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solutions 16 - Patel Kinal Homework 16 Due Nov 2 2007 7:00...

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Patel, Kinal – Homework 16 – Due: Nov 2 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 10 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 2) 10 points A particle is located at the vector position ~r = (1 . 6 m)ˆ ı + (4 . 3 m)ˆ and the force acting on it is ~ F = (2 . 3 N)ˆ ı + (3 . 9 N)ˆ  . What is the magnitude of the torque about the origin? Correct answer: 3 . 65 N m. Explanation: Basic Concept: ~ τ = ~r × ~ F Solution: Since neither position of the par- ticle, nor the force acting on the particle have the z -components, the torque acting on the particle has only z -component: ~ τ = [ x F y - y F x ] ˆ k = [(1 . 6 m) (3 . 9 N) - (4 . 3 m) (2 . 3 N)] ˆ k = [ - 3 . 65 N m] ˆ k . 002 (part 2 of 2) 10 points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(2 m) , (8 m)]? Correct answer: 6 . 95 N m. Explanation: Reasoning similarly as we did in the pre- vious section, but with the difference that relative to the point [(2 m) , (8 m)] the y - component of the particle is now [ y - (8 m)], we have ~ τ = { [ x - a ] F y - [ y - b ] F x } ˆ k = { [(1 . 6 m) - (2 m)] [3 . 9 N] - [(4 . 3 m) - (8 m)] [2 . 3 N] } ˆ k = { 6 . 95 N m } ˆ k . keywords: 003 (part 1 of 1) 10 points Consider vectors ~ A and ~ B with coordinate components shown in the illustration below. A B An isometric drawing: Each coordinate has a length of ± 5 units. To indicate the coordinates of each vector, a line is projected to the hor- izontal plane then two lines are pro- jected to the horizontal coordinates. As well, a line is directly projected to the vertical coordinate. What is the magnitude of the vector prod- uct k ~ A × ~ B k of these two vectors? Correct answer: 12 . 083 units 2 . Explanation: Basic Concept: ~ A × ~ B = fl fl fl fl fl fl ˆ ı ˆ ˆ k A x A y A z B x B y B z fl fl fl fl fl fl (1) = fl fl fl fl A y A z B y B z fl fl fl fl ˆ ı + fl fl fl fl A z A x B z B x fl fl fl fl ˆ + fl fl fl fl A x A y B x B y fl fl fl fl ˆ k = ( A y B z - A z B y ı + ( A z B x - A x B z + ( A x B y - A y B x ) ˆ k Let : A x = 5 units , A y = - 3 units , A z = 2 units ,
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