Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
Chapter 3. Technical Measurement and Vectors
Unit Conversions
3-1. What is the height in centimeters of a woman who is 5 feet and 6 inches tall? Ans. 168 cm.
5 feet, 6 inches = 66 in.;
 2.54 cm 
(66 in.) 
  168 cm
 1 in. 
3.2 A single floor tile measures 8 in. on each side. If the tiles are laid side by side, what distance
in meters can be covered by a single row of 20 tiles?
 8 in.  2.54 cm  1 m 
20 tile 


  4.06 m
 tile  1 in.  100 cm 
3-3. A soccer field is 100 m long and 60 m across. Find the length and width of the field in feet?
 100 cm   1 in.   1 ft 
(100 m) 


  328 ft
 1 m   2.54 cm   12 in. 
L = 328 ft
 100 cm  1 in.  1 ft 
(60 m) 


  197 ft
 1 m  2.54 cm  12 in. 
W = 197 ft
3-4. A wrench has a handle 8 in. long. What is the length of the handle in centimeters?
 2.54 cm 
(8 in.) 
  20.3 cm
 1 in. 
L = 20.3 cm
3-5. A 19-in. computer monitor has a viewable area that measures 18 in. diagonally. Express this
distance in meters.
.
 2.54 cm  1 m 
(18 in.) 

  0.457 m
 1 in.  100 cm 
L = 0.457 m
3-6. The length of a notebook is 234.5 mm and the width is 158.4 mm. Express the surface area
in square meters.
 1 m  1 m 
Area = (234.5 mm)(158.4 mm) 

  0.037
 1000 mm  1000 mm 
8
A = 0.0371 m2
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
3-7. A cube has 5 in. on a side. What is the volume of the cube in SI units and in fundamental
USCS units?
.
 2.54 cm 
V  (5 in.)  125 in. 

 1 in. 

3
3

3
3
 1m 
3

  0.00205 m
100
cm


V = 0.00205 m3
3
 1 ft 
3
V  (125 in. ) 
  0.0723 ft
 12 in. 
3
V = 0.0723 ft3
3-8. 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 

  121 km/h
h  1 mi 
(b) 75
mi  1 h   5280 ft 


 = 110 ft/s
h  3600 s   1 mi 
3-9. A Nissan engine has a piston displacement (volume) of 1600 cm3 and a bore diameter of 84
mm. Express these measurements in cubic inches and inches.
3
 1 in. 
(b) 84 mm = 
 = 3.31 in.
 25.4 mm 
 1 in. 
3
(a) 1600 cm 
 = 97.6 in.
2.54
cm



3

Ans. 97.6 in.3, 3.31 in.
3-10. 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?
 5280 ft 
1.2 mi 
  6340 ft
 1 mi 
L = 6340 ft
3-11. 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
 1 gal 
V  (3456 in.3 ) 
 15.0 gal
3 
 231 in. 
V = 15.0 gal
9
Ans. 15.0 gal.
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
3-12. The density of brass is 8.89 g/cm3. What is the density in kg/m3?
g   1 kg   100 cm 
kg


 8.89 3  
  8890 3
cm   1000 g   1 m 
m

3
 = 8890 kg/m3
Addition of Vectors by Graphical Methods
3-13. 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
8 km

Let 1 cm = 1 km; Then:
R = 8.94 km,  = 63.4
4 km
3-14. A land-rover, on the surface of Mars, moves a distance of 38 m at an angle of 1800. It then
0
turns and moves a distance of 66 m at an angle of 2700. What is the displacement from
the starting position?
Choose a scale, e.g., 1 cm = 10 m
38 m, 1800
Draw each vector to scale as shown.
Measure R = 7.62 cm or

67 m
R = 76.2 m
R
Measure angle  = 60.1 S of W
0
 = 1800 + 60.10 = 240.10
R = 76.2 m, 240.10
3-15. 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
displacement from the starting point?
.
Choose a scale, 1 cm = 100 m
B
C
Draw each vector tail to tip until all are drawn.
Construct resultant from origin to finish.
R = 500 m,  = 53.10 N of E or  = 126.90.
10
A
D
R

Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
3-16. A downward force of 200 N acts simultaneously with a 500-N force directed to the left.
500 N

