Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.
SOLUTION (a) (b) We measure: R 8.4 kN 19 D q 8.4 kN R 19q W
The cable stays AB and AD help support pole AC. Knowing that the tension is 500 N in AB and 160 N in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
We measure: D 51.3 ,q E 59q
(a)
(b)
Two forces P and Q are applied as shown at point A of a hook support. Knowing that P 15 lb and Q 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.
SOLUTION (a) (b) We measure: R 37 lb,D 76q 37 lb R 76q W
Two forces P and Q are applied as shown at point A of a hook support. Knowing that P 45 lb and Q 15 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.
SOLUTION (a) (b) We measure: R 61.5 lb,D 86.5q 61.5 lb R 86.5q W
Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the left-hand rod is F1 120 N, determine (a) the required force F2 in the right-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Graphically, by the triangle law
We measure: F2 #108 N
77 N R # By trigonometry: Law of Sines
2 120
sin sin 38 sin
F R D q E 90 28 62 , 180 62 38 80 D q q q E q q q q Then: 2 120 N
sin 62 sin 38 sin 80
F R
q q q
or (a) F2 107.6 NW (b) R 75.0 NW
Two control rods are attached at A to lever AB. Using trigonometry and knowing that the force in the right-hand rod is F2 80 N, determine (a) the required force F1 in the left-hand rod if the resultant R of the forces exerted by the rods on the lever is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the Law of Sines
1 80
sin sin 38 sin
F R D q E 90 10 80 , 180 80 38 62 D q q q E q q q q Then: 1 80 N
sin 80 sin 38 sin 62
F R
q q q
or (a) F1 89.2 NW (b) R 55.8 NW
The 50-lb force is to be resolved into components along lines -a ac and - .
b bc (a) Using trigonometry, determine the angle D knowing that the component along -a ac is 35 lb. (b) What is the corresponding value of the component along - ?b bc
SOLUTION
Using the triangle rule and the Law of Sines
(a) sin sin 40
35 lb 50 lb E q sinE 0.44995 26.74 E q Then: D E 40q 180q 113.3 D q W
(b) Using the Law of Sines:
50 lb sin sin 40 bb F Dc q 71.5 lb bb F c W
The 50-lb force is to be resolved into components along lines -a ac and - .
b bc (a) Using trigonometry, determine the angle D knowing that the component along -b bc is 30 lb. (b) What is the corresponding value of the component along - ?a ac
SOLUTION
Using the triangle rule and the Law of Sines
(a) sin sin 40
30 lb 50 lb D q sinD 0.3857 22.7 D q W (b) D E 40q 180q 117.31 E q 50 lb sin sin 40 aa F Ec q sin 50 lb sin 40 E c §¨ q·¸ © ¹ aa F W
To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that D 25q, determine (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the Law of Sines
Have: D 180q
35q 25q
120q
Then: 360 N
sin 35 sin120 sin 25
P R
q q q
or (a) P 489 NW (b) R 738 NW
To steady a sign as it is being lowered, two cables are attached to the sign at A. Using trigonometry and knowing that the magnitude of P is 300 N, determine (a) the required angle D if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the Law of Sines
(a) Have: 360 N 300 N sinD sin 35q sinD 0.68829 43.5 D q W (b) E 180
35q 43.5q
101.5q Then: 300 N sin101.5 sin 35 R q q or R 513 NW
Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (a) the required angleD if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the Law of Sines
(a) Have: 20 lb 14 lb sinD sin 30q sinD 0.71428 45.6 D q W (b) E 180q
30q 45.6q
104.4q Then: 14 lb sin104.4 sin 30 R q q 27.1 lb R W
For the hook support of Problem 2.3, using trigonometry and knowing that the magnitude of P is 25 lb, determine (a) the required magnitude of the force Q if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.
Problem 2.3: Two forces P and Q are applied as shown at point A of a hook support. Knowing that P 15 lb and Q 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
Using the triangle rule and the Law of Sines
(a) Have: 25 lb sin15 sin 30 Q q q 12.94 lb Q W (b) E 180q
15q 30q
135q Thus: 25 lb sin135 sin 30 R q q sin135 25 lb 35.36 lb sin 30 R §¨ q·¸ q © ¹ 35.4 lb R W
For the hook support of Problem 2.11, determine, using trigonometry, (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R.
Problem 2.11: Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (a) the required angle D if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
(a) The smallest force P will be perpendicular to R, that is, vertical
20 lb sin 30
P q
10 lb P 10 lb W
(b) R
20 lb cos 30
q
As shown in Figure P2.9, two cables are attached to a sign at A to steady the sign as it is being lowered. Using trigonometry, determine (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R.
SOLUTION
We observe that force P is minimum when is 90 ,D q that is, P is horizontal
Then: (a) P
360 N sin 35
q
or P 206 N W
And: (b) R
360 N cos 35
q
For the hook support of Problem 2.11, determine, using trigonometry, the magnitude and direction of the resultant of the two forces applied to the support knowing that P 10 lb and D 40q.
