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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

(a)

(b)

We measure: 8.4 kNR =

19 =

8.4 kN=R 19

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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

atA using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

We measure: 51.3 , 59 = =

(a)

(b)

We measure: 575 N, 67= = R

575 N=R 67

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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 themagnitude and direction of their resultant using (a) the parallelogram law,

(b) the triangle rule.

SOLUTION

(a)

(b)

We measure: 37 lb, 76= = R

37 lb=R 76

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PROBLEM 2.4

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 themagnitude and direction of their resultant using (a) the parallelogram law,

(b) the triangle rule.

SOLUTION

(a)

(b)

We measure: 61.5 lb, 86.5= = R

61.5 lb=R 86.5

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PROBLEM 2.5

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 ofR.

SOLUTION

Graphically, by the triangle law

We measure: 2 108 NF

77 NR

By trigonometry: Law of Sines

2 120

sin sin 38 sin

F R

= =

90 28 62 , 180 62 38 80 = = = =

Then:

2 120 N

sin 62 sin 38 sin80

F R= =

or (a) 2 107.6 NF =

(b) 75.0 NR =

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PROBLEM 2.6

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 ofR.

SOLUTION

Using the Law of Sines

1 80

sin sin 38 sin

F R

= =

90 10 80 , 180 80 38 62 = = = =

Then:

1 80 N

sin80 sin 38 sin 62

F R= =

or (a) 1 89.2 NF =

(b) 55.8 NR =

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PROBLEM 2.7

The 50-lb force is to be resolved into components along lines -a a and

- .b b (a) Using trigonometry, determine the angle knowing that thecomponent along -a a is 35 lb. (b) What is the corresponding value ofthe component along - ?b b

SOLUTION

Using the triangle rule and the Law of Sines

(a)sin sin 40

35 lb 50 lb

=

sin 0.44995 =

26.74 =

Then: 40 180 + + =

113.3 =

(b) Using the Law of Sines:

50 lb

sin sin 40

bbF

=

71.5 lbbbF =

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PROBLEM 2.8

The 50-lb force is to be resolved into components along lines -a a and

- .b b (a) Using trigonometry, determine the angle knowing that thecomponent along -b b is 30 lb. (b) What is the corresponding value ofthe component along - ?a a

SOLUTION


(a)sin sin 40

30 lb 50 lb

=

sin 0.3857 =

22.7 =

(b) 40 180 + + =

117.31 =

50 lb

sin sin 40

aaF

=

sin50 lb

sin 40

=

aaF

69.1lbaaF =

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PROBLEM 2.9

To steady a sign as it is being lowered, two cables are attached to the sign

at A. Using trigonometry and knowing that = 25, determine (a) therequired magnitude of the force P if the resultant R of the two forces

applied atA is to be vertical, (b) the corresponding magnitude ofR.

SOLUTION


Have: ( )180 35 25 = +

120=

Then:360 N

sin 35 sin120 sin 25

P R= =

or (a) 489 NP =

(b) 738 NR =

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PROBLEM 2.10

To steady a sign as it is being lowered, two cables are attached to the sign

atA. Using trigonometry and knowing that the magnitude ofP is 300 N,

determine (a) the required angle if the resultant R of the two forces

applied atA is to be vertical,(b) the corresponding magnitude ofR.

SOLUTION


(a) Have:360 N 300 N

sin sin 35=

sin 0.68829 =

43.5 =

(b) ( )180 35 43.5 = +

101.5=

Then:300 N

sin101.5 sin 35

R=

or 513 NR =

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PROBLEM 2.11

Two forces are applied as shown to a hook support. Using trigonometry

and knowing that the magnitude ofP is 14 lb, determine (a) the required

angle if the resultant R of the two forces applied to the support is to be

horizontal, (b) the corresponding magnitude ofR.

SOLUTION


(a) Have:20 lb 14 lb

sin sin 30=

sin 0.71428 =

45.6 =

(b) ( )180 30 45.6 = +

104.4=

Then:14 lb

sin104.4 sin 30

R=

27.1 lbR =

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PROBLEM 2.12

For the hook support of Problem 2.3, using trigonometry and knowing

that the magnitude ofP 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 ofR.

Problem 2.3: Two forces P and Q are applied as shown at point A of ahook support. Knowing that P = 15 lb and Q = 25 lb, determinegraphically the magnitude and direction of their resultant using (a) the

parallelogram law, (b) the triangle rule.

