3.1. Solve: (a) If one component of the vector is zero, then the other component must not be zero (unless the whole vector is zero). Thus the magnitude of the vector will be the value of the other component. For example, if Ax = 0 m and Ay = 5 m, then the magnitude of the vector is A = ( 0 m ) 2 + (5 m ) 2 = 5 m (b) A zero magnitude says that the length of the vector is zero, thus each component must be zero.
3.2.
Visualize:
r r Solve: (a) C = Ar+ B only r if A and B are in the same direction. Size does not matter. (b) C = A − B if A and B are in the opposite direction to each other. Size matters only in that A > B because C as a magnitude can only be positive.
3.3. Visualize:
r r Solve: (a) C = A − B if B is directed in the same direction as A . Size does not matter, except that A > B because C as a magnitude rcannot be negative. r (b) C = A + B if B is directed opposite to the direction of A . Size does not matter.
3.4. Visualize:
r r r r r Solve: (a) To find Ar + B , we place the tail of vector B on the tip of vector A and connect the tail of vector A with the tiprof vector r r r r B. r (b) Since A − B = A + ( − B) , we place the tail of the vector ( − B) on the tip of vector A and then connect the tail of r r vector A with the tip of vector ( − B) .
3.5. Visualize:
r r r r r A’s A B A + B Solve: (a) To find , we place the tail of vector on the tip of vector and then connect vector tail r with vector B’s r tip. r r r r r r r (b) To find A − B , we note that A − B = A + ( − B) . We place the tail of vector − B on the tip of vector A and then r r connect vector A’s tail with the tip of vector − B.
r
3.6. Visualize: The position vector r whose magnitude r is 10 m has an x-component of 6 m. It makes an angle θ with the +x -axis in the first quadrant.
Solve: Using trigonometry, rx = r cosθ , or 6 m = (10 m )cosθ . This gives θ = 53.1°. Thus the y-component of the r position vector r is ry = r sin θ = (10 m )sin 53.1° = 8 m. Assess: The y-component is positive since the position vector is in the first quadrant.
3.7. Visualize: The figure shows the components vx and vy, and the angle θ.
Solve:
We have,
v y = − v sin 40°,
or
−10 m /s = − vsin 40°,
or
v = 15.56 m/s.
Thus the x-component is v x = v cos 40° = (15.56 m /s ) cos 40° = 11.9 m /s . Assess: Note that we had to insert the minus sign manually with v y since the vector is in the fourth quadrant.
3.8. Visualize:
r Solve: Vector E points to the left and up, so the components E x and E y are negative and positive, respectively, according to the Tactics Box 3.1. (a)
E x = − E cosθ
and
E y = E sin θ .
(b) E x = − E sin φ and E y = E cos φ . Assess: Note that the role of sine and cosine are reversed because we are using a different angle. θ and φ are complementary angles.
3.9. Visualize:
We will follow rules in the Tactics Box 3.1. r Solve: (a) Vector r points to the right and down, so the components rx and ry are positive and negative, respectively:
ry = − r sin θ = −(100 m )sin 45° = −70.7 m rx = r cosθ = (100 m )cos 45° = 70.7 m r (b) Vector v points to the right and up, so the components v x and v y are both positive:
v x = v cosθ = (300 m /s ) cos 20° = 282 m /s r (c) Vector a has the following components:
v y = v sin θ = (300 m /s)sin 20° = 103 m /s
a y = − a sin θ = −(5.0 m /s 2 )sin 90° = −5.0 m /s 2 a x = − a cosθ = −(5.0 m /s 2 )cos 90° = 0 m /s 2 r (d) Vector F points to the left and up, so the components Fx and Fy are negative and positive respectively:
Fx = − F cosθ = −(50 N )cos 36.9° = −40.0 N
Fy = + F sin θ = +(50 N )sin 36.9° = 30.0 N
Assess: The components have same units as the vectors. Note the minus signs we have manually inserted according to the Tactics Box 3.1.