Use the polygon method to find the resultant force.
Chose scale, measure: R = 539 N,  = 21.80 S. of E.
200 N
R
3-17. The following three forces act simultaneously on the same object. A = 300 N, 300 N of E;
A
B = 600 N, 2700; and C = 100 N due east. Find the resultant force
using the polygon method.
Choose a scale, draw and measure:
R
B
R = 576 N,  = 51.40 S of E
C
3-18. A boat travels west a distance of 200 m, then north for 400 m, and finally 100 m at 300 S
C
of E. What is the net displacement? (Set 1 cm = 100 N)
Draw and measure:
R = 368 N,  = 108.00
B
R

A
3-19. 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.
Draw and measure:
R
R = 174 lb
B
A
3-20. Two forces A and B act on the same object producing a resultant force of 50 lb at 36.90 N
of W. The force A = 40 lb due west. Find the magnitude and direction of force B?
40 lb
Draw R = 50 lb, 36.90 N of W first, then draw 40 lb, W.
R
F = 30 lb, 900
36.90
11
F
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
Trigonometry and Vectors
3-21. Find the x and y-components of: (a) a displacement of 200 km, at 340. (b) a velocity of 40
km/h, at 1200; and (c) A force of 50 N at 330o.
(a) Dx = 200 cos 340 = 166 km
300
Dy = 200 sin 340 = 112 km
600
340
(b) vx = -40 cos 600 = -20.0 km/h
(c) Fx = 50 cos 300 = 43.3 N;
(c)
(b)
(a)
vy = 40 sin 600 = +34.6 km/h
Fy = - 50 sin 300 = -25.0 N
3-22. A sled is pulled with a force of 540 N at an angle of 400 with the horizontal. What are the
540 N
horizontal and vertical components of this force?
Fx = 540 cos 400 = 414 N
400
Fy = 540 sin 400 = 347 N
(a)
3-23. The hammer in Fig. 3-26 applies a force of 260 N at an angle of 150 with the vertical. What
is the upward component of the force on the nail?
F = 260 lb,  = 750;
F
Fxy = 260 sin  Fy = 251 N.

3-24. A boy attempts to lift his sister from the pavement. If the vertical component of his pull F
has a magnitude of 110 N and the horizontal component has a magnitude of 214 N, what
are the magnitude and direction of the force F?
R  (214)2  (110)2  241 N;
tan =
Fy
110 N
110
;  = 27.20 N of W
214
F

Fx = 214 N
R = 53.9 km/h, 21.80 S
of E
12
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
3-25. 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
R  (50) 2  (20) 2  53.9 km / h;
20
tan = ;  = 21.80 S of W
50
20 km/h
50 km/h
0
R = 53.9 km/h, 21.8 S
of E
* 3-26. A rope, making an angle of 300 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?
Fx = F cos 300;
F
F
Fx
40 lb

; F = 46.2 N
0
cos30
cos300
300
Fx
* 3-27. 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 340 with the wall?
Fy = F sin 300;
F
Fy
sin 34
0

40 lb
;
sin 34 0
F = 96.5 N
80 N
340
F
* 3-28. The resultant of two forces A and B is 400 N at 2100. If force A is 200 N at 2700, what are
the magnitude and direction of force B? ( = 210 - 1800 = 300)
0
B = -400 N cos 30 = -346 N:
B = 346 N, 180
13
0
200 N
B
300
400 N
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
The Component Method of Vector Addition
3-29. Find the resultant of the following perpendicular forces: (a) 400 N, 00; (b) 820 N, 2700;
and (b) 500 N, 900.
Ax = +400 N; Bx = 0; Cx = 0:
Rx = +400 N
Ay = 0; By = -820 N; Cy = +500 N;
R
A

Draw each vector, then find R:
 400   320
2
2
tan  
;
R
B
Ry = 0 – 820 N + 500 N = -320 N
320
;
400
C
R = 512 N, 38.70 S of E
3-30. 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.
Ax = +40 lb; Bx = 0; Cx = -70 lb Dx = 0:
C = 70 lb, W
D = 20 lb, S
Rx = +40 lb – 70 lb = -30 lb
R