SOLUTION


(a) Have:25 lb

sin15 sin 30

Q=

12.94 lbQ =

(b) ( )180 15 30 = +

135=

Thus:25 lb

sin135 sin 30

R=

sin13525 lb 35.36 lb

sin30R

= =

35.4 lbR =

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PROBLEM 2.13

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 ofR.

Problem 2.11: Two forces are applied as shown to a hook support. Usingtrigonometry and knowing that the magnitude of P is 14 lb, determine

(a) the required angle if the resultant R of the two forces applied to the

support is to be horizontal, (b) the corresponding magnitude ofR.

SOLUTION

(a) The smallest force Pwill be perpendicular to R, that is, vertical

( )20 lb sin30P =

10 lb= 10 lb=P

(b) ( )20 lb cos30R =

17.32 lb= 17.32 lbR =

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PROBLEM 2.14

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 atA is vertical, (b) the corresponding magnitude

ofR.

SOLUTION

We observe that force P is minimum when is 90 , that is, P is horizontal

Then: (a) ( )360 N sin35P =

or 206 N=P

And: (b) ( )360 N cos35R =

or 295 NR =

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PROBLEM 2.15

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 = 40.

Problem2.11: 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 if the resultant R of the two forces applied to the

support is to be horizontal, (b) the corresponding magnitude ofR.

SOLUTION

Using the force triangle and the Law of Cosines

( ) ( ) ( )( )2 22 10 lb 20 lb 2 10 lb 20 lb cos110R = +

( ) 2100 400 400 0.342 lb = +

2636.8 lb=

25.23 lbR =

Using now the Law of Sines

10 lb 25.23 lb

sin sin110=

10 lbsin sin110

25.23 lb

=

0.3724=

So: 21.87 =

Angle of inclination ofR, is then such that:

30 + =

8.13 =

Hence: 25.2 lb=R 8.13

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PROBLEM 2.16

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: ( )180 50 25 = +

105=

Then: ( ) ( ) ( )( )2 22 4.5 kN 6 kN 2 4.5 kN 6 kN cos105R = +

270.226 kN=

or 8.3801 kNR =

Now:8.3801 kN 6 kN

sin105 sin=

6 kNsin sin105

8.3801 kN

=

0.6916=

43.756 =

8.38 kN=R 18.76

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PROBLEM 2.17


Problem 2.2: The cable staysAB andAD help support poleAC. 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 atA using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

From the geometry of the problem:

1 2tan 38.662.5

= =

1 1.5tan 30.962.5

= =

Now: ( )180 38.66 30.96 110.38 = + =

And, using the Law of Cosines:

( ) ( ) ( )( )2 22 500 N 160 N 2 500 N 160 N cos110.38R = +

2331319 N=

575.6 NR =

Using the Law of Sines:

160 N 575.6 N

sin sin110.38=

160 Nsin sin110.38

575.6 N

=

0.2606=

15.1 =

( )90 66.44 = + =

576 N=R 66.4

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PROBLEM 2.18


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, determinegraphically the magnitude and direction of their resultant using (a) theparallelogram law, (b) the triangle rule.

SOLUTION

Using the force triangle and the Laws of Cosines and Sines

We have:

( )180 15 30 = +

135=

Then: ( ) ( ) ( ) ( )2 22 15 lb 25 lb 2 15 lb 25 lb cos135R = +

21380.3 lb=

or 37.15 lbR =

and

25 lb 37.15 lb

sin sin135=

25 lbsin sin135

37.15 lb

=

0.4758=

28.41 =

Then: 75 180 + + =

76.59 =

37.2 lb=R 76.6

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PROBLEM 2.19

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, usingtrigonometry, the magnitude and direction of the resultant of the forces

applied to the bracket by membersA andB.

SOLUTION


We have: ( )180 45 25 110 = + =

Then: ( ) ( ) ( )( )2 22

30 kN 20 kN 2 30 kN 20 kN cos110R = +

21710.4 kN=

41.357 kNR =

and

20 kN 41.357 kN

sin sin110=

20 kNsin sin110

41.357 kN

=

0.4544=

27.028 =

Hence: 45 72.028 = + =

41.4 kN=R 72.0

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PROBLEM 2.20

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, usingtrigonometry, the magnitude and direction of the resultant of the forces

applied to the bracket by membersA andB.