Problem 2.11: Two forces are applied as shown to a hook support. Using trigonometry and knowing that the magnitude of P is 14 lb, determine (a) the required angle D if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the force triangle and the Law of Cosines
2
2
2 10 lb 20 lb 2 10 lb 20 lb cos110 R q
2 100 400 400 0.342 lb ª º ¬ ¼ 2 636.8 lb 25.23 lb R
Using now the Law of Sines
10 lb 25.23 lb sinE sin110q 10 lb sin sin110 25.23 lb E §¨ ·¸ q © ¹ 0.3724 So: E 21.87q
Angle of inclination of R, I is then such that:
30 I E q
8.13
Solve Problem 2.1 using trigonometry
Problem 2.1: Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
Using the force triangle, the Law of Cosines and the Law of Sines
We have: D 180q
50q 25q
105q Then: R2
4.5 kN
2 6 kN
22 4.5 kN 6 kN cos105
q 2 70.226 kN or R 8.3801 kN Now: 8.3801 kN 6 kN sin105q sinE 6 kN sin sin105 8.3801 kN E §¨ ·¸ q © ¹ 0.6916 43.756 E q 8.38 kN R 18.76q W
Solve Problem 2.2 using trigonometry
Problem 2.2: The cable stays AB and AD help support pole AC. Knowing that the tension is 500 N in AB and 160 N in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
From the geometry of the problem: 1 2 tan 38.66 2.5 D q 11.5 tan 30.96 2.5 E q Now: T 180q
38.6630.96q
110.38 And, using the Law of Cosines:
2
2
2 500 N 160 N 2 500 N 160 N cos110.38 R q 2 331319 N 575.6 N R
Using the Law of Sines:
160 N 575.6 N sinJ sin110.38q 160 N sin sin110.38 575.6 N J §¨ ·¸ q © ¹ 0.2606 15.1 J q
90
66.44 I q D J q 576 N R 66.4q W
Solve Problem 2.3 using trigonometry
Problem 2.3: Two forces P and Q are applied as shown at point A of a hook support. Knowing that P 15 lb and Q 25 lb, determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
Using the force triangle and the Laws of Cosines and Sines We have:
180 15 30 J q q q 135q Then: R2
15 lb
2 25 lb
2 2 15 lb 25 lb cos135
q 2 1380.3 lb or R 37.15 lb and 25 lb 37.15 lb sinE sin135q 25 lb sin sin135 37.15 lb E §¨ ·¸ q © ¹ 0.4758 28.41 E q Then: D E 75q 180q 76.59 D q 37.2 lb R 76.6q W
Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 30 kN in member A and 20 kN in member B, determine, using trigonometry, the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.
SOLUTION
Using the force triangle and the Laws of Cosines and Sines We have: J 180q
45q 25q
110q Then: R2
30 kN
2 20 kN
2 2 30 kN 20 kN cos110
q 2 1710.4 kN 41.357 kN R and 20 kN 41.357 kN sinD sin110q 20 kN sin sin110 41.357 kN D §¨ ·¸ q © ¹ 0.4544 27.028 D q Hence: I D 45q 72.028q 41.4 kN R 72.0q W
Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 20 kN in member A and 30 kN in member B, determine, using trigonometry, the magnitude and direction of the resultant of the forces applied to the bracket by members A and B.
SOLUTION
Using the force triangle and the Laws of Cosines and Sines We have: J 180q
45q 25q
110q Then: R2
30 kN
2 20 kN
2 2 30 kN 20 kN cos110
q 2 1710.4 kN 41.357 kN R and 30 kN 41.357 kN sinD sin110q 30 kN sin sin110 41.357 kN D §¨ ·¸ q © ¹ 0.6816 42.97 D q Finally: I D 45q 87.97q 41.4 kN R 88.0q W
Determine the x and y components of each of the forces shown. SOLUTION 20 kN Force:
20 kN cos 40 ,
x F q Fx 15.32 kNW
20 kN sin 40 ,
y F q Fy 12.86 kNW 30 kN Force:
30 kN cos 70 ,
x F q Fx 10.26 kNW
30 kN sin 70 ,
y F q Fy 28.2 kNW 42 kN Force:
42 kN cos 20 ,
x F q Fx 39.5 kNW
42 kN sin 20 ,
y F q Fy 14.36 kNW
Determine the x and y components of each of the forces shown. SOLUTION 40 lb Force:
40 lb sin 50 ,
x F q Fx 30.6 lbW
40 lb cos 50 ,
y F q Fy 25.7 lbW 60 lb Force:
60 lb cos 60 ,
x F q Fx 30.0 lbW
60 lb sin 60 ,
y F q Fy 52.0 lbW 80 lb Force:
80 lb cos 25 ,
x F q Fx 72.5 lbW
80 lb sin 25 ,
y F q Fy 33.8 lbW
Determine the x and y components of each of the forces shown.
SOLUTION
We compute the following distances:
2 2 2 2 2 2 48 90 102 in. 56 90 106 in. 80 60 100 in. OA OB OC Then: 204 lb Force:
48 102 lb , 102 x F Fx 48.0 lbW
90 102 lb , 102 y F Fy 90.0 lbW 212 lb Force:
56 212 lb , 106 x F Fx 112.0 lbW
90 212 lb , 106 y F Fy 180.0 lbW 400 lb Force:
80 400 lb , 100 x F Fx 320 lbW
Determine the x and y components of each of the forces shown.
SOLUTION
We compute the following distances:
2 2 70 240 250 mm OA
2 2 210 200 290 mm OB
2 2 120 225 255 mm OC 500 N Force: 70 500 N 250 x F §¨ ·¸ © ¹ Fx 140.0 NW 240 500 N 250 y F §¨ ·¸ © ¹ Fy 480 NW 435 N Force: 210 435 N 290 x F §¨ ·¸ © ¹ Fx 315 NW 200 435 N 290 y F §¨ ·¸ © ¹ Fy 300 NW 510 N Force: 120 510 N 255 x F §¨ ·¸ © ¹ Fx 240 NW
While emptying a wheelbarrow, a gardener exerts on each handle AB a force P directed along line CD. Knowing that P must have a 135-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component. SOLUTION (a) cos 40 x P P q 135 N cos 40q or P 176.2 NW (b) Py Pxtan 40q Psin 40q
135 N tan 40
q or Py 113.3 NW
Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 960-N vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.
SOLUTION (a) sin 35 y P P q 960 N sin 35q or P 1674 NW (b) tan 35 y x P P q 960 N tan 35q or Px 1371 NW
Member CB of the vise shown exerts on block B a force P directed along line CB. Knowing that P must have a 260-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.