3.10. Visualize:
We will follow the rules given in the Tactics Box 3.1. Solve: (a) rx = −(2 km) sin 30° = −1 km ry = (2 km) cos 30° = 1.73 km (b) vx = −(5 cm/s)sin 90° = −5 cm/s vy = (5 cm/s)cos 90° = 0 cm/s (c) a x = −(10 m /s 2 )sin 40° = −6.43 m /s 2 a y = −(10 m /s 2 )cos 40° = −7.66 m /s 2 (d) Fx = (50 N) sin 36.9° = 30.0 N Fy = (50 N) cos 36.9° = 40.0 N Assess: The components have the same units as the vectors. Note the minus signs we have manually inserted according to the Tactics Box 3.1.
3.11. Visualize:
r r The components v of the vector C and D , and the angles θ are shown. v Solve: For C we have Cx = −(3.15 m )cos15° = −3.04 m and Cy = (3.15 m )sin 15° = 0.815 m. For D we have Dx = 25.67 sin 30° = 12.84 and Dy = −25.67 cos 30° = −22.23. r r r Assess: The components of the vector C have the same units as C itself. Dx and Dy are unitless because D is without units. Note the minus signs we have manually inserted following rules of the Tactics Box 3.1.
3.12. Visualize:
The magnitude of the vector is E = ( E x ) 2 + ( E y ) 2 = (125 V/ m ) 2 + ( −250 V/ m ) 2 = 280 V/ m. In the r v expression for E, the − jˆ and +iˆ means that E is in quadrant IV. The angle θ is below the positive x-axis. We have:
Solve:
θ = tan −1
| Ey | Ex
250 V/ m −1 = tan −1 = tan 2 = 63.4° 125 V/ m
Assess: Since | E y | > | E x | , the angle θ made with the +x-axis is larger than 45°. θ = 45° for | E y | = | E x | .
3.13. Visualize:
Solve:
(a) Using the formulas for the magnitude and direction of a vector, we have:
A = ( 4)2 + ( −6)2 = 7.21
θ = tan −1
6 = tan −1 1.5 = 56.3° 4
r 80 m = 58.0° θ = tan −1 y = tan −1 50 m rx 40 (c) v = ( −20 m /s)2 + ( 40 m /s)2 = 44.7 m /s θ = tan −1 = tan −1 2 = 63.4° 20 2 (d) a = (2 m /s2 )2 + ( −6 m /s2 )2 = 6.3 m /s2 θ = tan −1 = tan −1 0.33 = 18.4° 6 Assess: Note that the angle θ made with the x-axis is smaller than 45° whenever | E y | < | E x | , θ = 45° for | E y | = | E x |, and θ > 45° for | E y | > | E x | . But, of course, θ in part (d) is with the y-axis, where the opposite of this rule applies. (b)
r = (50 m ) 2 + (80 m ) 2 = 94.3 m
3.14. Visualize:
Solve:
(a) Using the formulas for the magnitude and direction of a vector, we have:
4 = tan −1 1 = 45° 4 1 (b) r = ( −2 cm )2 + ( −1 cm )2 = 2.24 cm θ = tan −1 = tan −1 0.5 = 26.6° 2 100 2 2 (c) v = ( −10 mph ) + ( −100 mph ) = 100.5 mph θ = tan −1 = tan −1 10 = 84.3° 10 10 (d) a = (10 m /s 2 ) 2 + (20 m /s 2 ) 2 = 22.4 m /s 2 θ = tan −1 = tan −1 0.5 = 26.6° 20 Assess: Note that θ ≤ 45° when | E y | ≤ | E x |, where θ is the angle made with the x-axis. On the other hand, θ > 45° when | Ey | > | E x | .
B = ( − 4) 2 + ( 4) 2 = 5.66
θ = tan −1
3.15. Visualize:
r r r r r r r v v We have C = A − B or C = A + ( − B), where − B = ( B, direction opposite B). Look back at Tactics Box 1.2 which shows how to perform vector graphically. r subtraction r r r r Solve: To obtain vector C from A and B , we place the tail of − B on the tip of A , and then use the tip-to-tail rule of graphical addition.
r r
r
r
r
3.16. Visualize: The vectors A, B, and C = A + B are shown.
r r r r r (a) We have A = 5iˆ + 2 jˆ and B = −3iˆ − 5 jˆ. Thus, C = A + B = (5iˆ + 2 jˆ ) + ( −3iˆ − 5 jˆ ) = 2iˆ − 3 jˆ. r r r (b) Vectors rA , B , and C are shown with their tails together. v (c) Since C = 2iˆ − 3 jˆ = Cx iˆ + Cy jˆ, Cx = 2, and Cy = −3. Therefore, the magnitude and direction of C are Solve:
C = (2) 2 + ( −3) 2 = 3.61
θ = tan −1
| Cy | Cx
3 = tan −1 = 56.3° 2
r Assess: The vector C is to the right and down, thus implying a negative y-component and positive x-component, as obtained above. Also θ > 45° since | Cy | > | Cx | .