SOLUTION


We have: ( )180 45 25 110 = + =

Then: ( ) ( ) ( )( )2 22

30 kN 20 kN 2 30 kN 20 kN cos110R = +

21710.4 kN=

41.357 kNR =

and

30 kN 41.357 kN

sin sin110=

30 kNsin sin110

41.357 kN

=

0.6816=

42.97 =

Finally: 45 87.97 = + =

41.4 kN=R 88.0

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PROBLEM 2.21

Determine thex andy components of each of the forces shown.

SOLUTION

20 kN Force:

( )20 kN cos40 ,xF = + 15.32 kNxF =

( )20 kN sin 40 ,yF = + 12.86 kNyF =

30 kN Force:

( )30 kN cos70 ,xF = 10.26 kNxF =

( )30 kN sin 70 ,yF = + 28.2 kNyF =

42 kN Force:

( )42 kN cos20 ,

xF = 39.5 kN

xF =

( )42 kN sin 20 ,yF = + 14.36 kNyF =

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PROBLEM 2.22


SOLUTION

40 lb Force:

( )40 lb sin 50 ,xF = 30.6 lbxF =

( )40 lb cos50 ,yF = 25.7 lbyF =

60 lb Force:

( )60 lb cos60 ,xF = + 30.0 lbxF =

( )60 lb sin 60 ,yF = 52.0 lbyF =

80 lb Force:

( )80 lb cos25 ,xF = + 72.5 lbxF =

( )80 lb sin 25 ,yF = + 33.8 lbyF =

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PROBLEM 2.23


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 ,102xF = 48.0 lbxF =

( )90

102 lb ,102

yF = + 90.0 lbyF =

212 lb Force:

( )56

212 lb ,106

xF = + 112.0 lbxF =

( )90

212 lb ,106

yF = + 180.0 lbyF =

400 lb Force:

( )80

400 lb ,100

xF = 320 lbxF =

( )60

400 lb ,100

yF = 240 lbyF =

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PROBLEM 2.24


SOLUTION

We compute the following distances:

( ) ( )2 2

70 240 250 mmOA = + =

( ) ( )2 2

210 200 290 mmOB = + =

( ) ( )2 2

120 225 255 mmOC = + =

500 N Force:

70500 N250

xF =

140.0 NxF =

240500 N

250yF

= +

480 NyF =

435 N Force:

210435 N

290xF

= +

315 NxF =

200435 N

290yF

= +

300 NyF =

510 N Force:

120510 N

255xF

= +

240 NxF =

225510 N

255yF

=

450 NyF =

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PROBLEM 2.25

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)cos40

xPP =

135 N

cos40=

or 176.2 NP =

(b) tan 40 sin 40y xP P P= =

( )135 N tan40=

or 113.3 NyP =

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PROBLEM 2.26

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)sin35

yPP =

960 N

sin35=

or 1674 NP =

(b)tan35

yx

PP =

960 N

tan35=

or 1371NxP =

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PROBLEM 2.27

Member CB of the vise shown exerts on blockB 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 onB along CB, and the horizontal component ofP is 260 lb.xP =

Then:

(a) sin50xP P=

sin50xPP =

260 lb

sin50=

339.4 lb= 339 lbP =

(b) tan50x yP P=

tan50

xy

PP =

260 lb

tan50=

218.2 lb= 218 lby =P

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PROBLEM 2.28

Activator rod AB exerts on crankBCD a force P directed along line AB.

Knowing that P must have a 25-lb component perpendicular to armBCof

the crank, determine (a) the magnitude of the force P, (b) its component

along line BC.

SOLUTION

Using the x andy axes shown.

(a) 25 lbyP =

Then:sin75

yPP =

25 lb

sin75=

or 25.9 lbP =

(b)tan75

yx

PP =

25 lb

tan 75=

or 6.70 lbxP =

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PROBLEM 2.29

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 ACis 450 N.

Then:

(a)450 N

549.3 Ncos35

P = =

549 NP =

(b) ( )450 N tan35xP =

315.1 N=

315 NxP =

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PROBLEM 2.30

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)sin38

xPP =

200 N

sin38=

324.8 N= or 325 NP =

(b)tan38

xy

PP =

200 N

tan38=

255.98 N=

or 256 NyP =

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PROBLEM 2.31

Determine the resultant of the three forces of Problem 2.24.

Problem 2.24: Determine thex andy 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= = +R F i j

Then:

1 330tan 38.5415

= =

( ) ( )2 2

415 N 330 N 530.2 NR = + =

Thus: 530 N=

R 38.5

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PROBLEM 2.32



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= = +R F i j

Then:

1 55.42tan 58.134.44

= =

( ) ( )2 2

55.42 kN 34.44 N 65.2 kNR = + =

65.2 kNR = 58.2

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PROBLEM 2.33


Problem 2.22: Determine the x andy components of each of the forces

shown.