SOLUTION
We note:
CB exerts force P on B along CB, and the horizontal component of P is Px 260 lb. Then: (a) Px Psin 50q sin 50 x P P q 260 lb sin50q 339.4 lb P 339 lbW (b) Px Pytan 50q tan 50 x y P P q 260 lb tan 50q 218.2 lb Py 218 lb W
Activator rod AB exerts on crank BCD a force P directed along line AB. Knowing that P must have a 25-lb component perpendicular to arm BC of the crank, determine (a) the magnitude of the force P, (b) its component along line BC.
SOLUTION
Using the x and y axes shown.
(a) Py 25 lb Then: sin 75 y P P q 25 lb sin 75q or P 25.9 lbW (b) tan 75 y x P P q 25 lb tan 75q or Px 6.70 lbW
The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 450-N component along line AC, determine (a) the magnitude of the force P, (b) its component in a direction perpendicular to AC.
SOLUTION
Note that the force exerted by BD on the pole is directed along BD, and the component of P along AC is 450 N. Then: (a) 450 N 549.3 N cos 35 P q 549 N P W (b) Px
450 N tan 35
q 315.1 N 315 N x P W
The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P has a 200-N perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC. SOLUTION (a) sin 38 x P P q 200 N sin 38q 324.8 N or P 325 NW (b) tan 38 x y P P q 200 N tan 38q
Determine the resultant of the three forces of Problem 2.24.
Problem 2.24: Determine the x and y components of each of the forces
shown. SOLUTION From Problem 2.24:
500 140 N 480 N F i j
425 315 N 300 N F i j
510 240 N 450 N F i j
415 N
330 N
6 R F i j Then: 1330 tan 38.5 415 D q
2
2 415 N 330 N 530.2 N R Thus: R 530 N 38.5q W
Determine the resultant of the three forces of Problem 2.21.
Problem 2.21: Determine the x and y components of each of the forces
shown. SOLUTION From Problem 2.21:
20 15.32 kN 12.86 kN F i j
30 10.26 kN 28.2 kN F i j
42 39.5 kN 14.36 kN F i j
34.44 kN
55.42 kN
6 R F i j Then: 1 55.42 tan 58.1 34.44 D q
2
2 55.42 kN 34.44 N 65.2 kN R 65.2 kN R 58.2q W
Determine the resultant of the three forces of Problem 2.22.
Problem 2.22: Determine the x and y components of each of the forces
shown.
SOLUTION
The components of the forces were determined in 2.23.
x y R R R i j
71.9 lb
i 43.86 lb
j 43.86 tan 71.9 D 31.38 D q
2
2 71.9 lb 43.86 lb R 84.23 lb
Force xcomp. (lb) ycomp. (lb) 40 lb 30.6 25.7 60 lb 30 51.96 80 lb 72.5 33.8 71.9 x R Ry 43.86
Determine the resultant of the three forces of Problem 2.23.
Problem 2.23: Determine the x and y components of each of the forces
shown.
SOLUTION
The components of the forces were determined in Problem 2.23.
204 48.0 lb 90.0 lb F i j
212 112.0 lb 180.0 lb F i j
400 320 lb 240 lb F i j Thus x y R R R
256 lb
30.0 lb
R i j Now: 30.0 tan 256 D 130.0 tan 6.68 256 D q and
2
2 256 lb 30.0 lb R
Knowing that D 35 ,q determine the resultant of the three forces shown. SOLUTION 300-N Force:
300 N cos 20
281.9 N x F q
300 N sin 20
102.6 N y F q 400-N Force:
400 N cos 55
229.4 N x F q
400 N sin 55
327.7 N y F q 600-N Force:
600 N cos 35
491.5 N x F q
600 N sin 35
344.1 N y F q and 1002.8 N x x R 6F 86.2 N y y R 6F
2
2 1002.8 N 86.2 N 1006.5 N R Further: 86.2 tan 1002.8 D 1 86.2 tan 4.91 1002.8 D q 1007 N R 4.91q W
Knowing that D 65 ,q determine the resultant of the three forces shown. SOLUTION 300-N Force:
300 N cos 20
281.9 N x F q
300 N sin 20
102.6 N y F q 400-N Force:
400 N cos85
34.9 N x F q
400 N sin 85
398.5 N y F q 600-N Force:
600 N cos 5
597.7 N x F q
600 N sin 5
52.3 N y F q and 914.5 N x x R 6F 448.8 N y y R 6F
2
2 914.5 N 448.8 N 1018.7 N R Further: 448.8 tan 914.5 D 1448.8 tan 26.1 914.5 D q
Knowing that the tension in cable BC is 145 lb, determine the resultant of the three forces exerted at point B of beam AB.
SOLUTION Cable BC Force:
84 145 lb 105 lb 116 x F
80 145 lb 100 lb 116 y F 100-lb Force:
3 100 lb 60 lb 5 x F
4 100 lb 80 lb 5 y F 156-lb Force:
12 156 lb 144 lb 13 x F
5 156 lb 60 lb 13 y F and 21 lb, 40 lb x x y y R 6F R 6F
2
2 21 lb 40 lb 45.177 lb R Further: 40 tan 21 D 140 tan 62.3 21 D q
Knowing that D 50 ,q determine the resultant of the three forces shown.
SOLUTION
The resultant force R has the x- and y-components:
140 lb cos 50
60 lb cos85
160 lb cos 50
x x R 6F q q q 7.6264 lb x R and
140 lb sin 50
60 lb sin 85
160 lb sin 50
y y R 6F q q q 289.59 lb y R Further: 290 tan 7.6 D 1290 tan 88.5 7.6 D q Thus: R 290 lb 88.5q W
Determine (a) the required value of D if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant.
SOLUTION
For an arbitrary angle ,D we have:
140 lb cos
60 lb cos
35
160 lb cos
x x
R 6F D D q D
(a) So, for R to be vertical:
140 lb cos
60 lb cos
35
160 lb cos
0
x x
R 6F D D q D
Expanding,
cosD 3 cos cos 35D sin sin 35D 0
q q Then: 1 3 cos 35 tan sin 35 D q q or 1 1 cos 35 3 tan 40.265 sin 35 D §¨ q ·¸ q ¨ q ¸ © ¹ D 40.3q W (b) Now:
140 lb sin 40.265
60 lb sin 75.265
160 lb sin 40.265
y y
R R 6F q q q
252 lb
For the beam of Problem 2.37, determine (a) the required tension in cable BC if the resultant of the three forces exerted at point B is to be vertical, (b) the corresponding magnitude of the resultant.