3.17. Visualize:
r r r r r v (a) We have A = 5iˆ + 2 jˆ, B = −3iˆ − 5 jˆ, and − B = +3iˆ + 5 jˆ. Thus, D = A + ( − B) = 8iˆ + 7 jˆ. r r r (b) Vectorsr A , B and D are shown in the above figure. r (c) Since D = 8iˆ + 7 jˆ = Dx iˆ + Dy jˆ, Dx = 8 and Dy = 7. Therefore, the magnitude and direction of D are Solve:
D = (8) 2 + (7) 2 = 10.6
D 7 θ = tan −1 y = tan −1 = 41.2° 8 Dx
Assess: Since | Dy | < | Dx | , the angle θ is less than 45°, as it should be.
3.18. Visualize:
r r r r Solve: (a) We have A = 5iˆ + 2 jˆ and B = −3iˆ − 5 jˆ. This means 2 A = 10iˆ + 4 jˆ and 3 B = −9iˆ − 15 jˆ. Hence, r r r E = 2 A + 3 B = 1iˆ − 11 jˆ. r r r (b) Vectors A , B, and E are shown in the above figure. r v (c) From the E vector, E x = 1 and E y = −11 . Therefore, the magnitude and direction of E are
E = (1) 2 + ( −11) 2 = 11.05
E 1 φ = tan −1 x = tan −1 = 5.19° | | 11 E y
Assess: Note that φ is the angle made with the y-axis, and that is why φ = tan −1 ( E x /| E y |) rather than tan −1 (| E y |/ E x ) which would be the case if φ were the angle made with the x-axis.
3.19. Visualize:
r r r r r r Solve: (a) We have A = 5iˆ + 2 jˆ and B = −3iˆ − 5 jˆ. This means − 4 B = +12iˆ + 20 jˆ. Hence, F = A − 4 B = 17iˆ + 22 jˆ = Fx iˆ + Fy jˆ with Fx = 17 and Fy = 22. r r r (b) The vectors A , B, and F are shown in the above figure. v (c) The magnitude and direction of F are F = Fx2 + Fy2 = (17) 2 + (22) 2 = 27.8
F 22 θ = tan −1 y = tan −1 = 52.3° 17 Fx Assess:
Fy > Fx implies θ > 45°, as is observed.
3.20. Solve: A different coordinate system can only mean a different orientation of the grid and a different origin of the grid. (a) False, because the size of a vector is fixed. (b) False, because the direction of a vector in space is independent of any coordinate system. (c) True, because the orientation of the vector relative to the axes can be different.
3.21. Visualize:
r r In coordinate system I, A = −( 4 m ) jˆ, so Ax = 0 m and Ay = −4 m. The vector B makes an angle of r v 120° clockwise from A, or it makes an angle of θ = 30° with the –x-axis. Since B points to the left and up, it has a negative x-component and a positive y-component. That is, Bx = −(5.0 m )cos 30° = −4.33 m and By = r + (5.0 m )sin 30° = 2.50 m. Thus, B = −( 4.33 m )iˆ + (2.50 m ) jˆ. r In coordinate system II, A points to the left and down, and makes an angle of 30° with the –y-axis. r Therefore, Ax = −(4.0 m ) sin 30° = −2.00 m and Ay = −( 4.0)cos 30° = −3.46 m. This implies A = −(2.00 m ) iˆ − r (3.46 m ) jˆ. The vector B makes an angle of 30° with the +y-axis and is to the left and up. This means, we have to Solve:
manually insert a minus sign with the x-component. Bx = − B sin 30° = −(5.0 m ) sin 30° = −2.50 m, and By = r + Bcos 30° = (5.0 m )cos 30° = 4.33 m. Thus B = −(2.50 m )iˆ + ( 4.33 m ) jˆ.
r
3.22. Visualize: Refer to Figure Ex. 3.22. The velocity vector v points west and makes an angle of 30° with r
the –x-axis. v points to the left and up, implying that v x is negative and v y is positive. Solve: We have v x = − v cos30° = −(100 m /s)cos 30° = −86.6 m /s and v y = + v sin 30° = (100 m /s)sin 30° = 50.0 m /s. r Assess: v x and v y have the same units as v.