SOLUTION

The components of the forces were determined in 2.23.

x yR R= +R i j

( ) ( )71.9 lb 43.86 lb= i j

43.86tan

71.9 =

31.38 =

( ) ( )2 2

71.9 lb 43.86 lbR = +

84.23 lb=

84.2 lb=R 31.4

Force comp. (lb)x comp. (lb)y

40 lb 30.6 25.7

60 lb 30 51.96

80 lb 72.5 33.8

71.9xR = 43.86yR =

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PROBLEM 2.34



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:

30.0tan

256 =

1 30.0tan 6.68256

= =

and

( ) ( )2 2

256 lb 30.0 lbR = +

257.75 lb=

258 lb=R 6.68

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PROBLEM 2.35

Knowing that 35 , = determine the resultant of the three forces

shown.

SOLUTION

300-N Force:

( )300 N cos 20 281.9 NxF = =

( )300 N sin 20 102.6 NyF = =

400-N Force:

( )400 N cos 55 229.4 NxF = =

( )400 N sin 55 327.7 NyF = =

600-N Force:

( )600 N cos 35 491.5 NxF = =

( )600 N sin 35 344.1 NyF = =

and

1002.8 Nx xR F= =

86.2 Ny yR F= =

( ) ( )2 2

1002.8 N 86.2 N 1006.5 NR = + =

Further:

86.2

tan 1002.8 =

1 86.2tan 4.911002.8

= =

1007 N=R 4.91

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PROBLEM 2.36


shown.

SOLUTION

300-N Force:

( )300 N cos 20 281.9 NxF = =

( )300 N sin 20 102.6 NyF = =

400-N Force:

( )400 N cos85 34.9 NxF = =

( )400 N sin 85 398.5 NyF = =

600-N Force:

( )600 N cos 5 597.7 NxF = =

( )600 N sin 5 52.3 NyF = =

and

914.5 Nx xR F= =

448.8 Ny yR F= =

( ) ( )2 2

914.5 N 448.8 N 1018.7 NR = + =

Further:

448.8tan

914.5 =

1 448.8tan 26.1914.5

= =

1019 N=R 26.1

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PROBLEM 2.37

Knowing that the tension in cableBCis 145 lb, determine the resultant of

the three forces exerted at pointB of beamAB.

SOLUTION

Cable BC Force:

( )84

145 lb 105 lb116x

F = =

( )80

145 lb 100 lb116

yF = =

100-lb Force:

( )3

100 lb 60 lb5

xF = =

( )4

100 lb 80 lb5

yF = =

156-lb Force:

( )12

156 lb 144 lb13

xF = =

( )5

156 lb 60 lb13

yF = =

and

21 lb, 40 lbx x y yR F R F= = = =

( ) ( )2 2

21 lb 40 lb 45.177 lbR = + =

Further:

40tan

21 =

1 40tan 62.321

= =

Thus: 45.2 lb=R 62.3

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PROBLEM 2.38


shown.

SOLUTION

The resultant forceR has thex- andy-components:

( ) ( ) ( )140 lb cos50 60 lb cos85 160 lb cos50x xR F= = +

7.6264 lbxR =

and

( ) ( ) ( )140 lb sin 50 60 lb sin85 160 lb sin 50y yR F= = + +

289.59 lbyR =

Further:

290tan

7.6 =

1 290tan 88.57.6

= =

Thus: 290 lb=R 88.5

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PROBLEM 2.39

Determine (a) the required value of if the resultant of the three forcesshown is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION

For an arbitrary angle , we have:

( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cosx xR F = = + +

(a) So, forR to be vertical:

( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cos 0x xR F = = + + =

Expanding,

( )cos 3 cos cos35 sin sin 35 0 + =

Then:

13

cos35

tan sin35

=

or

11 3

cos35tan 40.265

sin35

= = 40.3 =

(b) Now:

( ) ( ) ( )140 lb sin 40.265 60 lb sin 75.265 160 lb sin 40.265y yR R F= = = + +

252 lbR R= =

39

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PROBLEM 2.40

For the beam of Problem 2.37, determine (a) the required tension in cableBCif 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 cableBCis 145 lb, determinethe resultant of the three forces exerted at pointB of beamAB.