Problem 2.37: Knowing that the tension in cable BC is 145 lb, determine
the resultant of the three forces exerted at point B of beam AB.
SOLUTION We have:
84 12 3 156 lb 100 lb 116 13 5 x x BC R 6F T or Rx 0.724TBC 84 lb and
80 5 4 156 lb 100 lb 116 13 5 y y BC R 6F T 0.6897 140 lb y BC R T
(a) So, for R to be vertical,
0.724 84 lb 0 x BC R T 116.0 lb BC T W (b) Using 116.0 lb BC T
0.6897 116.0 lb 140 lb 60 lb y R R 60.0 lb R R W
Boom AB is held in the position shown by three cables. Knowing that the tensions in cables AC and AD are 4 kN and 5.2 kN, respectively, determine (a) the tension in cable AE if the resultant of the tensions exerted at point A of the boom must be directed along AB, (b) the corresponding magnitude of the resultant.
SOLUTION
Choose x-axis along bar AB. Then
(a) Require
0: 4 kN cos 25 5.2 kN sin 35 sin 65 0
y y AE R 6F q q T q or TAE 7.2909 kN 7.29 kN AE T W (b) R 6Fx
4 kN sin 25
5.2 kN cos 35
7.2909 kN cos 65
q q q
9.03 kN
9.03 kN
For the block of Problems 2.35 and 2.36, determine (a) the required value of D of the resultant of the three forces shown is to be parallel to the incline, (b) the corresponding magnitude of the resultant.
Problem 2.35: Knowing that D 35 ,q determine the resultant of the three forces shown.
Problem 2.36: Knowing that D 65 ,q determine the resultant of the three forces shown.
SOLUTION
Selecting the x axis along aac,we write
300 N 400 N cos 600 N sin x x R 6F D D (1)
400 N sin
600 N cos
y y R 6F D D (2)
(a) Setting Ry in Equation (2): 0
Thus tan 600 1.5
400 D
56.3 D q W (b) Substituting for D in Equation (1):
300 N 400 N cos 56.3 600 N sin 56.3 x R q q 1021.1 N x R 1021 N x R R W
Two cables are tied together at C and are loaded as shown. Determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram
From the geometry, we calculate the distances:
2
2
16 in. 12 in. 20 in.
AC
2
2
20 in. 21 in. 29 in.
BC
Then, from the Free Body Diagram of point C:
16 21 0: 0 20 29 x AC BC F T T 6 or 29 4 21 5 BC AC T u T and 0: 12 20 600 lb 0 20 29 y AC BC F T T 6 or 12 20 29 4 600 lb 0 20TAC 29 21 5TAC § · ¨ u ¸ © ¹ Hence: TAC 440.56 lb (a) TAC 441 lbW (b) TBC 487 lbW
Knowing that D 25 ,q determine the tension (a) in cable AC, (b) in rope BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of Sines:
5 kN sin115 sin 5 sin 60
AC BC T T q q q (a) 5 kN sin115 5.23 kN sin 60 AC T q q TAC 5.23 kNW (b) 5 kN sin 5 0.503 kN sin 60 BC T q q TBC 0.503 kNW
Knowing that D 50q and that boom AC exerts on pin C a force directed long line AC, determine (a) the magnitude of that force, (b) the tension in cable BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of Sines:
400 lb sin 25 sin 60 sin 95
AC BC F T q q q (a) 400 lbsin 25 169.69 lb sin 95 AC F q q FAC 169.7 lbW (b) 400 sin 60 347.73 lb sin 95 BC T q q TBC 348 lbW
Two cables are tied together at C and are loaded as shown. Knowing that 30 ,
D q determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram Force Triangle
Law of Sines:
2943 N sin 60 sin 55 sin 65
AC BC T T q q q (a) 2943 Nsin 60 2812.19 N sin 65 AC T q q TAC 2.81 kNW (b) 2943 Nsin 55 2659.98 N sin 65 BC T q q TBC 2.66 kNW
A chairlift has been stopped in the position shown. Knowing that each chair weighs 300 N and that the skier in chair E weighs 890 N, determine that weight of the skier in chair F.
SOLUTION
Free-Body Diagram Point B
Force Triangle
Free-Body Diagram Point C
Force Triangle
In the free-body diagram of point B, the geometry gives: 1 9.9 tan 30.51 16.8 AB T q 1 12 tan 22.61 28.8 BC T q
Thus, in the force triangle, by the Law of Sines: 1190 N sin 59.49 sin 7.87 BC T q q 7468.6 N BC T
In the free-body diagram of point C (with W the sum of weights of chair and skier) the geometry gives:
11.32
tan 10.39
7.2
CD
T q
Hence, in the force triangle, by the Law of Sines: 7468.6 N sin12.23 sin100.39 W q q 1608.5 N W
Finally, the skier weight 1608.5 N300 N 1308.5 N
A chairlift has been stopped in the position shown. Knowing that each chair weighs 300 N and that the skier in chair F weighs 800 N, determine the weight of the skier in chair E.
SOLUTION
Free-Body Diagram Point F
Force Triangle
Free-Body Diagram Point E
Force Triangle
In the free-body diagram of point F, the geometry gives: 1 12 tan 22.62 28.8 EF T q 11.32 tan 10.39 7.2 DF T q
Thus, in the force triangle, by the Law of Sines: 1100 N sin100.39 sin12.23 EF T q q 5107.5 N BC T
In the free-body diagram of point E (with W the sum of weights of chair and skier) the geometry gives:
1 9.9
tan 30.51
16.8
AE
T q
Hence, in the force triangle, by the Law of Sines: 5107.5 N sin 7.89 sin 59.49 W q q 813.8 N W
Finally, the skier weight 813.8 N300 N 513.8 N
Four wooden members are joined with metal plate connectors and are in equilibrium under the action of the four fences shown. Knowing that FA 510 lb and FB 480 lb, determine the magnitudes of the other two
forces.
