F y + R y = W , so (2) R y = W - F y = W - F sin φ = (9 . 8 N) - (7 . 99541 N)sin68 = 2 . 38678 N , and k ~ R k = q R 2 x + R 2 y = q ( - 2 . 99513 N) 2 + (2 . 38678 N) 2 = 3 . 82983 N . 003 (part 3 of 3) 10 points What is the angle ( - 180 θ +180 ) the reactive force on the pivot at point O makes with the positive x -axis? Correct answer: 141
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## This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw27 - Overton Mays Homework 27 Due 4:00 am Inst Turner...

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