SOL31 - Thevalingam, Donald Homework 31 Due: May 24 2007,...

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Thevalingam, Donald – Homework 31 – Due: May 24 2007, midnight – Inst: Eslami 1 This print-out should have 15 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 1) 10 points ±ind the magnitude oF the torque produced by a 4.9 N Force applied to a door at a perpen- dicular distance oF 0.39 m From the hinge. Correct answer: 1 . 911 N · m. Explanation: Given : F = 4 . 9 N , d = 0 . 39 m , and θ = 90 . 0 . τ = F d (sin θ ) τ = (4 . 9 N) (0 . 39 m) sin 90 . 0 = 1 . 911 N · m . keywords: 002 (part 1 oF 2) 10 points A simple pendulum consists oF a 2.1 kg point mass hanging at the end oF a 4.7 m long light string that is connected to a pivot point. The acceleration oF gravity is 9 . 81 m / s 2 . a) Calculate the magnitude oF the torque (due to the Force oF gravity) around this pivot point when the string makes a 2 . 4 angle with the vertical. Correct answer: 4 . 0546 N · m. Explanation: Basic Concept: τ = F d (sin θ ) = mgd (sin θ ) Given: m = 2 . 1 kg d = 4 . 7 m θ = 2 . 4 g = 9 . 81 m / s 2 Solution: τ = (2 . 1 kg) ( 9 . 81 m / s 2 ) (4 . 7 m)(sin 2 . 4 ) = 4 . 0546 N · m 003 (part 2 oF 2) 10 points b) Repeat this calculation For an angle oF 13 . 3 . Correct answer: 22 . 2745 N · m. Explanation: Given: θ = 13 . 3 Solution: τ = (2 . 1 kg) ( 9 . 81 m / s 2 ) (4 . 7 m)(sin 13 . 3 ) = 22 . 2745 N · m keywords: 004 (part 1 oF 1) 10 points Given: A circular shaped object with an inner radius oF 14 cm and an outer radius oF 26 cm. There are three Forces (acting perpendicular to the axis oF rotation) whose magnitudes are 10 N, 26 N, and 14 N acting on the object, as shown in the fgure. The Force oF magnitude 26 N is 23 below horizontal. 10 N 14 N 26 N 6 . 5 kg 23 ω 14 cm 26 cm ±ind the magnitude oF the net torque on the wheel about the axle through the center oF the object. Correct answer: 2 . 6 N m. Explanation: Let : a = 14 cm , b = 26 cm , F 1 = 10 N , F 2 = 26 N , F 3 = 14 N , and θ = 23 .
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Thevalingam, Donald – Homework 31 – Due: May 24 2007, midnight – Inst: Eslami 2 F 1 F 3 F 2 M θ ω a b The total torque is τ = a F 2 - b F 1 - b F 3 = (14 cm) (26 N) - (26 cm) h (10 N) + (14 N) i = - 2 . 6 N m k ~ τ k = 2 . 6 N m . keywords:
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This note was uploaded on 03/04/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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SOL31 - Thevalingam, Donald Homework 31 Due: May 24 2007,...

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