SOLUTION
Free-Body Diagram
Resolving the forces into x and y components:
0: 510 lb sin15 480 lb cos15 0 x C F F 6 q q or FC 332 lbW
0: 510 lb cos15 480 lb sin15 0 y D F F 6 q q or FD 368 lbW
Four wooden members are joined with metal plate connectors and are in equilibrium under the action of the four fences shown. Knowing that FA 420 lb and FC 540 lb, determine the magnitudes of the other two
forces.
SOLUTION
Resolving the forces into x and y components:
0: cos15 540 lb 420 lb cos15 0 or 671.6 lb x B B F F F 6 q q 672 lb B F W
0: 420 lb cos15 671.6 lb sin15 0 y D F F 6 q q or FD 232 lbW
Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and the P 400 lb and Q 520 lb, determine the magnitudes of the forces exerted on the rods A and B.
SOLUTION
Free-Body Diagram Resolving the forces into x and y directions: 0 A B R P Q F F Substituting components:
400 lb
ª
520 lb cos 55
º ª
520 lb sin 55
º ¬ q ¼ ¬ q¼ R j i j
cos 55
sin 55
0 B A A F F F i q i q j
In the y-direction (one unknown force)
400 lb 520 lb sin 55 FAsin 55 0 q q Thus,
400 lb 520 lb sin 55 1008.3 lb sin 55 A F q q 1008 lb A F W In the x-direction:
520 lb cos 55
q FB FAcos 55q 0 Thus,
cos 55 520 lb cos 55 B A F F q q
1008.3 lb cos 55
q
520 lb cos 55
q 280.08 lb
Two forces P and Q are applied as shown to an aircraft connection. Knowing that the connection is in equilibrium and that the magnitudes of the forces exerted on rods A and B are FA 600 lb and FB 320 lb,
determine the magnitudes of P and Q.
SOLUTION
Free-Body Diagram Resolving the forces into x and y directions: 0 A B R P Q F F Substituting components:
320 lb
»
600 lb cos 55
q ¼º ¬ª
600 lb sin 55
qº¼ R i i j
cos 55
sin 55
0 P Q Q i q i q j
In the x-direction (one unknown force)
320 lb 600 lb cos 55q Qcos 55q 0 Thus,
320 lb 600 lb cos 55 42.09 lb cos 55 Q q q 42.1 lb Q W In the y-direction:
600 lb sin 55
q PQsin 55q 0 Thus,
600 lb sin 55
sin 55 457.01 lb P q Q q 457 lb P W
Two cables tied together at C are loaded as shown. Knowing that W 840 N, determine the tension (a) in cable AC, (b) in cable BC.
SOLUTION
Free-Body Diagram From geometry:
The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. Thus:
3 15 15 0: 680 N 0 5 17 17 x CA CB F T T 6 or 1 5 200 N 5TCA 17TCB (1) and
4 8 8 0: 680 N 840 N 0 5 17 17 y CA CB F T T 6 or 1 2 290 N 5TCA17TCB (2)
Solving Equations (1) and (2) simultaneously:
(a) TCA 750 NW
Two cables tied together at C are loaded as shown. Determine the range of values of W for which the tension will not exceed 1050 N in either cable.
SOLUTION
Free-Body Diagram From geometry:
The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. Thus:
3 15 15 0: 680 N 0 5 17 17 x CA CB F T T 6 or 1 5 200 N 5TCA 17TCB (1) and
4 8 8 0: 680 N 0 5 17 17 y CA CB F T T W 6 or 1 2 1 80 N 5TCA17TCB 4W (2) Then, from Equations (1) and (2)
17 680 N 28 25 28 CB CA T W T W Now, withT d1050 N 25 : 1050 N 28 CA CA T T W or W 1176 N
The cabin of an aerial tramway is suspended from a set of wheels that can roll freely on the support cable ACB and is being pulled at a constant speed by cable DE. Knowing that D 40q and E 35q, that the combined weight of the cabin, its support system, and its passengers is 24.8 kN, and assuming the tension in cable DF to be negligible, determine the tension (a) in the support cable ACB, (b) in the traction cable DE.
SOLUTION
Note: In Problems 2.55 and 2.56 the cabin is considered as a particle. If considered as a rigid body (Chapter 4) it would be found that its center of gravity should be located to the left of the centerline for the line CD to be vertical.
Now
0: cos 35 cos 40 cos 40 0
x ACB DE F T T 6 q q q or 0.0531TACB 0.766TDE 0 (1) and
0: sin 40 sin 35 sin 40 24.8 kN 0
y ACB DE F T T 6 q q q or 0.0692TACB 0.643TDE 24.8 kN (2) From (1) 14.426 ACB DE T T Then, from (2)
0.0692 14.426TDE 0.643TDE 24.8 kN and (b) TDE 15.1 kNW
The cabin of an aerial tramway is suspended from a set of wheels that can roll freely on the support cable ACB and is being pulled at a constant speed by cable DE. Knowing that D 42q and E 32q, that the tension in cable DE is 20 kN, and assuming the tension in cable DF to be negligible, determine (a) the combined weight of the cabin, its support system, and its passengers, (b) the tension in the support cable ACB.
SOLUTION
Free-Body Diagram
First, consider the sum of forces in the x-direction because there is only one unknown force:
0: cos 32 cos 42 20 kN cos 42 0
x ACB F T 6 q q q or 0.1049TACB 14.863 kN (b) TACB 141.7 kNW Now
0: sin 42 sin 32 20 kN sin 42 0
y ACB
F T W
6 q q q
or
A block of weight W is suspended from a 500-mm long cord and two springs of which the unstretched lengths are 450 mm. Knowing that the constants of the springs are kAB 1500 N/m and kAD 500 N/m,
determine (a) the tension in the cord, (b) the weight of the block.