Ay = 0; By = +80 lb; Cy = 0; Dy = -20 lb ;
A = 40 lb, E
Ry = 0 + 80 lb - 20 lb = +60 lb
R
 30   60
2
2
;
tan  
B = 80 lb, E
60
;
30
R = 67.1 N, 116.60
*3-31. Two forces act on the car in Fig. 3-27. Force A is 120 N, west and force B is 200 N at 600
N of W. What are the magnitude and direction of the resultant force on the car?
Ax = -120;
Bx = - (200 N) cos 600 = -100 N
Rx = – 120 N - 100 N; Rx = -220 N
0
Ay = 0, By = 200 sin 60 = +173 N;
B
Ry = 0 + 173 N = 173 N;
Thus, Rx = -100 N, Ry = +173 N and
Resultant direction:
tan  
R
600 
A
R  (220) 2  (173) 2  280 N
173
;   38.20 N of W ;
210
14
R = 280 N, 141.80
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
*3-32. Suppose the direction of force B in Problem 3-29 is reversed (+1800) and other parameters
are unchanged. What is the new resultant? (This result is the vector difference A – B).
The vector B will now be 600 S of E instead of N of E.
Ax = -120 N;
A
Bx = +(200 N) cos 600 = +100 N
600 
Rx = – 120 N + 100 N; Rx = -20 N
Ay = 0, By = -200 sin 600 = -173 N;
Thus, Rx = -20 N, Ry = -173 N and
Resultant direction:
tan  
-B
Ry = 0 - 173 N = -173 N;
R
R  (20) 2  (173) 2  174 N
173
;   83.4 0 S of W ;
20
R = 174 N, 253.40
*3-33. Determine the resultant force on the bolt in Fig. 3-28. ( Ax = 0 )
600
Bx = -40 cos 200 = -37.6 lb; Bx = -50 cos 600 = -25.0 lb
Rx = 0 – 37.6 lb – 25.0 lb;
Rx = -62.6 lb
C
50 lb
200
B
40 lb
A = 60 lb
R

Ay = +60 lb; By = 40 sin 200 = 13.7 lb; Cy = 50 sin 60 = -43.3 lb
Ry = 60 lb – 13.7 lb – 43.3 lb; Ry = +30.4 lb
R  (62.6) 2  (30.4) 2
R = 69.6 lb
tan  
30.4
;
62.6
 = 25.90 N of W
*3-34. Determine the resultant of the following forces by the component method of vector
addition: A = (200 N, 300); B = (300 N, 3300; and C = (400 N, 2500).
Ax = 200 cos 300 = 173 N; Bx = 300 cos 300 = 260 N
Cx = -400 cos 700 = -137 N; Rx = Fx = 296 N
Ay = 200 sin 300 = 100 N; By = 300 sin 300 = -150 N
Cy = -400 sin 700 = -376 N; Ry = Fy = -430 N
R  (296) 2  (430) 2 ;
tan  
426
;
296
15
R = 519 N, 55.20 S of E
A
300

R
0
B 30
700
C
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
*3-35. Three boats exert forces on a mooring hook as shown in Fig. 3-29. Find the resultant of
these three forces.
C = 500 N
Ax = 420 cos 600 = +210 N; Cx = -500 cos 400 = -383 N
Bx = 0;
400
Rx = 210 N + 0 –383 N; Rx = -173 N
Ay = 420 sin 600 = 364 N; By = 150;
R  ( 173) 2  (835) 2 ;

Ry = Fy = 835 N;
Cy = 500 sin 400 = 321 N
tan  
835
;
173
150 N
B
R
420 N
A
600
R = 853 N, 78.30 N of W
Vector Difference
3-36. Two displacements are: A = 9 m, E and B = 12 m, S. Compare the magnitude and
direction of (A + B) and (A - B).
A
Ax = 9 m; Bx = 0 m;
Rx = Fx = 9 m
tan  
-B
+B
Ay = 0 m; By = -12 m; Ry = Fy = -12 m
R  (9) 2  (12) 2  15.0 m ;
A-B
12
;
9
For A – B, Ry = Fy = -(-80 N) or + 80 N;
A+B
A
A + B = 15.0 m, 53.1 S of E
A - B = 15.0 m, 53.10 N of E
3-37. Given that A = 24 m, W; B = 50 m, N. Find the magnitude and direction of (a) A + B
-A
and (b) B – A.
Ax = -24 N; Bx = 0;
Rx = Fx = -24 N
+B
Ay = -0 ; By = +50 N; Ry = Fy = +50 N
R  (24) 2  (50) 2  55.5 m ;
tan  
50
;
24
For B – A, Rx = Fx = 0 – (-24 m) or + 24 N;
16
A+B

