Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
Chapter 3. Technical Measurement and Vectors
Unit Conversions
3-1. A soccer field is 100 m long and 60 m across. What are the length and width of the field in feet?
(100 m)
100 cm 1 in.
1 ft
1 m 2.54 cm 12 in.
328 ft L = 328 ft
(60 m)
100 cm
1 m
1 in.
1 ft
2.54 cm 12 in.
197 ft W = 197 ft
3-2. A wrench has a handle 8 in. long. What is the length of the handle in centimeters?
(8 in.)
2.54 cm
1 in.
20.3 cm L = 20.3 cm
3-3. A 19-in. computer monitor has a viewable area that measures 18 in. diagonally. Express this distance in meters. .
(18 in.)
2.54 cm 1 m
1 in.
100 cm
0.457 m L = 0.457 m
3-4. The length of a notebook is 234.5 mm and the width is 158.4 mm. Express the surface area in square meters.
Area = (234.5 mm)(158.4 mm)
1 m 1 m
1000 mm 1000 mm
0.037
A = 0.0371 m
2
3-5. A cube has 5 in. on a side. What is the volume of the cube in SI units and in fundamental
USCS units? .
V
3
(5 in.) 125 in.
3
2.54 cm
1 in.
3
1 m
100 cm
3
0.00205 m
3
V = 0.00205 m
3
V
3
(125 in. )
1 ft
12 in.
3
0.0723 ft
3
V = 0.0723 ft
3
8

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
3-6. The speed limit on an interstate highway is posted at 75 mi/h. (a) What is this speed in kilometers per hour? (b) In feet per second?
(a) 75 mi 1.609 km h 1 mi
121 km/h (b) 75 mi 1 h 5280 ft h 3600 s 1 mi
= 110 ft/s
3-7. A Nissan engine has a piston displacement (volume) of 1600 cm
3
and a bore diameter of 84 mm. Express these measurements in cubic inches and inches. Ans. 97.6 in.
3
, 3.31 in.
(a) 1600 cm
3
1 in.
2.54 cm
3
= 97.6 in.
3
(b) 84 mm =
1 in.
25.4 mm
= 3.31 in.
3-8. An electrician must install an underground cable from the highway to a home located 1.20 mi into the woods. How many feet of cable will be needed?
1.2 mi
5280 ft
1 mi
6340 ft L = 6340 ft
3-9. One U.S. gallon is a volume equivalent to 231 in.
3
. How many gallons are needed to fill a tank that is 18 in. long, 16 in. wide, and 12 in. high?
V = (18 in.)(16 in.)(12 in.) = 3456 in.
3
Ans. 15.0 gal.
V
3
(3456 in. )
1 gal
231 in.
3
15.0 gal V = 15.0 gal
3-10. The density of brass is 8.89 g/cm
3
. What is the density in kg/m
3
? g 1 kg 100 cm
8.89
cm 3 1000 g 1 m
3 kg
8890 m 3
= 8890 kg/m
3
Addition of Vectors by Graphical Methods
3-11. A woman walks 4 km east and then 8 km north. (a) Use the polygon method to find her resultant displacement. (b) Verify the result by the parallelogram method.
R
Let 1 cm = 1 km; Then: R = 8.94 km, = 63.4
0
4 km
8 km
9

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
3-12. A land-rover, on the surface of Mars, moves a distance of 38 m at an angle of 180
0
. It then turns and moves a distance of 66 m at an angle of 270
0
. What is the displacement from the starting position?
Choose a scale, e.g., 1 cm = 10 m
38 m, 180
0
Draw each vector to scale as shown.
Measure R = 7.62 cm or R = 76.2 m
Measure angle = 60.1
0
S of W
= 180
0
+ 60.1
0
= 240.1
0
R = 76.2 m, 240.1
0
67 m
3-13. A surveyor starts at the southeast corner of a lot and charts the following displacements:
A = 600 m, N; B = 400 m, W; C = 200 m, S; and D = 100 m, E. What is the net
. B displacement from the starting point?
Choose a scale, 1 cm = 100 m
Draw each vector tail to tip until all are drawn.
C
D
R
A
Construct resultant from origin to finish.
R = 500 m, = 53.1
0
N of E or = 126.9
0
.
3-14. A downward force of 200 N acts simultaneously with a 500-N force directed to the left.
Use the polygon method to find the resultant force.
500 N
Chose scale, measure: R = 539 N, = 21.8
0
S. of E.
200 N
R
3-15. The following three forces act simultaneously on the same object. A = 300 N, 30
0
N of E;
B = 600 N, 270
0
; and C = 100 N due east. Find the resultant force
A using the polygon method. Choose a scale, draw and measure:
R = 576 N, = 51.4
0
S of E
B
C
R
10