SOLUTION

We have:

( ) ( )84 12 3156 lb 100 lb116 13 5

x x BCR F T= = +

or 0.724 84 lbx BCR T= +

and

( ) ( )80 5 4

156 lb 100 lb116 13 5

y y BCR F T= =

0.6897 140 lby BCR T=

(a) So, forR to be vertical,

0.724 84 lb 0x BCR T= + =

116.0 lbBCT = (b) Using

116.0 lbBCT =

( )0.6897 116.0 lb 140 lb 60 lbyR R= = =

60.0 lbR R= =

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PROBLEM 2.41

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

Choosex-axis along bar AB.

Then

(a) Require

( ) ( )0: 4 kN cos 25 5.2 kN sin 35 sin 65 0y y AER F T= = + =

or 7.2909 kNAET =

7.29 kNAET =

(b) xR F=

( ) ( ) ( )4 kN sin 25 5.2 kN cos35 7.2909 kN cos 65=

9.03 kN=

9.03 kNR =

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PROBLEM 2.42

For the block of Problems 2.35 and 2.36, determine (a) the required value

of 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 35 , = determine the resultant of the

three forces shown.

Problem 2.36: Knowing that 65 , = determine the resultant of the

three forces shown.

SOLUTION

Selecting the x axis along ,aa we write

( ) ( )300 N 400 N cos 600 N sinx xR F = = + + (1)

( ) ( )400 N sin 600 N cosy yR F = = (2)

(a) Setting 0yR = in Equation (2):

Thus600

tan 1.5400

= =

56.3 =

(b) Substituting for in Equation (1):

( ) ( )300 N 400 N cos56.3 600 N sin 56.3xR = + +

1021.1 NxR =

1021 NxR R= =

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PROBLEM 2.43

Two cables are tied together at Cand 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:

16 210: 0

20 29x AC BCF T T = + =

or29 4

21 5BC ACT T=

and12 20

0: 600 lb 020 29

y AC BCF T T = + =

or12 20 29 4

600 lb 020 29 21 5

AC ACT T

+ =

Hence: 440.56 lbACT =

(a) 441 lbACT =

(b) 487 lbBCT =

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PROBLEM 2.44

Knowing that 25 , = determine the tension (a) in cable AC, (b) in

ropeBC.

SOLUTION

Free-Body Diagram Force Triangle

Law of Sines:

5 kN

sin115 sin 5 sin 60

AC BCT T= =

(a)5 kN

sin115 5.23 kNsin60

ACT = =

5.23 kNACT =

(b)5 kN

sin 5 0.503 kNsin60

BCT = =

0.503 kNBCT =

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PROBLEM 2.45

Knowing that 50 = 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 cableBC.

SOLUTION


Law of Sines:

400 lb

sin 25 sin 60 sin 95

AC BCF T= =

(a)400 lb

sin 25 169.69 lbsin95

ACF = =

169.7 lbACF =

(b)400

sin 60 347.73 lbsin95

BCT = =

348 lbBCT =

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PROBLEM 2.46

Two cables are tied together at Cand are loaded as shown. Knowing that

30 , = determine the tension (a) in cableAC, (b) in cableBC.

SOLUTION


Law of Sines:

2943 N

sin 60 sin 55 sin 65

AC BCT T= =

(a)2943 N

sin 60 2812.19 Nsin 65

ACT = =

2.81 kNACT =

(b)2943 N

sin 55 2659.98 Nsin65

BCT = =

2.66 kNBCT =

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PROBLEM 2.47

A chairlift has been stopped in the position shown. Knowing that each

chair weighs 300 N and that the skier in chairEweighs 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 pointB, the geometry gives:

1 9.9tan 30.5116.8

AB

= =

1 12tan 22.6128.8

BC

= =

Thus, in the force triangle, by the Law of Sines:

1190 N

sin 59.49 sin 7.87

BCT=

7468.6 NBC

T =

In the free-body diagram of point C(with Wthe sum of weights of chairand skier) the geometry gives:

1 1.32tan 10.397.2

CD

= =

Hence, in the force triangle, by the Law of Sines:

7468.6 N

sin12.23 sin100.39

W=

1608.5 NW =

Finally, the skier weight 1608.5 N 300 N 1308.5 N= =

skier weight 1309 N=

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PROBLEM 2.48

A chairlift has been stopped in the position shown. Knowing that each

chair weighs 300 N and that the skier in chair Fweighs 800 N, determine

the weight of the skier in chairE.

