HW5 - Pickering, Tracey – Homework 5 – Due: Feb 1 2006,...

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Unformatted text preview: Pickering, Tracey – Homework 5 – Due: Feb 1 2006, noon – Inst: Drummond 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 1) 10 points Four vectors, each of magnitude 50 m, lie along the sides of a parallelogram as shown in the figure. The angle between vector A and B is 71 ◦ . Hint: Remember you are adding four vec- tors. A 5 m C 50 m B 50 m D 5 m 71 ◦ What is the magnitude of the vector sum of the four vectors? Correct answer: 162 . 823 m. Explanation: α = 71 ◦ and ‘ = 50 m . Basic Concepts: The components of a vector ~ R = ~ A + ~ B + ~ C + ~ D are obtained by adding up the respective com- ponents of each vector in the sum. Solution: ~ B = ~ C = (50 m) ˆ ı ~ A = ~ D = (50 m) (cos A ) ˆ ı + (50 m)(sin α ) ˆ Thus, ~ A + ~ B + ~ C + ~ D = 2 [ ‘ ı + ‘ cos α ˆ ı ] + 2 [ ‘ sin α ˆ ] = 2 [(50 m) ı + (50 m) cos(71 ◦ ) ˆ ı ] + 2 [(50 m) sin(71 ◦ ) ˆ ] = (132 . 557 m) ı + (94 . 5519 m) ˆ , where R x = 132 . 557 m and R y = 94 . 5519 m . The magnitude of the vector sum is R = q R 2 x + R 2 y = q (132 . 557 m) 2 + (94 . 5519 m) 2 = 162 . 823 m . Alternative Solution: Rotate your figue clockwise by one-half the angle between vec- tors A and B . The resulant vector must lie along the horizontal line connecting the tails of vectors A and B and the heads of vectors C and D . The four components along this line are ‘ times the cosine of the half angle. R = 4 ‘ cos ‡ α 2 · = 4 (50 m) cos (35 . 5 ◦ ) = 162 . 823 m . 002 (part 1 of 2) 10 points A person walks the path shown. The total trip consists of four straight-line paths. S N W E 101 m 379m 52 . ◦ 1 5 9 m 2 5 2 m 24 . ◦ Note: Figure is not drawn to scale. a) At the end of the walk, what is the mag- nitude of the person’s resultant displacement measured from the starting point? Correct answer: 461 . 537 m. Explanation: Pickering, Tracey – Homework 5 – Due: Feb 1 2006, noon – Inst: Drummond 2 d θ 101 m 379m- 128 ◦ 1 5 9 m 2 5 2 m 156 ◦ Note: Figure is not drawn to scale. Basic Concepts: Δ x = d (cos θ ) Δ y = d (sin θ ) Δ x tot = Δ x 1 + Δ x 3 + Δ x 4 since Δ x 2 = 0 m....
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This note was uploaded on 06/23/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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HW5 - Pickering, Tracey – Homework 5 – Due: Feb 1 2006,...

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