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
3-16. A boat travels west a distance of 200 m, then north for 400 m, and finally 100 m at 30
0
S of E. What is the net displacement? (Set 1 cm = 100 N)
C
B R
Draw and measure: R = 368 N, = 108.0
0
A
3-17. Two ropes A and B are attached to a mooring hook so that an angle o 60 exists between the two ropes. The tension in rope A is 80 lb and the tension in rope B is 120 lb. Use the parallelogram method to find the resultant force on the hook.
R
Draw and measure: R = 174 lb
B
A
3-18. Two forces A and B act on the same object producing a resultant force of 50 lb at 36.9
0
N of W. The force A = 40 lb due west. Find the magnitude and direction of force B?
40 lb
Draw R = 50 lb, 36.9
0
N of W first, then draw 40 lb, W.
R
F = 30 lb, 90
0
36.9
0
F
Trigonometry and Vectors
3-19. Find the x and y-components of: (a) a displacement of 200 km, at 34
0
. (b) a velocity of 40 km/h, at 120
0
; and (c) A force of 50 N at 330 o
.
(a) D x
= 200 cos 34
0
= 166 km
D y
= 200 sin 34
0
= 112 km
(b) v x
= -40 cos 60
0
= -20.0 km/h
34
0
(a) v y
= 40 sin 60
0
= +34.6 km/h
(c) F x
= 50 cos 30
0
= 43.3 N; F y
= - 50 sin 30
0
= -25.0 N
60
0
(b)
30
0
(c)
11

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
3-20. A sled is pulled with a force of 540 N at an angle of 40
0
with the horizontal. What are the
540 N horizontal and vertical components of this force?
40
0
F x
= 540 cos 40
0
= 414 N F y
= 540 sin 40
0
= 347 N
(a)
3-21. The hammer in Fig. 3-26 applies a force of 260 N at an angle of 15
0
with the vertical. What is the upward component of the force on the nail?
F = 260 lb, = 75
0
; F xy
= 260 sin F y
= 251 N.
F
3-22. A jogger runs 2.0 mi west and then 6.0 mi north. Find the magnitude and direction of the resultant displacement.
2
R ( ) 6
2
6.32 mi tan =
6
2
; = 71.6
0
N of W
R
* 3-23. A river flows south with a velocity of 20 km/h. A boat has a maximum speed of 50 km/h in still water. In the river, at maximum throttle, the boat heads due west. What is the resultant speed and direction of the boat?
R
20 km/h
R (50)
2
( 20 )
2
.
tan =
20
;
50
0
= 21.8 S of W R = 53.9 km/h, 21.8
0
S of E
50 km/h
* 3-24. A rope, making an angle of 30
0
with the horizontal, drags a crate along the floor. What must be the tension in the rope, if a horizontal force of 40 lb is required to drag the crate?
F x
= F cos 30
0
; F
F x cos30
0
40 lb cos30
0
; F = 46.2 N
F
30
0
F x
12