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 12tan 22.6228.8

EF

= =

1 1.32tan 10.397.2

DF

= =

Thus, in the force triangle, by the Law of Sines:

1100 N

sin100.39 sin12.23

EFT=

5107.5 NBCT =

In the free-body diagram of point E(with Wthe sum of weights of chairand skier) the geometry gives:

1 9.9tan 30.5116.8

AE

= =

Hence, in the force triangle, by the Law of Sines:

5107.5 N

sin 7.89 sin 59.49

W=

813.8 NW =

Finally, the skier weight 813.8 N 300 N 513.8 N= =

skier weight 514 N=

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PROBLEM 2.49

Four wooden members are joined with metal plate connectors and are in

equilibrium under the action of the four fences shown. Knowing thatFA = 510 lb and FB = 480 lb, determine the magnitudes of the other two

forces.

SOLUTION

Free-Body Diagram

Resolving the forces intox andy components:

( ) ( )0: 510 lb sin15 480 lb cos15 0x CF F = + =

or 332 lbCF =

( ) ( )0: 510 lb cos15 480 lb sin15 0y DF F = + =

or 368 lbDF =

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PROBLEM 2.50

Four wooden members are joined with metal plate connectors and are in

equilibrium under the action of the four fences shown. Knowing thatFA = 420 lb and FC = 540 lb, determine the magnitudes of the other two

forces.

SOLUTION

Resolving the forces intox andy components:

( ) ( )0: cos15 540 lb 420 lb cos15 0 or 671.6 lbx B BF F F = + + = =

672 lbBF =

( ) ( )0: 420 lb cos15 671.6 lb sin15 0y DF F = + =

or 232 lbDF =

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PROBLEM 2.51

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 andQ = 520 lb, determine the magnitudes of the forces exerted on the rodsA andB.

SOLUTION

Free-Body Diagram Resolving the forces intox andy directions:

0A B= + + + =R P Q F F

Substituting components:

( ) ( ) ( )400 lb 520 lb cos55 520 lb sin 55 = + R j i j

( ) ( )cos55 sin 55 0B A AF F F+ + =i i j

In they-direction (one unknown force)

( )400 lb 520 lb sin 55 sin 55 0AF + =

Thus,

( )400 lb 520 lb sin 551008.3 lb

sin55AF

+ = =

1008 lbAF =

In thex-direction:

( )520 lb cos55 cos55 0B AF F + =

Thus,

( )cos55 520 lb cos55B AF F=

( ) ( )1008.3 lb cos 55 520 lb cos 55=

280.08 lb=

280 lbBF =

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PROBLEM 2.52

Two forces P and Q are applied as shown to an aircraft connection.

Knowing that the connection is in equilibrium and that the magnitudes ofthe forces exerted on rods A and B are FA = 600 lb and FB = 320 lb,determine the magnitudes ofP and Q.

SOLUTION

Free-Body Diagram Resolving the forces intox andy directions:

0A B= + + + =R P Q F F

Substituting components:

( ) ( ) ( )320 lb 600 lb cos55 600 lb sin 55 = + R i i j

( ) ( )cos55 sin 55 0P Q Q+ + =i i j

In thex-direction (one unknown force)

( )320 lb 600 lb cos55 cos55 0Q + =

Thus,

( )320 lb 600 lb cos5542.09 lb

cos55Q

+ = =

42.1lbQ =

In they-direction:

( )600 lb sin 55 sin 55 0P Q =

Thus,

( )600 lb sin 55 sin 55 457.01 lbP Q= =

457 lbP =

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PROBLEM 2.53

Two cables tied together at C are loaded as shown. Knowing that

W = 840 N, determine the tension (a) in cableAC, (b) in cableBC.

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 05 17 17

x CA CBF T T = + =

or

1 5200 N

5 17CA CBT T + = (1)

and

( )4 8 8

0: 680 N 840 N 05 17 17y CA CBF T T = + =

or

1 2290 N

5 17CA CBT T+ = (2)

Solving Equations (1) and (2) simultaneously:

(a) 750 NCAT =

(b) 1190 NCBT =

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PROBLEM 2.54

Two cables tied together at Care loaded as shown. Determine the range

of values of W for which the tension will not exceed 1050 N in eithercable.

