This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 3 1. The x and the y components of a vector a lying on the xy plane are given by cos , sin x y a a a a θ θ = = where   a a = is the magnitude and θ is the angle between a and the positive x axis. (a) The x component of a is given by a x = 7.3 cos 250° = – 2.5 m. (b) and the y component is given by a y = 7.3 sin 250° = – 6.9 m. In considering the variety of ways to compute 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° to the left 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. 2. The angle described by a full circle is 360° = 2 π rad, which is the basis of our conversion factor. (a) ( 29 2 rad 20.0 20.0 0.349 rad 360 π = = . (b) ( 29 2 rad 50.0 50.0 0.873 rad 360 π = = (c) ( 29 2 rad 100 100 1.75 rad 360 π = = (d) ( 29 360 0.330 rad = 0.330 rad 18.9 2 rad π = (e) ( 29 360 2.10 rad = 2.10 rad 120 2 rad π = (f) ( 29 360 7.70 rad = 7.70 rad 441 2 rad π = 79 CHAPTER 3 3. A vector a can be represented in the magnitudeangle notation ( a , θ ), where 2 2 x y a a a = + is the magnitude and 1 tan y x a a θ = is the angle a makes with the positive x axis. (a) Given A x =  25.0 m and A y = 40.0 m, 2 2 ( 25.0 m) (40.0 m) 47.2 m A = + = (b) Recalling that tan θ = tan ( θ + 180°), tan –1 [40/ (– 25)] = – 58° or 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. 4. (a) With r = 15 m and θ = 30°, the x component of r is given by r x = r cos θ = 15 cos 30° = 13 m. (b) Similarly, the y component is given by r y = r sin θ = 15 sin 30° = 7.5 m. 5. The vector sum of the displacements d storm and . d new must give the same result as its originally intended displacement o . =120j d where east is i , north is j, and the assumed length unit is km. Thus, we write storm new ˆ ˆ ˆ 100i, i + j. d d A B = = r r (a) The equation storm new o d d d + = readily yields A = –100 km and B = 120 km. The magnitude of d new is therefore 2 2 156km A B + = . (b) And its direction is tan –1 ( B / A ) = –50.2° or 180° + ( –50.2°) = 129.8°. We choose the latter value since it indicates a vector pointing in the second quadrant, which is what we expect here. The answer can be phrased several equivalent ways: 129.8° counterclockwise from east, or 39.8° west from north, or 50.2° north from west....
View
Full
Document
This note was uploaded on 04/07/2008 for the course PHYSICS AN 1100, 2101 taught by Professor Giammanico during the Spring '07 term at LSU.
 Spring '07
 Giammanico

Click to edit the document details