+B

B-A
+A
A + B = 55.5 m, 64.40 N of W
B - A = 55.5 N, 64.40 N of E
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
3-38. Velocity has a magnitude and a direction that can be represented by a vector. Consider a
boat moving initially with a velocity of 30 m/s directly west. At some later instant the
boat is found to have a velocity of 12 m/s at 300 S of W. What is the change in velocity?
Solution. Each velocity is a Vector: A = 30 m/s, W and B = 12 m/s, 300 S of W
Ax = -30 m/s; (-B)x = 12 cos 300 = +10.4 m/s
Ay = 0; (-B)y = 12 sin 300 = 6.00 m/s
-B30
300
+B
Rx = Fx = -30 m/s + 10.4 m/s; Rx = -19.6 m/s
A-B
0

300
+A
Ry = Fy = 0 + 6.00 m/s;
Ry = +6.00 m/s
R  (19.6) 2  (6) 2  20.5 m/s ; tan  
6
;
19.6
B - A = 20.5 m/s, 17.00 N of W
3-39. Consider four vectors: A = 450 N, W; B = 160 N, 440 N of W; C = 800 N, E; and D =
100 N, 349 N. of E. Determine the magnitude and direction of A – B + C – D. Draw the
vector polygon.
Ans. 21.30 S of W
Ax = -450 N; Ay = 0:
Cx = +800 N; Cy = 0:
A
B
D
440
340
0
Bx = -(160 N)(cos 44 ) = -115 N;
C
By = +(160 N)(sin 440) = 111 N;
Dx = (100 N)(cos 340) = 82.9 N;
Dy = (100 N)(sin 340) = 55.9 N;
(Note vectors –B and –D as drawn in polygon below.)
A
A–B+C-D
440
-B
Rx = Ax – (Bx) + Cx – (Dx) Be careful of signs
Rx = -450 N – (- 115 N) + 800 N – (+82.9 N)
C
340
-D
Rx = -450 N + 115 N + 800 N – 82.9 N
Ry = Ay – (By) + Cy – (Dy) = 0 – (111 N) + 0 – (55.9 N);
R  (382) 2  (167) 2  417 N ; tan  
17
167
;
382
Rx = 382 N
Ry = -167 N
R = 417 N, 17.00 S of E
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
Additional Problems
3-40. Find the horizontal and vertical components of the following vectors: A = (400 N, 370);
B = 90 m, 3200); and C = (70 km/h, 1500).
Ax = 400 cos 370 = 319 N;
Ay = 400 sin 370 = 241 N
Bx = 90 cos 400 = 68.9 N;
By = 90 sin 400 = 57.9
Cx = -70 cos 300 = -60.6 N;
400 N
70 N
A
C
370
300
400
90 N
Cy = 70 sin 300 = 25.0 N
B
3-41. A cable is attached to the end of a beam. What pull at an angle of 400 with the horizontal
is needed to produce an effective horizontal force of 200 N?
P
400
0
P cos 40 = 200 N;
P = 261 N
3-42. A fishing dock runs north and south. What must be the speed of a boat heading at an
angle of 400 E of N if its velocity component along the dock is to be 30 km/h?
v cos 400 = 30 km/h;
v = 39.2 km/h
3-43. 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  ( 269) 2  (520) 2  585 N
B
(a)

A
tan  
520
; R = 585 N,  = 62.60 S of W
295
(b) R  ( 15) 2  (18) 2 ;
tan  
18
;
15
18
R
R
520 N

B
15 m/s
R = 23.4 N, 50 .20 N of W
A
18 m/s
(b)
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
*3-44. Determine the vector difference (A – B) for the pairs of forces in Problem 3-37.
(a) R  (269) 2  (520) 2  585 N
tan  
(a)

520
;
295
520 N
tan  
18
;
15
18 m/s
R

A
R = 585 N, 62.60 S of E
(b) R  (15) 2  (18) 2 ;
A
-B = 269 N
R
(b) -B =15 m/s
R = 23.4 N, 50.20 N of E
*3-45. A traffic light is attached to the midpoint of a rope so that each segment makes an angle
of 100 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?
Rx = Fx = 0; T sin 100 + T sin 100 – W = 0;
2(200) sin 100 = W:
T
T
W = 69.5 N
W
*3-46. Determine the resultant of the forces shown in Fig. 3-30
C = 410 lb
Rx = 420 N – 200 cos 700 – 410 cos 530 = 105 lb
530
B = 200 lb
70
Ry = 0 + 200 sin 700 – 410 sin 700 = -139.5 lb
0
A = 420 lb
R
R = 175 lb;  = 306.90
R  Rx2  Ry2
*3-47. Determine the resultant of the forces shown in Fig. 3-34.
Rx = 150 cos 300 – 200 cos 450 – 240 cos 550 = -281 N
B = 300 N
270
C = 155 N
Ry = 150 sin 55 + 200 sin 40 – 240 sin 27 = 142 N;
0
R  Rx2  Ry2
0
400
0
R = 315 N; = 26.90of W,  = 153.10
19
R

