Physics Solution Manual for 1100 and 2101

6 as tan 1 0281 nm l since l is the length of the

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Unformatted text preview: m, its value can be determined from the Pythagorean theorem: L= ( 0.281 nm )2 + ( 0.281 nm )2 = 0.397 nm Chapter 1 Problems 11 Thus, the angle is 0.281 nm −1 0.281 nm = tan = 35.3° L 0.397 nm ______________________________________________________________________________ θ = tan −1 20. REASONING There are two right triangles in the drawing. Each H2 contains the common side that is 21° 52° shown as a dashed line and is D labeled D, which is the distance between the buildings. The H1 hypotenuse of each triangle is one of the lines of sight to the top and base of the taller building. The remaining (vertical) sides of the triangles are labeled H1 and H2. Since the height of the taller building is H1 + H2 and the height of the shorter building is H1, the ratio that we seek is (H1 + H2)/H1. We will use the tangent function to express H1 in terms of the 52° angle and to express H2 in terms of the 21° angle. The unknown distance D will be eliminated algebraically when the ratio (H1 + H2)/H1 is calculated. SOLUTION The ratio of the building heights is H + H2 Height of taller building =1 Height of shorter building H1 Using the tangent function, we have that tan 52° = H1 D tan 21° = H2 D or H1 = D tan 52° or H 2 = D tan 21° Substituting these results into the expression for the ratio of the heights gives H + H 2 D tan 52° + D tan 21° Height of taller building =1 = Height of shorter building H1 D tan 52° = 1+ tan 21° = 1.30 tan 52° 12 INTRODUCTION AND MATHEMATICAL CONCEPTS If the taller building were half again as tall as the shorter building, this ratio would have been 1.50. Therefore, your friend is wrong . 21. SSM REASONING The drawing at the right shows the location of each deer A, B, and C. From the problem statement it follows that C c B b = 62 m β α a γ b 77° 51° c = 95 m A γ = 180° − 51° − 77° = 52° Applying the law of cosines (given in Appendix E) to the geometry in the figure, we have 2 2 2 a − 2 ab cos γ + (b − c ) = 0 which is an expression that is quadratic in a. It can be simplified to Aa 2 + Ba + C = 0 , with A=1 B = –2 b cos γ = – 2(62 m) cos 52 ° = –76 m 2 2 2 2 2 C = (b − c ) = (62 m) − (95 m) = –5181 m This quadratic equation can be solved for the desired quantity a. SOLUTION Suppressing units, we obtain from the quadratic formula a= − ( −76 ) ± ( −76 ) 2 − 4 (1)(–5181) = 1.2 × 10 2 m and – 43 m 2 (1) Discarding the negative root, which has no physical significance, we conclude that the 2 distance between deer A and C is 1.2 × 10 m . Chapter 1 Problems 22. REASONING The trapeze cord is L = 8.0 m long, so the trapeze is initially h1 = L cos 41° meters below the support. At the instant he releases the trapeze, it is h2 = L cos θ meters below the support. The difference in heights is d = h2 – h1 = 13 θ 41° h1 L h2 L 41° θ 0.75 m. Given that the trapeze is released at a lower elevation than the platform, we expect to find θ < 41°. 0.75 m SOLUTION Putting the above relationships together, we have d = h2 − h1 = L cos θ − L cos 41o cos θ = or d + L cos 41o = L cos θ d + cos 41o L d 0.75 m + cos 41o = cos−1 + cos 41o = 32o L 8.0 m θ = cos−1 23. SSM REASONING AND SOLUTION A single rope must supply the resultant of the two forces. Since the forces are perpendicular, the magnitude of the resultant can be found from the Pythagorean theorem. a. Applying the Pythagorean theorem, 475 N F = ( 475 N ) + (315 N) = 5.70 × 10 N 2 2 2 b. The angle θ that the resultant makes with the westward direction is θ = tan −1 θ W 315 F N I = 33.6° G NJ HK 475 315 N F S Thus, the rope must make an angle of 33.6° s outh of west . _____________________________________________________________________________________________ 14 INTRODUCTION AND MATHEMATICAL CONCEPTS 24. REASONING The Pythagorean theorem (Equation 1.7) can be used to find the magnitude of the resultant vector, and trigonometry can be employed to determine its direction. a. Arranging the vectors in tail-to-head fashion, we can see that the vector A gives the resultant a westerly direction and vector B gives the resultant a southerly direction. Therefore, the resultant A + B points south of west. b. Arranging the vectors in tail-to-head fashion, we can see that the vector A gives the resultant a westerly direction and vector –B gives the resultant a northerly direction. Therefore, the resultant A + (–B) points north of west. SOLUTION Using the Pythagorean theorem and trigonometry, we obtain the following results: a. Magnitude of A + B = ( 63 units )2 + ( 63 units )2 = 89 units 63 units = 45° south of west 63 units θ = tan −1 b. Magnitude of A − B = ( 63 units )2 + ( 63 units )2 = 89 units 63 units = 45° north of west 63 units ______________________________________________________________________________ θ = tan −1 25. SSM WWW REASONING a. Since the two force vectors A and B have directions due west and due north, they are perpendicular. Therefore, the resultant vector F = A + B has a magnitude give...
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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.

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