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Physics Solution Manual for 1100 and 2101

# 0 cm thus the magnitude of m is square m 440 cm 2

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Unformatted text preview: square M= ( 44.0 cm )2 + (12.0 cm )2 = 45.6 cm b. Similarly, the lengths of the horizontal and vertical distances of K = 2P − Q are 4 horizontal squares and 9 vertical squares, or 16.0 cm and 36.0 cm, respectively. The magnitude of K is then K= (16.0 cm )2 + ( 36.0 cm )2 = 39.4 cm 18 INTRODUCTION AND MATHEMATICAL CONCEPTS 29. REASONING At the turning point, the distance to the campground is labeled d in the drawing. Note that d is the length of the hypotenuse of a right triangle. Since we know the lengths of the other two sides of the triangle, the Pythagorean theorem can be used to find d. The direction that cyclist #2 must head during the last part of the trip is given by the angle θ. It can be determined by using the inverse tangent function. Turning point N θ θ d Campground 1950 m 1430 m Start SOLUTION E W 1080 m a. The two sides of the triangle have lengths of 1080 m S and 520 m (1950 m − 1430 m = 520 m). The length d of the hypotenuse can be determined from the Pythagorean theorem, Equation (1.7), as d= (1080 m )2 + ( 520 m )2 = 1200 m b. Since the lengths of the sides opposite and adjacent to the angle θ are known, the inverse tangent function (Equation 1.6) can be used to find θ. 520 m = 26° south of east 1080 m ______________________________________________________________________________ θ = tan −1 30. REASONING a. Since the two displacement vectors A and B have directions due south and due east, they are perpendicular. Therefore, the resultant vector R = A + B has a 2 2 2 magnitude given by the Pythagorean theorem: R = A + B . Knowing the magnitudes of R and A, we can calculate the magnitude of B. The direction of the resultant can be obtained using trigonometry. b. For the vector R′ = A – B we note that the subtraction can be regarded as an addition in the following sense: R′ = A + (–B). The vector –B points due west, opposite the vector B, so the two vectors are once again perpendicular and the magnitude of R′ again is given by the Pythagorean theorem. The direction again can be obtained using trigonometry. SOLUTION a. The drawing shows the two vectors and the resultant vector. According to the Pythagorean theorem, we have R 2 = A2 + B 2 B= ( 3.75 km ) or 2 B = R 2 − A2 − ( 2.50 km ) = 2.8 km 2 North North A θ R R′ B –B (a) (b) θ A Chapter 1 Problems 19 Using trigonometry, we can see that the direction of the resultant is tan θ = B A or θ = tan −1 F2.8 km I = G.50 km J HK 2 48 ° east of south b. Referring to the drawing and following the same procedure as in part a, we find bg R′ 2 = A2 + − B 2 tan θ = B= or B A or 3 2 b.75 km g− b.50 kmg = 2.8 F km I = 48° west of south G.50 km J HK 2 R′ 2 − A2 = θ = tan −1 2 2 2 .8 km _____________________________________________________________________________________________ 31. SSM WWW REASONING AND SOLUTION The single force needed to produce the same effect is equal to the resultant of the forces provided by the two ropes. The figure below shows the force vectors drawn to scale and arranged tail to head. The magnitude and direction of the resultant can be found by direct measurement using the scale factor shown in the figure. 2900 N 2900 N 1000 N Scale: Resultant a. From the figure, the magnitude of the resultant is 5600 N . b. The single rope should be directed along the dashed line in the text drawing. ____________________________________________________________________________________________ 32. REASONING a. and b. The drawing shows the two vectors A and B, as well as the resultant vector A + B. The three vectors form a right triangle, of which two of the sides are known. We can employ the Pythagorean theorem, Equation 1.7, to find the length of the third side. The angle θ in the drawing can be determined by using the inverse cosine function, Equation 1.5, since the side adjacent to θ and the length of the hypotenuse are known. N B 15.0 units A+B W A θ E 12.3 units S 20 INTRODUCTION AND MATHEMATICAL CONCEPTS c. and d. The drawing illustrates the two vectors A and −B, as well as the resultant vector A − B. The three vectors form a right triangle, which is identical to the one above, except for the orientation. Therefore, the lengths of the hypotenuses and the angles are equal. N 12.3 units W A A−B E θ 15.0 units −B S SOLUTION a. Let R = A + B. The Pythagorean theorem (Equation 1.7) states that the square of the 2 2 2 hypotenuse is equal to the sum of the squares of the sides, so that R = A + B . Solving for B yields 2 (15.0 units )2 − (12.3 units )2 2 B= R −A = = 8.6 units b. The angle θ can be found from the inverse cosine function, Equation 1.5: 12.3 units = 34.9° north of west 15.0 units θ = cos −1 c. Except for orientation, the triangles in the two drawings are the same. Thus, the value for B is the same as that determined in part (a) above: B = 8.6 units d. The angle θ is the same as that found in part (a), except the resultant vector points south of west, rather than north of west: θ = 34.9° south o...
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