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
* 3-25. An vertical lift of 80 N is needed to lift a window. A long pole is used to lift the window.
What force must be exerted along the pole if it makes an angle of 34
0
with the wall?
F y
= F sin 30
0
; F
F y sin 34
0
40 lb sin 34
0
; F = 96.5 N
80 N
34
0
F
* 3-26. The resultant of two forces A and B is 400 N at 210
0
. If force A is 200 N at 270
0
, what are the magnitude and direction of force B? ( = 210 - 180
0
= 30
0
)
B = -400 N cos 30
0
= -346 N: B = 346 N, 180
0 200 N
B
30
0
400 N
The Component Method of Vector Addition
3-27. Find the resultant of the following perpendicular forces: (a) 400 N, 0
0
; (b) 820 N, 270
0
; and (b) 500 N, 90
0
. Draw each vector, then find R: A
A x
= +400 N; B x
= 0; C x
= 0: R x
= +400 N
A y
= 0; B y
= -820 N; C y
= +500 N; R y
= 0 820 N + 500 N = -320 N
R
C
B
R 400
2
320
2
; tan
320
;
400
R = 512 N, 38.7
0
S of E
3-28. Four ropes, all at right angles to each other, pull on a ring. The forces are A = 40 lb, E;
B = 80 lb, N; C = 70 lb, W; and D = 20 lb, S. Find the resultant force on the ring.
A x
= +40 lb; B x
= 0; C x
= -70 lb D x
= 0:
R x
= +40 lb 70 lb = -30 lb
A y
= 0; B y
= +80 lb; C y
= 0; D y
= -20 lb ;
R y
= 0 + 80 lb - 20 lb = +60 lb
R 30
2
60
2
; tan
60
30
; R = 67.1 N, 116.6
0
A = 40 lb, E
13

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
*3-29. Two forces act on the car in Fig. 3-27. Force A is 120 N, west and force B is 200 N at 60
0
N of W. What are the magnitude and direction of the resultant force on the car?
A x
= -120; B x
= - (200 N) cos 60
0
= -100 N
B
R x
= 120 N - 100 N; R x
= -220 N
A y
= 0, B y
= 200 sin 60
0
= +173 N; R y
= 0 + 173 N = 173 N;
60
A
0
Thus, R x
= -100 N, R y
= +173 N and R
2 2
(220) (173) 280 N
Resultant direction: tan
173
210
;
0
38.2 N of W ; R = 280 N, 141.8
0
*3-30. Suppose the direction of force B in Problem 3-29 is reversed (+180
0
) and other parameters are unchanged. What is the new resultant? (This result is the vector difference A B).
The vector B will now be 60
0
S of E instead of N of E.
A x
= -120 N; B x
= +(200 N) cos 60
0
= +100 N
A
60
B
0
R x
= 120 N + 100 N; R x
= -20 N
A y
= 0, B y
= -200 sin 60
0
= -173 N; R y
= 0 - 173 N = -173 N;
Thus, R x
= -20 N, R y
= -173 N and R
2 2
( 20) (173) 174 N
Resultant direction:
R
2 tan
( 62.6) (30.4)
2
173
;
20
R = 69.6 lb
0 tan
S of W ; R = 174 N, 253.4
0
*3-31. Determine the resultant force on the bolt in Fig. 3-28. ( A x
B x
= -40 cos 20
0
= -37.6 lb; B x
= -50 cos 60
0
= -25.0 lb
R x
= 0 37.6 lb 25.0 lb; R x
= -62.6 lb
A y
= +60 lb; B y
= 40 sin 20
0
= 13.7 lb; C y
= 50 sin 60 = -43.3 lb
B = 40 lb
20
0
60
0
C = 50 lb
R y
= 60 lb 13.7 lb 43.3 lb; R y
= +30.4 lb
30.4
62.6
; = 25.9
0
N of W
14

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
*3-32. Determine the resultant of the following forces by the component method of vector addition: A = (200 N, 30
0
); B = (300 N, 330
0
; and C = (400 N, 250
0
).
A x
= 200 cos 30
0
= 173 N; B x
= 300 cos 30
0
= 260 N
C x
= -400 cos 70
0
= -137 N; R x
= F x
= 296 N
70
0
A y
= 200 sin 30
0
= 100 N; B y
= 300 sin 30
0
= -150 N
C
C y
= -400 sin 70
0
= -376 N; R y
= F y
= -430 N
A
30
0
30
0
B
R
2
(296) ( 430)
2
; tan
426
;
296
R = 519 N, 55.2
0
S of E
*3-33. Three boats exert forces on a mooring hook as shown in Fig. 3-29. Find the resultant of these three forces.
A x
= 420 cos 60
0
= +210 N; C x
= -500 cos 40
0
= -383 N
B x
= 0; R x
= 210 N + 0 383 N; R x
= -173 N
A y
= 420 sin 60
0
= 364 N; B y
= 150;
Cy = 500 sin 40
0
= 321 N R y
= F y
= 835 N;
500 N
C
40
0
B
150 N
A
420 N
60
0
R (
2
173 ) (835)
2
; tan
835
173
; R = 853 N, 78.3
0
N of W
Challenge Problems
3-34. Find the horizontal and vertical components of the following vectors: A = (400 N, 37
0
);
B = 90 m, 320
0
); and C = (70 km/h, 150
0
).
A
B x x
= 400 cos 37
= 90 cos 40
0
0
= 319 N; A y
= 400 sin 37
= 68.9 N; B y
= 90 sin 40
0
= 241 N
0
= 57.9
C x
= -70 cos 30
0
= -60.6 N; C y
= 70 sin 30
0
= 25.0 N
70 N
30
0
C
400 N
A
37
0
40
0
B
90 N
15