3.23. Visualize:
r r r Solve: (a) Vector C is the sum of vectors A and B which is obtained using the tip-to-tip rule of graphical addition. Its magnitude is measured to be 4.7 and its angle made with the +x-axis is measured to be 33°. (b) Using the law of cosines, C 2 = A 2 + B 2 − 2 AB cos φ , and the geometry of parallelograms which shows that φ = 180° − (θ B − θ A ) = 180° − (60° − 20°) = 140°, we obtain
C = (3)2 + (2)2 − 2(3)(2)cos(140°) = 4.71 Using the law of sines:
sin α sin 140° ⇒ α = 15.8° = 2 4.71 Thus, θ C = α + 20° = 35.8°. (c) We have: Ax = A cos θ A = 3 cos 20° = 2.82
Ay = A sin θ A = 3 sin 20° = 1.03
Bx = B cos θ B = 2 cos 60° = 1.00
By = B sin θ B = 2 sin 60° = 1.73
This means: Cx = Ax + Bx = 3.82 and Cy = Ay + By = 2.76. The magnitude and direction of C are given by
C = Cx2 + Cy2 = (3.82) 2 + (2.76) 2 = 4.71
C 2.76 = 35.8° θ C = tan −1 y = tan −1 3.82 Cx
Assess: Using the method of vector components and their algebraic addition to find the resultant vector yields the same results as using the graphical addition of vectors.
3.24. Visualize:
Solve: (a) θ E = tan −1 11 = 45°
θ F = tan −1 21 = 63.4° Thus φ = 180° − θ E − θ F = 71.6° (b) From the figure, E = 2 and F = 5 . Using G 2 = E 2 + F 2 − 2 EF cos φ = ( 2 ) 2 + ( 5 ) 2 − 2( 2 )( 5 ) cos(180° − 71.6°) ⇒ G = 3.00. sin α sin(180° − 71.6°) Furthermore, using = ⇒ α = 45° 2.975 5 r Since θ E = 45° , the angle made by the vector G with the +x-axis is θ G = (α + θ E ) = 45° + 45° = 90°. (c) We have E x = +1.0,
and E y = +1.0
Fx = −1.0,
and Fy = +2.0
Gx = 0.0, ⇒G =
and Gy = 3.0
(0.0)2 + (3.0)2 = 3.0, and θ = tan −1
|Gy |
3.0 = tan −1 = 90° |G x | 0.0
r That is, the vector G makes an angle of 90° with the x-axis. Assess: The graphical solution and the vector solution give the same answer within the given significance of figures.
3.25. Visualize: Refer to Figure r P3.25 r r in your textbook. r
r r r Solve: (a) We are given that A + B + C = −2iˆ with A = 4iˆ, and C = −2 jˆ. This means A + C = 4iˆ − 2 jˆ. Thus, r r r r r r B = ( A + B + C ) − ( A + C ) = ( −2iˆ ) − ( 4iˆ − 2 jˆ ) = −6iˆ + 2 ˆj. r (b) We have B = Bx iˆ + By jˆ with Bx = −6 and By = 2. Hence, B = ( −6) 2 + (2) 2 = 6.32
θ = tan −1
By | Bx |
= tan −1
2 = 18.4° 6
r r Since B has a negative x-component and a positive y-component, the angle θ made by B is with the –x-axis and it is above the –x-axis. Assess: Since | By | < | Bx |, θ < 45° as is obtained above.
3.26. Visualize: Refer to Figure P3.26.
From the rules of trigonometry, we have Ax = 4 cos 40° = 3.06 and Ay = 4 sin 40° = 2.57. Also, Bx = r r r r r r r r r −2 cos10° = −1.97 and By = +2 sin 10°= 0.35. Since A + B + C = 0, C = − A − B = ( − A) + ( − B) = ( −3.06iˆ − 2.57 jˆ ) + s ( +1.97iˆ − 0.35 j ) = −1.09iˆ − 2.92 ˆj.
