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Unformatted text preview: 1 chapter 3 1. The vectors should be parallel to achieve a resultant 7 m long (the unprimed case shown below), antiparallel (in opposite directions) to achieve a resultant 1 m long (primed case shown), and perpendicular to achieve a resultant √ 3 2 + 4 2 = 5 m long (the doubleprimed case shown). In each sketch, the vectors are shown in a “headtotail” sketch but the resultant is not shown. The resultant would be a straight line drawn from beginning to end; the beginning is indicated by A (with or without primes, as the case may be) and the end is indicated by B . b A b B b A b B b A 00 b 6 B 00 2. A sketch of the dis placements is shown. The resultant (not shown) would be a straight line from start (Bank) to finish (Walpole). With a careful drawing, one should find that the resultant vector has length 29 . 5 km at 35 ◦ west of south. West East South Bank @ @ @ @ @ @ @ @ @ @ R H H H H H H H H H H H H H H H Y B B B B B B B N Walpole 3. The x component of ~a is given by a x = 7 . 3 cos 250 ◦ = 2 . 5 and the y component is given by a y = 7 . 3 sin 250 ◦ = 6 . 9. In considering the variety of ways to com pute these, we note that the vector is 70 ◦ below the x axis, so the components could also have been found from a x = 7 . 3 cos 70 ◦ and a y = 7 . 3 sin 70 ◦ . In a similar vein, we note that the vector is 20 ◦ from the y axis, so one could use a x = 7 . 3 sin 20 ◦ and a y = 7 . 3 cos 20 ◦ to achieve the same results. 4. The angle described by a full circle is 360 ◦ = 2 π rad, which is the basis of our conversion factor. Thus, (20 . ◦ ) 2 π rad 360 ◦ = 0 . 349 rad and (similarly) 50 . ◦ = 0 . 873 rad and 100 ◦ = 1 . 75 rad. Also, (0 . 330 rad) 360 ◦ 2 π rad = 18 . 9 ◦ and (similarly) 2 . 10 rad = 120 ◦ and 7 . 70 rad = 441 ◦ . 5. The textbook’s approach to this sort of problem is through the use of Eq. 3 6, and is illustrated in Sample Problem 33. However, most modern graphical calculators can produce the results quite efficiently using rectangular ↔ polar “shortcuts.” (a) q ( 25) 2 + 40 2 = 47 . 2 m (b) Recalling that tan ( θ ) = tan ( θ + 180 ◦ ), we note that the two possibilities for tan 1 (40 / 25) are 58 ◦ and 122 ◦ . Noting that the vector is in the third quadrant (by the signs of its x and y components) we see that 122 ◦ is the correct answer. The graphical calculator “shortcuts” mentioned above are designed to correctly choose the right possibility. 6. The x component of ~ r is given by 15 cos 30 ◦ = 13 m and the y component is given by 15 sin 30 ◦ = 7 . 5 m. 7. The point P is displaced vertically by 2 R , where R is the radius of the wheel. It is displaced horizontally by half the circumference of the wheel, or πR . Since R = 0 . 450 m, the horizontal component of the displacement is 1 . 414 m and the vertical component of the displacement is 0 . 900 m. If the x axis is horizontal and the y axis is vertical, the vector displacement (in meters) is ~ r...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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