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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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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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