Solve:
3.27. Visualize:
r r In the tilted coordinate system, the vectors A and B are expressed as: r r A = (2 sin 15° m )iˆ + (2 cos15° m ) jˆ and B = ( 4 cos15° m )iˆ − ( 4 sin 15° m ) jˆ. r r r Therefore, D = 2 A + B = ( 4 m )[(sin 15° + cos15°)iˆ + (cos15° − sin 15°) jˆ ] = ( 4.90 m )iˆ + (2.83 m ) jˆ. The magnitude of this vector is D = 5.66 m, and it makes an angle of θ = tan −1 (0.707 m /1.225 m ) = 30° with the +x-axis. Assess: The resultant vector can be obtained graphically by using the rule of tail-to-tip addition. Solve:
3.28. Visualize:
(a)
r r r (b) The components of the vectors A, B, and C are Ax = (3 m ) cos 20° = 2.82 m and Ay = −(3 m ) sin 20°= −1.03 m; Bx = 0 m and By = 2 m; Cx = −(5 m ) cos 70° = −1.71 m and Cy = −(5 m ) sin 70°= −470 m. This means the vectors can be written, r r r A = (2.82 m )iˆ − (1.03 m ) jˆ B = (2 m ) jˆ C = ( −1.71 m )iˆ − ( 4.70 m ) jˆ r r r r (c) We have D = A + B + C = (1.11 m )iˆ − (3.73 m ) jˆ. This means Solve:
D = (1.11 m ) 2 + (3.73 m ) 2 = 3.89 m
θ = tan −1
r The direction of D is south of east, 73.4° below the positive x-axis.
3.73 = tan −1 3.36 = 73.4° 1.11
3.29. Visualize:
r r r Solve: Using the of tail-to-tip graphical addition, r method r r r ther diagram shows the resultant for D + E + F in (a), the resultant for D + 2 E in (b), and the resultant for D â&#x2C6;&#x2019; 2 E + F in (c).
r
r
3.30. Solve: We have E = E x iˆ + Ey jˆ = 2iˆ + 3 jˆ, which means E x = 2 and Ey = 3. Also, F = Fx iˆ + Fy jˆ = 2iˆ − 2 jˆ, which means Fx = 2 and Fy = −2. r v (a) The magnitude of E is given by E = E x2 + E y2 = (2) 2 + (3) 2 = 3.61 and the magnitude of F is given by
F = Fx2 + Fy2 = (2) 2 + (2) 2 = 2.83. r r v v (b) Since E + F = 4iˆ + 1 jˆ, the magnitude of E + F is ( 4) 2 + (1) 2 = 4.12. r r r r r (c) Since − E − 2 F = −(2iˆ + 3 jˆ ) − 2(2iˆ − 2 j ) = −6iˆ + 1 jˆ, the magnitude of − E − 2 F is
( −6) 2 + (1) 2 = 6.08.
3.31. Visualize:
r The magnitude of the unknown vector is 1 and its direction is along iˆ + jˆ . Let A = iˆ + jˆ as shown in the diagram. r r That is, A = 1iˆ + 1 jˆ and the x- and y-components of A are both unity. Since θ = tan −1 ( Ay / Ax ) = 45° , the unknown vector must make an angle of 45° with the +x-axis and have unit magnitude. r Solve: Let the unknown vector be B = Biˆ + By jˆ where
Bx = B cos 45° = r We want the magnitude of B to be 1, so we have
B=
1 1 B and By = B sin 45° = B 2 2
Bx2 + By2 = 1 2
2
1 1 ⇒ B + B = 1 ⇒ B 2 = 1 ⇒ B = 1 2 2 Hence,
Bx = Finally,
1 1 ⋅1 = 2 2
and By =
1 1 ⋅1 = 2 2
r 1 ˆ 1 ˆ B = Bx iˆ + By jˆ = i+ j 2 2
r
r
3.32. Solve: We have r = (5iˆ + 4 jˆ )t 2 m. This means that r does not change the ratio of its components as t
r v increases, that is, the direction of r is constant. The magnitude of r is given by r = (5t 2 ) 2 + ( 4 t 2 ) 2 m = (6.40t 2 ) m. (a) The particle’s distance from the origin at t = 0 s, t = 2 s, and t = 5 s is 0 m, 25.6 m, and 160 m. r r dr dt 2 (b) The particle’s velocity is v = = (5iˆ + 4 jˆ ) m /s = (5iˆ + 4 jˆ )2t m/s = (10iˆ + 8 jˆ )t m/s dt dt (c) The magnitude of the particle’s velocity is given by v = (10 t ) 2 + (8t ) 2 = 12.8t m /s. The particle’s speed at t = 0 s, t = 2 s, and t = 5 s is 0 m/s, 25.6 m /s , and 64.0 m /s .