550
A = 150 N
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
*3-48. A 200-N block rests on a 300 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.
300
Wx = 200 sin 300;
Wy = 200 sin 600;
Wx = 173 N, down the plane.
W
300
Wx = 100 N, normal to the plane.
*3-49. Find the resultant of the following three displacements: A = 220 m, 600; B = 125 m, 2100;
and C = 175 m, 3400.
A
0
0
Ax = 220 cos 60 = 110 m; Ay = 220 sin 60 = 190.5 m
600
Bx = 125 cos 2100 = -108 m; By = 125 sin 2100 = -62.5 m
300
Rx = 110 m –108 m + 164.4 m; Ry = 190.5 m – 62.5 m – 59.9 m ;
B
300
R  (166.4) 2  (681
. ) 2  180 m
200
tan  
681
.
;   22.30 ;
166.4
C
B
Cx = 175 cos 3400 = 164.4 m; Cy = 175 sin 3400 = -59.9 m
Rx = 166.4 m; Ry = 68.1 m
200
A
C
R
R = 180 m,  = 22.30
600
Critical Thinking Questions
3-50. Consider three vectors: A = 100 m, 00; B = 400 m, 2700; and C = 200 m, 300. 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
A–C
C+B+A
20
C-A
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
3-51. 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.
*3-52. 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?
B = 40 N
C = 60 N
Since R = C = 60 N is smaller than 80 N, the angle

 between A and B must be > 900. Applying the law of
A = 80 N
cosines to the triangle, we can find  and then .
C2 = A2 + B2 – 2AB Cos ;

(60)2 = (80)2 + (40)2 – 2(80)(40) Cos ;
 = 46.60.
The angle between the direction of A and the direction of B is  = 1800 - ;
 = 133.40.
*3-53. What third force F must be added to the following two forces so that the resultant force is
zero: A = 120 N, 1100 and B = 60 N, 2000?
Components of A: Ax = 120 Cos 1100 = -40.0 N; Ay = 120 Sin 1100 = 113 N
Components of B: Bx = 60 Cos 2000 = -56.4 N; By = 60 Sin 2000 = -20.5 N
Rx = 0; Rx = Ax + Bx + Fx = 0;
Rx = -40.0 N –56.4 N + Fx = 0; Or
Fx = +97.4 N
Ry = 0; Ry = Ay + By + Fy = 0;
Ry = 113 N – 20.5 N + Fy = 0;
Fy = -92.2 N
F  (97.4) 2  ( 92.2) 2  131 N
tan  
92.2
;   43.30 And  = 3600 – 43.40
97.4
Thus, the force F has a magnitude and direction of:
21
Or
F = 134 N,  = 316.60
Physics, 7th Edition
Chapter 3 Technical Measurement and Vectors
*3-54. 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 300 S of W, what
direction should the aircraft be pointed and what will be its speed relative to the ground?
From the diagram, Rx = R, Ry = 0, So that Ay + By = 0.
A = 600 km/h
Ay = 600 sin y = -40 sin 300 = -20 km/h


R
600 sin - 20 = 0; 600 sin  = 20
B = 40 km/h
20
sin  
;   191
. 0 N of W (direction aircraft should be pointed)
600
Noting that R = Rx and that Ax +Bx = Rx, we need only find sum of x-components.
Ax = -600 cos 599.7 km/hx = 40 Cos 300 = -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, 00; and the plane must be pointed in a direction of 1.910 N of W.
*3-55. 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?
Rx = 400 lb and Ry = 0; Rx = Ax + Fx = 400 lb
F
200 Cos 200 + Fx = 400 lb

200
Fx = 400 lb – 200 Cos 20 = 212 lb
0
Ry = 0 = Ay + Fy;
Ay = -200 sin 200 = -68.4 lb
A = 200 lb
Fy = -Ay = +68.4 lb; So, Fx = 212 lb and Fy = +68.4 lb
F  (212) 2  (68.4) 2 ; tan =
68.4
;
212
R = 223 lb, 17.90 N of E
22
E