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 05 17 17

x CA CBF T T = + =

or

1 5200 N

5 17CA CBT T + = (1)

and

( )4 8 8

0: 680 N 05 17 17

y CA CBF T T W = + =

or

1 2 180 N

5 17 4+ = +CA CBT T W (2)

Then, from Equations (1) and (2)

17680 N

28

25

28

CB

CA

T W

T W

= +

=

Now, with 1050 NT

25: 1050 N 28

CA CAT T W= =

or 1176 NW =

and

17: 1050 N 680 N

28CB CBT T W= = +

or 609 NW = 0 609 N W

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PROBLEM 2.55

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 40 = and = 35, that thecombined 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

cableDE.

SOLUTION

Note: In Problems 2.55 and 2.56 the cabin is considered as a particle. Ifconsidered 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: cos35 cos40 cos 40 0x ACB DEF T T = =

or

0.0531 0.766 0ACB DET T = (1)

and

( )0: sin 40 sin 35 sin 40 24.8 kN 0y ACB DEF T T = + =

or

0.0692 0.643 24.8 kNACB DET T+ = (2)

From (1)

14.426ACB DET T=

Then, from (2)

( )0.0692 14.426 0.643 24.8 kNDE DET T+ =

and

(b) 15.1 kNDET =

(a) 218 kNACBT =

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PROBLEM 2.56

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 cableDE. Knowing that 42 = and = 32, 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 supportsystem, and its passengers, (b) the tension in the support cableACB.

SOLUTION

Free-Body Diagram

First, consider the sum of forces in thex-direction because there is only one unknown force:

( ) ( )0: cos32 cos42 20 kN cos 42 0x ACBF T = =

or

0.1049 14.863 kNACBT =

(b) 141.7 kNACBT =

Now

( ) ( )0: sin 42 sin 32 20 kN sin 42 0

y ACBF T W = + =

or

( )( ) ( )( )141.7 kN 0.1392 20 kN 0.6691 0W+ =

(a) 33.1 kNW =

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PROBLEM 2.57

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 theconstants 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 hypotenuseAD are in the ratio 8:15:17.

The sides of the triangle with hypotenuseAB are in the ratio 3:4:5.

The sides of the triangle with hypotenuseACare in the ratio 7:24:25.

Then:

( )AB AB AB oF k L L=

and

( ) ( )2 2

0.44 m 0.33 m 0.55 mABL = + =

So:

( )1500 N/m 0.55 m 0.45 mABF =

150 N=

Similarly,

( )AD AD AD oF k L L=

Then:

( ) ( )2 2

0.66 m 0.32 m 0.68 mADL = + =

( )1500 N/m 0.68 m 0.45 mADF =

115 N=

(a)

( ) ( )4 7 15

0: 150 N 115 N 05 25 17

x ACF T = + =

or

66.18 NACT = 66.2 NACT =

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PROBLEM 2.57 CONTINUED

(b) and

( ) ( ) ( )3 24 8

0: 150 N 66.18 N 115 N 05 25 17

yF W = + + =

or 208 N=W

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PROBLEM 2.58

A load of weight 400 N is suspended from a spring and two cords which

are attached to blocks of weights 3Wand Was shown. Knowing that theconstant of the spring is 800 N/m, determine (a) the value ofW, (b) the

unstretched length of the spring.

SOLUTION

Free-Body Diagram At AFirst note from geometry:

The sides of the triangle with hypotenuseAD are in the ratio 12:35:37.

The sides of the triangle with hypotenuseACare in the ratio 3:4:5.

The sides of the triangle with hypotenuseAB are also in the ratio

12:35:37.

Then:

( ) ( )4 35 12

0: 3 05 37 37

x sF W W F = + + =

or

4.4833sF W=

and

( ) ( )3 12 35

0: 3 400 N 05 37 37

y sF W W F = + + =

Then:

( ) ( ) ( )3 12 35

3 4.4833 400 N 05 37 37

W W W+ + =

or

62.841NW =

and

281.74 NsF =

or

(a) 62.8 NW =

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PROBLEM 2.58 CONTINUED

(b) Have spring force

( )s AB oF k L L=

Where

( )AB AB AB oF k L L=

and

( ) ( )2 2

0.360 m 1.050 m 1.110 mABL = + =

So:

( )0281.74 N 800 N/m 1.110 mL=

or 0 758 mmL =

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PROBLEM 2.59

For the cables and loading of Problem 2.46, determine (a) the value of

for which the tension in cable BC is as small as possible, (b) the

corresponding value of the tension.