3.33. Visualize: The coordinate system (x,y,z) is shown here. While +x denotes East and +y denotes North, the r r r r +z-direction is vertically up. The vectors Smorning (shortened as Sm ), Safternoon (shortened as Sa ), and the total r r r displacement vector Stotal = Sa + Sm are also shown.
r r Solve: Sm = (2000iˆ + 3000 jˆ + 200 kˆ ) m, and Sa = ( −1500iˆ + 2000 jˆ − 300 kˆ ) m (a) The sum of the z-components of the afternoon and morning displacements is Sza + Szm = −300 m + 200 m = −100 m, that is, 100 m lower. r r r ˆ (b) S = S + S = (500i + 5000 jˆ − 100 kˆ ) m, that is, (500 m east) + (5000 m north) – (100 m vertical). total
a
m
3.34. Visualize:
Only the minute hand ris shown in the figure.r Solve: (a) We have S8:00 = (2.0 cm ) jˆ and S8:20 = (2.0 cm )cos 30°iˆ − (2.0 cm )sin 30° jˆ. The displacement vector is r r r ∆r = S8:20 − S8:00
= (2.0 cm )[cos 30°iˆ − (sin 30° + 1) jˆ ] = (2.0 cm )[0.87iˆ − 1.50 jˆ ] = (1.74 cm )iˆ − (3.00 cm ) jˆ r r r r r (b) We have S8:00 = (2.0 cm ) jˆ and S9:00 = (2.0 cm ) jˆ. The displacement vector is ∆r = S9:00 − S8:00 = 0. Assess: The displacement vector in part (a) has positive x-component and negative y-component. The vector thus is to the right and points down, in the IV quadrant. This is what the vector drawn from the tip of the 8:00 a.m. arm to the tip of the 8:20 a.m. arm will be.
3.35. Visualize:
(a)
Note that +x is along the east and +y is along the north. r r r Solve: (b) We are given A = −(200 m ) j , and can use trigonometry to obtain B = −(283 m )iˆ − (283 m ) jˆ and r r r r r C = (100 m )iˆ + (173 m ) jˆ. We want A + B + C + D = 0. This means r r r r D = −A − B − C
= (200 m jˆ ) + (283 m iˆ + 283 m jˆ ) + ( −100 m iˆ − 173 m jˆ ) = 183 m iˆ + 310 m jˆ v The magnitude and direction of D are Dy 310 m D = (183 m ) 2 + (310 m ) 2 = 360 m, and θ = tan −1 = tan −1 = 59.4° Dx 183 m r This means D = (360 m, 59.4° north of east). r (c) The measured rlength of the vector D on the graph (with a ruler) is approximately 1.75 times the measured length of vector A . Since A = 200 m, this gives D = 1.75 × 200 m = 350 m. Similarly, the angle θ measured with the protractor is close to 60°. These answers are in close agreement to part (b).
3.36. Visualize:
Solve:
(a) The figure shows Sparky’s individual displacements and his net displacement.
r r r r (b) Dnet = D1 + D2 + D3 , where individual displacements are r D1 = (50 cos 45°iˆ + 50 sin 45° jˆ ) m = (35.4iˆ + 35.4 jˆ ) m r D2 = −70iˆ m r D3 = −20 jˆ m
Thus Sparky’s displacement is Dnet = ( −34.6i + 15.4 j ) m. (c) As a magnitude and angle,
(D ) θ net = tan −1 net y = 24.0°. |( Dnet ) x | Sparky’s net displacement is 37.9 m in a direction 24.0° north of west. Dnet = ( Dnet )2x + ( Dnet )2y = 37.9 m,
3.37. Visualize:
r r r Solve: We are given A = 5 m iˆ and C = ( −1 m )kˆ. Using trigonometry, B = (3 cos 45° m )iˆ − (3 sin 45° m ) jˆ. r r r r r The total displacement is r = A + B + C = (7.12 m )iˆ − (2.12 m ) jˆ − (1 m )kˆ. The magnitude of r is r =
(7.12) 2 + (2.12) 2 + (1) 2 m = 7.5 m. Assess: A displacement of 7.5 m is a reasonable displacement.