SOLUTION
Free-Body Diagram At A First note from geometry:
The sides of the triangle with hypotenuse AD are in the ratio 8:15:17. The sides of the triangle with hypotenuse AB are in the ratio 3:4:5. The sides of the triangle with hypotenuse AC are in the ratio 7:24:25. Then:
AB AB AB o F k L L and
2
2 0.44 m 0.33 m 0.55 m AB L So:
1500 N/m 0.55 m 0.45 m AB F 150 N Similarly,
AD AD AD o F k L L Then:
2
2 0.66 m 0.32 m 0.68 m AD L
1500 N/m 0.68 m 0.45 m AD F 115 N (a)
4 7 15 0: 150 N 115 N 0 5 25 17 x AC F T 6 or 66.18 N AC T TAC 66.2 NW
(b) and
3 24 8 0: 150 N 66.18 N 115 N 0 5 25 17 y F W 6 or W 208 NW
A load of weight 400 N is suspended from a spring and two cords which are attached to blocks of weights 3W and W as shown. Knowing that the constant of the spring is 800 N/m, determine (a) the value of W, (b) the unstretched length of the spring.
SOLUTION
Free-Body Diagram At A
First note from geometry:
The sides of the triangle with hypotenuse AD are in the ratio 12:35:37. The sides of the triangle with hypotenuse AC are in the ratio 3:4:5. The sides of the triangle with hypotenuse AB are also in the ratio 12:35:37. Then:
4 35 12 0: 3 0 5 37 37 x s F W W F 6 or 4.4833 s F W and
3 12 35 0: 3 400 N 0 5 37 37 y s F W W F 6 Then:
3 12 35 3 4.4833 400 N 0 5 W 37 W 37 W or 62.841 N W and 281.74 N s F or (a) W 62.8 NW
(b) Have spring force
s AB o F k L L Where
AB AB AB o F k L L and
2
2 0.360 m 1.050 m 1.110 m AB L So:
0
281.74 N 800 N/m 1.110L m or L0 758 mmW
For the cables and loading of Problem 2.46, determine (a) the value of D for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.
SOLUTION
The smallest TBC is when TBC is perpendicular to the direction of TAC
Free-Body Diagram At C Force Triangle
(a) D 55.0q W (b) TBC
2943 N sin 55
q 2410.8 N 2.41 kN BC T W
Knowing that portions AC and BC of cable ACB must be equal, determine the shortest length of cable which can be used to support the load shown if the tension in the cable is not to exceed 725 N.
SOLUTION Free-Body Diagram: C
ForT 725 N
6Fy 0: 2Ty 1000 N 0 500 N y T 2 2 2 x y T T T
2
2 2 500 N 725 N x T 525 N x T By similar triangles: 1.5 m 725BC 525 2.07 m ? BC
2 4.14 m L BC 4.14 m L W
Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension in each cable is 200 lb, determine (a) the magnitude of the largest force P which may be applied at C, (b) the corresponding value of D.
SOLUTION
Free-Body Diagram: C Force Triangle
Force triangle is isoceles with
2E 180q 85q 47.5
E q
(a) P 2 200 lb cos 47.5
q 270 lb
Since P !0, the solution is correct. P 270 lbW
Two cables tied together at C are loaded as shown. Knowing that the maximum allowable tension is 300 lb in cable AC and 150 lb in cable BC, determine (a) the magnitude of the largest force P which may be applied at C, (b) the corresponding value of D.
SOLUTION
Free-Body Diagram: C Force Triangle
(a) Law of Cosines:
2
2
2 300 lb 150 lb 2 300 lb 150 lb cos85
P q
323.5 lb P
Since P !300 lb, our solution is correct. P 324 lbW
(b) Law of Sines: sin sin 85 300 323.5 E q q sinE 0.9238 or E 67.49q 180 55 67.49 57.5 D q q q q 57.5 D q W
For the structure and loading of Problem 2.45, determine (a) the value of D for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.
SOLUTION
BC
T must be perpendicular to FAC to be as small as possible.
Free-Body Diagram: C Force Triangle is a right triangle
(a) We observe: D 55q D 55q W
(b) TBC
400 lb sin 60
q
Boom AB is supported by cable BC and a hinge at A. Knowing that the boom exerts on pin B a force directed along the boom and that the tension in rope BD is 70 lb, determine (a) the value of D for which the tension in cable BC is as small as possible, (b) the corresponding value of the tension.
SOLUTION
Free-Body Diagram: B (a) Have: TBD FAB TBC 0
where magnitude and direction of TBD are known, and the direction of FAB is known.
Then, in a force triangle:
By observation, TBC is minimum when D 90.0q W (b) Have TBC
70 lb sin 180
q 70q 30q
68.93 lb
68.9 lb
BC
Collar A shown in Figure P2.65 and P2.66 can slide on a frictionless vertical rod and is attached as shown to a spring. The constant of the spring is 660 N/m, and the spring is unstretched when h 300 mm. Knowing that the system is in equilibrium when h 400 mm, determine the weight of the collar.
SOLUTION
Free-Body Diagram: Collar A
Have: Fs k L
cAB LAB
where:
2
2 0.3 m 0.4 m 0.3 2 m AB AB Lc L 0.5 m Then: Fs 660 N/m 0.5
0.3 2 m
49.986 N
For the collar:
4 0: 49.986 N 0 5 y F W 6 or W 40.0 NW
The 40-N collar A can slide on a frictionless vertical rod and is attached as shown to a spring. The spring is unstretched when h 300 mm. Knowing that the constant of the spring is 560 N/m, determine the value of h for which the system is in equilibrium.