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
3-35. A cable is attached to the end of a beam. What pull at an angle of 40
0
with the horizontal is needed to produce an effective horizontal force of 200 N?
P cos 40
0
= 200 N; P = 261 N
P
40
0
3-36. A fishing dock runs north and south. What must be the speed of a boat heading at an angle of 40
0
E of N if its velocity component along the dock is to be 30 km/h?
v cos 40
0
= 30 km/h; v = 39.2 km/h
3-37. Find the resultant R = A + B for the following pairs of forces: A = (520 N, south); B =
269 N, west; (b) A = 18 m/s, north; B = 15 m/s, west.
269 N
(a) R (
2
269 ) (520)
2
585 N
B tan
520
295
; R = 585 N, = 62.6
0
S of W
(a)
R
A
520 N
R
A
18 m/s
B
15 m/s
(b)
(b)
2
R ( 15 ) ( 18 )
2
; tan
18
15
; R = 23.4 N, 50 .2
0
N of W
*3-38. Determine the vector difference (A B) for the pairs of forces in Problem 3-37.
(a)
2
R ( 269 ) (520)
2 tan
520
;
295
585 N
R = 585 N, 62.6
0
S of E
Change B into the vector B, then ADD :
(a) -B = 269 N A
R
18 m/s
A
520 N
R
(b) -B =15 m/s
(b)
2
R ( 15 ) ( 18 )
2
; tan
18
15
; R = 23.4 N, 50.2
0
N of E
*3-39. A traffic light is attached to the midpoint of a rope so that each segment makes an angle of 10
0
with the horizontal. The tension in each rope segment is 200 N. If the resultant force at the midpoint is zero, what must be the weight of the traffic light?
R x
= F x
= 0; T sin 10
0
+ T sin 10
0
W = 0;
T
T
2(200) sin 10
0
= W: W = 69.5 N W
16

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
*3-40. Determine the resultant of the forces shown in Fig. 3-30
B = 200 lb
R x
= 420 N 200 cos 70
0
410 cos 53
0
= 105 lb
R y
= 0 + 200 sin 70
0
410 sin 70
0
= -139.5 lb
70
0
53
0
A = 420 lb
2
R R x
R
2 y
R = 175 lb; = 306.9
0 C = 410 lb
*3-41. Determine the resultant of the forces shown in Fig. 3-31.
R x
= 200 cos 30
0
300 cos 45
0
155 cos 55
0
= 128 N
R y
= 0 + 200 sin 70
0
410 sin 70
0
= -185 N;
B = 300 N
A = 200 N
45
0
55
0
30
0
2
R R R x
2 y
R = 225 N; = 124.6
0 C = 155 N
*3-42. A 200-N block rests on a 30
0
inclined plane. If the weight of the block acts vertically downward, what are the components of the weight down the plane and perpendicular to the plane? Choose x-axis along plane and y-axis perpendicular.
30
0
W x
= 200 sin 30
0
; W x
= 173 N, down the plane.
W y
= 200 sin 60
0
; W x
= 100 N, normal to the plane.
30
0
W
*3-43. Find the resultant of the following three displacements: A = 220 m, 60
0
; B = 125 m, 210
0
; and C = 175 m, 340
0
.
A
A x
= 220 cos 60
0
= 110 m; A y
= 220 sin 60
0
= 190.5 m
B x
= 125 cos 210
0
= -108 m; B y
= 125 sin 210
0
= -62.5 m
C x
= 175 cos 340
0
= 164.4 m; C y
= 175 sin 340
0
= -59.9 m
R x
= 110 m 108 m + 164.4 m; R y
= 190.5 m 62.5 m 59.9 m ;
30
0
B
60
0
20
0
C
R x
= 166.4 m; R y
= 68.1 m R ( . )
2
( . )
2
180 m tan ; 22 3
0
; R = 180 m, = 22.3
0
17