3.38. Model: Carlos will be represented as a particle and the particle model will be used for motion under constant acceleration kinetic equations. Visualize:
Solve:
Substituting the known values into the equation from kinematics:
r r r 1r r1 = r0 + v0 (t1 − t 0 ) + a (t1 − t 0 ) 2 2 r r r r = 0 + v0 (600 s − 0 s) + 0 = (600 s)v0
r Now v0 = v0 cos 25° iˆ + v0 sin 25° jˆ = (5 cos 25°iˆ + 5 sin 25° jˆ ) m/s = ( 4.53 iˆ + 2.11 jˆ ) m/s. r Thus the position r1 becomes r r1 = (600 s)( 4.53iˆ + 2.11 jˆ ) m/s = (2720iˆ + 1270 jˆ ) m That is, Carlos ends up 1270 m north of his starting position. Assess: The choice of our coordinate system is such that the x-component of the displacement is along the east and the y-component is along the north. The magnitude of 1270 m is reasonable.
3.39. Visualize:
r Solve: We have v = v x iˆ + v y jˆ = v||iˆ + v⊥ jˆ = v cosθiˆ + v sin θjˆ. Thus, v|| = v cosθ = (100 m /s)cos 30° = 86.6 m /s. Assess: For the small angle of 30°, the obtained value of 86.6 m/s for the horizontal component is reasonable.
3.40. Visualize:
2.5 m /s (a) Since v x = v cosθ , we have 2.5 m /s = (3.0 m /s)cosθ ⇒ θ = cos −1 = 33.6°. 3.0 m /s (b) The vertical component is v y = v sin θ = (3.0 m /s) sin 33.6° = 1.7 m /s. Solve:
r
3.41. Visualize: Establish a coordinate system with origin at the tree and with the x-axis pointing east. Let A r be a displacement vector directly from the tree to the treasure. Vector A is r A = (100iˆ + 500 jˆ ) paces.
This describes the displacement you would undergo by walking north 500 paces, then east 100 paces. Instead, you follow the road for 300 paces and undergo displacement r B = (300 sin 60°iˆ + 300 cos 60° jˆ ) paces
= (260iˆ + 150 jˆ ) paces.
Solve:
r Now let C be the displacement vector from your position to the treasure. From the figure r r r A = B + C.
So the displacement you need to reach the treasure is r r r C = A − B = ( −160iˆ + 350 jˆ ) paces. r If θ is the angle measured between C and the y-axis,
160 θ = tan −1 = 24.6° 350 You should head 24.6° west of north. You need to walk distance C = Cx2 + Cy2 = 385 paces to get to the treasure.
3.42. Visualize:
Solve: (a) The river is 100 m wide. If Mary rows due north at a constant speed of 2.0 m/s, it will take her 50 s to row across. But while she’s doing so, the current sweeps her boat sideways a distance 1 m/s × 50 s = 50 m. Mary’s net displacement is the vector sum of the displacement due to her rowing plus the displacement due to the river’s current. She lands 50 m east of the point that was directly across the river from her when she started. (b) Mary’s net displacement is shown on the figure.
3.43. Visualize: A 3% grade rises 3 m for every 100 m horizontal distance. The angle of the ground is thus α = tan −1 (3 /100) = tan −1 (0.03) = 1.72°. Establish a tilted coordinate system with one axis parallel to the ground and the other axis perpendicular to the ground.
Solve:
r From the figure, the magnitude of the component vector of v perpendicular to the ground is v⊥ = v sin α = 15.0 m /s.
r But this is only the size. We also have to note that the direction of v⊥ is down, so the component is v⊥ = −15.0 m /s.