SOLUTION
Free-Body Diagram: Collar A
2 2 0: 0 0.3 y s h F W F h 6 or hFs 40 0.09h2 Now.. Fs k L
ABc LAB
where LABc
0.3 2h2 m LAB 0.3 2 m Then: hª«560
0.09h2 0.3 2
º» 40 0.09h2 ¬ ¼ or
14h1
0.09h2 4.2 2 h hm Solving numerically, 415 mm h W
A 280-kg crate is supported by several rope-and-pulley arrangements as shown. Determine for each arrangement the tension in the rope. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram of pulley
(a) (b) (c) (d) (e)
2
0: 2 280 kg 9.81 m/s 0 y F T 6
1 2746.8 N 2 T 1373 N T W
2
0: 2 280 kg 9.81 m/s 0 y F T 6
1 2746.8 N 2 T 1373 N T W
2
0: 3 280 kg 9.81 m/s 0 y F T 6
1 2746.8 N 3 T 916 N T W
2
0: 3 280 kg 9.81 m/s 0 y F T 6
1 2746.8 N 3 T 916 N T W
2
0: 4 280 kg 9.81 m/s 0 y F T 6
1 2746.8 N 4 T 687 N T W
Solve parts b and d of Problem 2.67 assuming that the free end of the rope is attached to the crate.
Problem 2.67: A 280-kg crate is supported by several rope-and-pulley
arrangements as shown. Determine for each arrangement the tension in the rope. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram of pulley and crate (b) (d)
2
0: 3 280 kg 9.81 m/s 0 y F T 6
1 2746.8 N 3 T 916 N T W
2
0: 4 280 kg 9.81 m/s 0 y F T 6
1 2746.8 N 4 T 687 N T W
A 350-lb load is supported by the rope-and-pulley arrangement shown. Knowing that E 25q, determine the magnitude and direction of the force P which should be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram: Pulley A
0: 2 sin 25 cos 0 x F P P D 6 q and cosD 0.8452 or D r32.3q For D 32.3q 0: 2 cos 25 sin 32.3 350 lb 0 y F P P 6 q q or P 149.1 lb 32.3q W For D 32.3q 0: 2 cos 25 sin 32.3 350 lb 0 y F P P 6 q q or P 274 lb 32.3q W
A 350-lb load is supported by the rope-and-pulley arrangement shown. Knowing that D 35 ,q determine (a) the angle E, (b) the magnitude of the force P which should be exerted on the free end of the rope to maintain equilibrium. (Hint: The tension in the rope is the same on each side of a simple pulley. This can be proved by the methods of Chapter 4.)
SOLUTION
Free-Body Diagram: Pulley A
0: 2 sin cos 25 0
x
F P E P
6 q
Hence:
(a) sin 1cos 25
2 E q or E 24.2q W (b) 6Fy 0: 2 cosP E Psin 35q 350 lb 0 Hence: 2 cos 24.2P q Psin 35q 350 lb 0 or P 145.97 lb P 146.0 lbW
A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P 800 N, determine (a) the tension in cable ACB, (b) the magnitude of load Q.
SOLUTION
Free-Body Diagram: Pulley C
(a) 6Fx 0: TACB
cos 30q cos 50q
800 N cos 50
q 0
Hence TACB 2303.5 N
2.30 kN
ACB
T W
(b) 6Fy 0: TACB
sin 30q sin 50q
800 N sin 50
q Q 0
2303.5 N sin 30
q sin 50q
800 N sin 50
q Q 0
A 2000-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in the cable ACB, (b) the magnitude of load P.
SOLUTION
Free-Body Diagram: Pulley C
0: cos 30 cos 50 cos 50 0
x ACB
F T P
6 q q q
or P 0.3473TACB (1)
0: sin 30 sin 50 sin 50 2000 N 0
y ACB
F T P
6 q q q
or 1.266TACB 0.766P 2000 N (2)
(a) Substitute Equation (1) into Equation (2):
1.266TACB 0.766 0.3473TACB 2000 N Hence: TACB 1305.5 N 1306 N ACB T W (b) Using (1)
0.3473 1306 N 453.57 N P 454 N P W
Determine (a) the x, y, and z components of the 200-lb force, (b) the anglesTx,Ty, and Tz that the force forms with the coordinate axes.
SOLUTION
(a) Fx
200 lb cos 30 cos 25
q q 156.98 lb
157.0 lb x F W
200 lb sin 30
100.0 lb y F q 100.0 lb y F W
200 lb cos 30 sin 25
73.1996 lb z F q q 73.2 lb z F W (b) cos 156.98 200 x T or Tx 38.3q W cos 100.0 200 y T or Ty 60.0q W 73.1996 cos 200 z T or Tz 111.5q W
Determine (a) the x, y, and z components of the 420-lb force, (b) the anglesTx,Ty, and Tz that the force forms with the coordinate axes.
SOLUTION
(a) Fx
420 lb sin 20 sin 70
q q 134.985 lb
135.0 lb x F W
420 lb cos 20
394.67 lb y F q 395 lb y F W
420 lb sin 20 cos 70
49.131 lb z F q q 49.1 lb z F W (b) cos 134.985 420 x T 108.7 x T q W 394.67 cos 420 y T 20.0 y T q W 49.131 cos 420 z T 83.3 z T q W
To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in cable AB is 4.2 kN, determine (a) the components of the force exerted by this cable on the tree, (b) the angles Tx,Ty, and Tz that the force forms with axes at A which
are parallel to the coordinate axes.
SOLUTION
(a) Fx
4.2 kN sin 50 cos 40
q q 2.4647 kN
2.46 kN x F W Fy
4.2 kN cos 50
q 2.6997 kN 2.70 kN y F W Fz
4.2 kN sin 50 sin 40
q q 2.0681 kN 2.07 kN z F W (b) cos 2.4647 4.2 x T 54.1 x T q W
cos 2.7 4.2 y T 130.0 y T q W 2.0681 cos 4.0 z T 60.5 z T q W
To stabilize a tree partially uprooted in a storm, cables AB and AC are attached to the upper trunk of the tree and then are fastened to steel rods anchored in the ground. Knowing that the tension in cable AC is 3.6 kN, determine (a) the components of the force exerted by this cable on the tree, (b) the angles Tx,Ty, and Tz that the force forms with axes at A which
are parallel to the coordinate axes.