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
Critical Thinking Questions
3-44. Consider three vectors: A = 100 m, 0
0
; B = 400 m, 270
0
; and C = 200 m, 30
0
. Choose an appropriate scale and show graphically that the order in which these vectors is added does not matter, i.e., A + B + C = C + B + A. Is this also true for subtracting vectors? Show graphically how A C differs from C A.
A + B + C C + B + A A C C - A
3-45. Two forces A = 30 N and B = 90 N can act on an object in any direction desired. What is the maximum resultant force? What is the minimum resultant force? Can the resultant force be zero?
Maximum resultant force occurs when A and B are in same direction.
A + B = 30 N + 90 N = 120 N
Minimum resultant force occurs when A and B are in opposite directions.
B - A = 90 N 30 N = 60 N
No combination gives R = 0.
*3-46. Consider two forces A = 40 N and B = 80 N. What must be the angle between these two forces in order to produce a resultant force of 60 N?
Since R = C = 60 N is smaller than 80 N, the angle
between A and B must be > 90
0
. Applying the law of
C = 60 N
B = 40 N
A = 80 N cosines to the triangle, we can find and then .
18

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
C
2
= A
2
+ B
2
2AB Cos ; (60)
2
= (80)
2
+ (40)
2
2(80)(40) Cos ; = 46.6
0
.
The angle between the direction of A and the direction of B is = 180
0
- ; = 133.4
0
.
*3-47. What third force F must be added to the following two forces so that the resultant force is zero: A = 120 N, 110
0
and B = 60 N, 200
0
?
Components of A: A x
= 120 Cos 110
0
= -40.0 N; A y
= 120 Sin 110
0
= 113 N
Components of B: B x
= 60 Cos 200
0
= -56.4 N; B y
= 60 Sin 200
0
= -20.5 N
R x
= 0; R x
= A x
+ B x
+ F x
= 0; R x
= -40.0 N 56.4 N + F x
= 0; Or F x
= +97.4 N
R y
= 0; R y
= A y
+ B y
+ F y
= 0; R y
= 113 N 20.5 N + F y
= 0; Or F y
= -92.2 N
F ( . )
2
( . )
2
131 N tan ; 43 3
0
And = 360
0
43.4
0
Thus, the force F has a magnitude and direction of: F = 134 N, = 316.6
0
*3-48. An airplane needs a resultant heading of due west. The speed of the plane is 600 km/h in still air. If the wind has a speed of 40 km/h and blows in a direction of 30
0
S of W, what direction should the aircraft be pointed and what will be its speed relative to the ground?
From the diagram, R x
= R, R y
= 0, So that A y
+ B y
= 0.
A y
= 600 sin y
= -40 sin 30
0
= -20 km/h
A = 600 km/h
R
600 sin - 20 = 0; 600 sin = 20
1 91
0
N of W (direction aircraft should be pointed)
B = 40 km/h sin
20
600
;
Noting that R = R x
and that A x
+B x
= R x
, we need only find sum of x-components.
A x
= -600 cos 599.7 km/h x
= 40 Cos 30
0
= -34.6 km/h
R = -599.7 km/h 34.6 km/h; R = -634 km/h. Thus, the speed of the plane relative to the ground is 634 km/h, 0
0
; and the plane must be pointed in a direction of 1.91
0
N of W.
19

Chapter 3 Technical Measurement and Vectors Physics, 6 th
Edition
*3-49. What are the magnitude F and direction of the force needed to pull the car of Fig. 3-32 directly east with a resultant force of 400 lb?
R x
= 400 lb and R y
= 0; R x
= A x
+ F x
= 400 lb
200 Cos 20
0
+ F x
= 400 lb
F x
= 400 lb 200 Cos 20
0
= 212 lb
R y
= 0 = A y
+ F y
; A y
= -200 sin 20
0
= -68.4 lb
F y
= -A y
= +68.4 lb; So, F x
= 212 lb and F y
= +68.4 lb
20
0
A = 200 lb
2
F ( 212 ) ( 68 4
2
; tan =
68.4
; R = 223 lb, 17.9
0
N of E
212
F
E
20

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