3.44. Visualize:
r r r r r The resulting velocity is given by v = vfly + vwind , where vwind = 6 m/s iˆ and vfly = − v sin θ iˆ − v cosθ jˆ. r Substituting the known values we get v = −8 m/s sin θ iˆ − 8 m/s cosθ jˆ + 6 m/s iˆ. We need to have v x = 0. This 6 means 0 = −8 m /s sin θ + 6 m /s, so sin θ = and θ = 48.6°. Thus the ducks should head 48.6° west of south. 8 Solve:
3.45. Visualize:
The coordinate system used here is tilted with x-axis along the slope. Solve: The component of the velocity parallel to the x-axis is v|| = − v cos 70° = − v sin 20° = −10 m /s (.34) = −3.4 m /s . This is the speed down the slope. The component of the velocity perpendicular v⊥ = − v sin 70° = − v cos 20° = −10 m /s (.94) = −9.4 m /s. This is the speed toward the ground. Assess: A speed of approximately 10 m/s implies a fall time of approximately 1 second under free-fall. Note that g = −9.8 m/s2. This time is reasonable for a drop of approximately 5 m or 16 feet.
3.46. Model: The car is treated as a particle in this problem. Visualize:
Solve: (a) The tangential component is a|| = a sin 30° = (2 m /s 2 )(0.5) = 1.00 m /s 2 . (b) The perpendicular component is aâ&#x160;Ľ = a cos 30° = (2 m /s 2 )(0.866) = 1.73 m /s 2 . Assess: Magnitudes of the tangential and perpendicular components of acceleration are reasonable.
3.47. Model: We will treat the knot in the rope as a particle in static equilibrium. Visualize:
r r Expressing the vectors using unit vectors, we have F1 = 3iˆ and F2 = −5 sin 30°iˆ + 5 cos 30° jˆ. r r r r r r r Since F1 + F2 + F3 = 0, we can write F3 = − F1 − F2 = −0.5iˆ − 4.33 jˆ . r r The magnitude of F3 is given by F3 = ( −0.5) 2 + ( −4.33) 2 = 4.36 units. The angle F3 makes is θ = tan −1 ( 4.33 / 0.5) = 83.4° and is below the negative x-axis. Assess: The resultant vector has both components negative, and is therefore in quadrant rIII. Its magnitude and direction are reasonable. Note the minus sign that we have manually inserted with the force F2 .
Solve:
3.48. Visualize:
Note that the tilted coordinate system is such that x-axis is down the slope. r r r Solve: Expressing all three forces in terms of unit vectors, we have F1 = −(3 N )iˆ, F2 = +(6 N ) jˆ, and F3 = (5 N )sin θ iˆ − (5 N )cosθ jˆ. r r r r r (a) The component of Fnet parallel to the floor is ( Fnet ) x = ( F1 + F2 + F3 ) x = −(3 N )iˆ + (5 N )sin 30°iˆ = −(0.5 N ) iˆ. r r r r r (b) The component of Fnet perpendicular to the floor is ( Fnet ) y = ( F1 + F2 + F3 ) y = (6 N ) jˆ − (5 N )cos 30° jˆ = (1.67 N ) jˆ. r r (c) The magnitude of Fnet is Fnet = ( Fnet ) x + ( Fnet ) y = ( −0.5 N ) 2 + (1.67 N ) 2 = 1.74 N. The angle Fnet makes is
(F ) 1.67 N θ = tan −1 rnet y = tan −1 = 73.3° 0.5 N |( Fnet ) x |
r Fnet is 73.3° above the –x-axis in quadrant II.
3.49. Visualize:
Using trigonometry to calculate θ, we get θ = tan −1 (100 cm/141 cm) = 35.3°. r r r Expressing the three forces in unit vectors, FB = −(3 N )iˆ, FC = −(6 N ) jˆ, and FD = +(2 N )cos 35.3iˆ − (2 N )sin 35.3 jˆ = (1.63 N )iˆ − (1.16 N ) jˆ. r r r r r The total force is F = F + F + F = −1.37 N iˆ − 7.16 N jˆ. The magnitude of F is F =
Solve:
net
B
C
D
net
(1.37 N ) + (7.16 N ) = 7.29 N. 2
2
θ net = tan −1 r Fnet = (7.29 N,79.2° below −x in quadrant III).
|( Fnet ) y | |( Fnet ) x |
7.16 N = tan −1 = 79.2° 1.37 N
net