SOLUTION
(a) Fx
3.6 kN cos 45 sin 25
q q 1.0758 kN
1.076 kN x F W
3.6 kN sin 45
2.546 kN y F q 2.55 kN y F W
3.6 kN cos 45 cos 25
2.3071 kN z F q q 2.31 kN z F W (b) cos 1.0758 3.6 x T
2.546 cos 3.6 y T 135.0 y T q W 2.3071 cos 3.6 z T 50.1 z T q W
A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30q angles with the vertical. Knowing that the x component of the force exerted by wire AD on the plate is 220.6 N, determine (a) the tension in wire AD, (b) the angles Tx,
Ty, and Tz that the force exerted at A forms with the coordinate axes.
SOLUTION
(a) Fx Fsin 30 sin 50q q 220.6 N (Given) 220.6 N 575.95 N sin30 sin50q q F 576 N F W (b) cos 220.6 0.3830 575.95 T x x F F 67.5 x T q W cos 30 498.79 N y F F q 498.79 cos 0.86605 575.95 y y F F T 30.0 y T q W Fz Fsin 30 cos 50q q
575.95 N sin 30 cos 50
q q 185.107 N 185.107 cos 0.32139 575.95 z z F F T
A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30q angles with the vertical. Knowing that the z component of the force exerted by wire BD on the plate is –64.28 N, determine (a) the tension in wire BD, (b) the angles Tx,
Ty, and Tz that the force exerted at B forms with the coordinate axes.
SOLUTION
(a) Fz Fsin 30 sin 40q q 64.28 N (Given)
64.28 N 200.0 N sin30 sin40q q F F 200 NW (b) Fx Fsin 30 cos 40q q
200.0 N sin 30 cos 40
q q 76.604 N 76.604 cos 0.38302 200.0 x x F F T Tx 112.5q W Fy Fcos 30q 173.2 N cos 173.2 0.866 200 y y F F T Ty 30.0q W Fz 64.28 N 64.28 cos 0.3214 200 z z F F T Tz 108.7q W
A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30q angles with the vertical. Knowing that the tension in wire CD is 120 lb, determine (a) the components of the force exerted by this wire on the plate, (b) the angles Tx,Ty, and Tz that the force forms with the coordinate axes.
SOLUTION
(a) Fx
120 lb sin 30 cos 60
q q 30 lb
30.0 lb x F W
120 lb cos 30
103.92 lb y F q 103.9 lb y F W
120 lb sin 30 sin 60
51.96 lb z F q q 52.0 lb z F W (b) cos 30.0 0.25 120 x x F F T 104.5 x T q W 103.92 cos 0.866 120 y y F F T 30.0 y T q W 51.96 cos 0.433 120 z z F F T 64.3 z T q W
A horizontal circular plate is suspended as shown from three wires which are attached to a support at D and form 30q angles with the vertical. Knowing that the x component of the forces exerted by wire CD on the plate is –40 lb, determine (a) the tension in wire CD, (b) the angles Tx,Ty,
andTz that the force exerted at C forms with the coordinate axes.
SOLUTION
(a) Fx Fsin 30 cos 60q q 40 lb (Given) 40 lb 160 lb sin30 cos60q q F 160.0 lb F W (b) cos 40 0.25 160 x x F F T 104.5 x T q W
160 lb cos 30
103.92 lb y F q cos 103.92 0.866 160 y y F F T 30.0 y T q W
160 lb sin 30 sin 60
69.282 lb z F q q 69.282 cos 0.433 160 z z F F T 64.3 T q W
Determine the magnitude and direction of the force
800 lb
260 lb
320 lb
. F i j k SOLUTION
2
2
2 2 2 2 800 lb 260 lb 320 lb x y z F F F F F 900 lbW 800 cos 0.8889 900 x x F F T Tx 27.3q W cos 260 0.2889 900 y y F F T Ty 73.2q W cos 320 0.3555 900 z z F F T Tz 110.8q W
Determine the magnitude and direction of the force
400 N
1200 N
300 N
. F i j k SOLUTION
2
2
2 2 2 2 400 N 1200 N 300 N x y z F F F F F 1300 NW 400 cos 0.30769 1300 x x F F T Tx 72.1q W cos 1200 0.92307 1300 y y F F T Ty 157.4q W cos 300 0.23076 1300 z z F F T Tz 76.7q W
A force acts at the origin of a coordinate system in a direction defined by the angles Tx 64.5q and Tz 55.9q. Knowing that the y component of
the force is –200 N, determine (a) the angle Ty, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
2
2
2
2
2
2
cosTx cosTy cosTz 1 cosTy 1 cosTy cosTz Since Fy we must have cos0 Ty 0
Thus, taking the negative square root, from above, we have:
2
2
cosTy 1 cos 64.5q cos 55.9q 0.70735 Ty 135.0q W (b) Then: 200 N 282.73 N cos 0.70735 y y F F T
and Fx FcosTx
282.73 N cos 64.5
q Fx 121.7 NW
Fz FcosTz
282.73 N cos 55.9
q Fy 158.5 NW
A force acts at the origin of a coordinate system in a direction defined by the angles Tx 75.4q and Ty 132.6q. Knowing that the z component of
the force is –60 N, determine (a) the angle Tz, (b) the other components
and the magnitude of the force.
SOLUTION
(a) We have
2
2
2
2
2
2
cosTx cosTy cosTz 1 cosTy 1 cosTy cosTz Since Fz we must have cos0 Tz 0
Thus, taking the negative square root, from above, we have:
2
2
cosTz 1 cos 75.4q cos132.6q 0.69159 Tz 133.8q W (b) Then: 60 N 86.757 N cos 0.69159 z z F F T F 86.8 NW
and Fx FcosTx