SOLUTION

The smallest BCT is when BCT is perpendicular to the direction of ACT

Free-Body Diagram At C Force Triangle

(a) 55.0 =

(b) ( )2943 N sin55BCT =

2410.8 N=

2.41kNBCT =

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PROBLEM 2.60

Knowing that portionsACandBCof cableACB 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

( )For 725 NT = 0: 2 1000 N 0y yF T = =

500 NyT =

2 2 2x yT T T+ =

( ) ( )2 22 500 N 725 NxT + =

525 NxT =

By similar triangles:

1.5 m

725 525

BC=

2.07 m =BC

( )2 4.14 mL BC= =

4.14 mL =

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PROBLEM 2.61

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.

SOLUTION

Free-Body Diagram: C Force Tr iangle

Force triangle is isoceles with

2 180 85 =

47.5 =

(a) ( )2 200 lb cos 47.5 270 lbP = =

Since 0,P > the solution is correct. 270 lbP =

(b) 180 55 47.5 77.5 = = 77.5 =

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PROBLEM 2.62

Two cables tied together at C are loaded as shown. Knowing that the

maximum allowable tension is 300 lb in cable ACand 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.

SOLUTION

Fr ee-Body Diagram: C Force Tr iangle

(a) Law of Cosines:

( ) ( ) ( )( )2 22 300 lb 150 lb 2 300 lb 150 lb cos85P = +

323.5 lbP =

Since 300 lb,P > our solution is correct. 324 lbP =

(b) Law of Sines:

sin sin85

300 323.5

=

sin 0.9238 =

or 67.49 =

180 55 67.49 57.5 = =

57.5 =

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PROBLEM 2.63

For the structure and loading of Problem 2.45, determine (a) the value of

for which the tension in cable BC is as small as possible, (b) the

corresponding value of the tension.

SOLUTION

BCT must be perpendicular to ACF to be as small as possible.

Fr ee-Body Diagram: C Force Tr iangle is

a r ight tr iangle

(a) We observe: 55 = 55 =

(b) ( )400 lb sin 60BCT =

or 346.4 lbBCT = 346 lbBCT =

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PROBLEM 2.64

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 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: 0BD AB BC+ + =T F T

where magnitude and direction of BDT are known, and the direction

of ABF is known.

Then, in a force triangle:

By observation, BCT is minimum when 90.0 = (b) Have ( ) ( )70 lb sin 180 70 30BCT =

68.93 lb=

68.9 lbBCT=

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PROBLEM 2.65

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: ( )s AB ABF k L L=

where:

( ) ( )2 2

0.3 m 0.4 m 0.3 2 mAB ABL L = + =

0.5 m=

Then: ( )660 N/m 0.5 0.3 2 msF =

49.986 N=

For the collar:

( )4

0: 49.986 N 05

yF W = + =

or 40.0 NW =

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PROBLEM 2.66

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

ofh for which the system is in equilibrium.

SOLUTION

Free-Body Diagram: Collar A( )

2 20: 0

0.3y s

hF W F

h

= + =+

or 240 0.09shF h= +

Now.. ( )s AB ABF k L L=

where

( )

2 20.3 m 0.3 2 mAB AB

L h L = + =

Then: ( )2 2560 0.09 0.3 2 40 0.09h h h + = +

or ( ) 214 1 0.09 4.2 2 mh h h h + =

Solving numerically,

415 mmh =

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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. Thiscan be proved by the methods of Chapter 4.)

SOLUTION

Fr ee-Body Diagram of pulley

(a)

(b)

(c)

(d)

(e)

( )( )20: 2 280 kg 9.81 m/s 0yF T = =

( )1

2746.8 N2

T =

1373 NT =

( )( )20: 2 280 kg 9.81 m/s 0yF T = =

( )1

2746.8 N2

T =

1373 NT =

( )( )20: 3 280 kg 9.81 m/s 0yF T = =

( )1

2746.8 N

3

T =

916 NT =

( )( )20: 3 280 kg 9.81 m/s 0yF T = =

( )1

2746.8 N3

T =

916 NT =

( )( )20: 4 280 kg 9.81 m/s 0

y

F T = =

( )1

2746.8 N4

T =

687 NT =

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PROBLEM 2.68

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 inthe 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

Fr ee-Body Diagram of pulley

and crate

(b)

(d)

( )( )20: 3 280 kg 9.81 m/s 0yF T = =

( )1

2746.8 N3

T =

916 NT =

( )

( )

20: 4 280 kg 9.81 m/s 0yF T = =

( )1

2746.8 N4

T =

687 NT =

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