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PROBLEM 2.119

A force P is applied as shown to a uniform cone which is supported by

three cords, where the lines of action of the cords pass through the vertex

A of the cone. Knowing that the cone weighs 2.4 lb and that P = 0,determine the tension in each cord.

SOLUTION

Note that because the line of action of each of the cords passes through the vertex A of the cone, the cords all

have the same length, and the unit vectors lying along the cords are parallel to the unit vectors lying along the

generators of the cone.

Thus, for example, the unit vector along BEis identical to the unit vector along the generatorAB.

Hence:

cos45 8 sin 45

65AB BE

+

= =i j k

It follows that:

cos45 8 sin 45

65BE BE BE BE

T T +

= =

i j kT

cos30 8 sin 3065

CF CF CF CF T T + +

= =

i j kT

cos15 8 sin15

65DG DG DG DG

T T +

= =

i j kT

At A: 0: 0BE CF DG

= + + + + =F T T T W P

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PROBLEM 2.119 CONTINUED

Then, isolating the factors if , , andi j kwe obtain three algebraic equations:

: cos45 cos30 cos15 065 65 65

BE CF DGT T T

+ =i

or cos45 cos30 cos15 0BE CF DG

T T T + = (1)

8 8 8: 0

65 65 65BE CF DG

T T T W + + =j

or2.4

65 0.3 658

BE CF DGT T T+ + = = (2)

: sin 45 sin 30 sin15 065 65 65

BE CF DGT T T

P + =k

or sin 45 sin 30 sin15 65BE CF DG

T T T P + = (3)

With 0,P = the tension in the cords can be found by solving the resulting Equations (1), (2), and (3) usingconventional methods in Linear Algebra (elimination, matrix methods or iterationwith MATLAB or Maple,

for example). We obtain

0.299 lbBE

T =

1.002 lbCF

T =

1.117 lbDG

T =

133

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PROBLEM 2.120

A force P is applied as shown to a uniform cone which is supported by

three cords, where the lines of action of the cords pass through the vertex

A of the cone. Knowing that the cone weighs 2.4 lb and that P = 0.1 lb,determine the tension in each cord.

SOLUTION

See Problem 2.121 for the analysis leading to the linear algebraic Equations (1), (2), and (3) below:

cos 45 cos30 cos15 0BE CF DGT T T + = (1)

0.3 65BE CF DG

T T T+ + = (2)

sin 45 sin 30 sin15 65BE CF DG

T T T P + = (3)

With 0.1lb,=P solving (1), (2), and (3), using conventional methods in Linear Algebra (elimination, matrixmethods or iterationwith MATLAB or Maple, for example), we obtain

1.006 lbBE

T =

0.357 lbCF

T =

1.056 lbDG

T =

134

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PROBLEM 2.121

Using two ropes and a roller chute, two workers are unloading a 200-kg

cast-iron counterweight from a truck. Knowing that at the instant shown

the counterweight is kept from moving and that the positions of points A,B, and Care, respectively, A(0, 0.5 m, 1 m), B(0.6 m, 0.8 m, 0), and

C(0.7 m, 0.9 m, 0), and assuming that no friction exists between thecounterweight and the chute, determine the tension in each rope. (Hint:

Since there is no friction, the force exerted by the chute on the

counterweight must be perpendicular to the chute.)

SOLUTION

From the geometry of the chute:

( ) ( )2 0.8944 0.44725

NN= + = +N j k j k

As in Problem 2.11, for example, the force in each rope can be written as

the product of the magnitude of the force and the unit vector along the

cable. Thus, with

( ) ( ) ( )0.6 m 1.3 m 1 mAB = + +i j kJJJG

( ) ( ) ( )2 2 2

0.6 m 1.3 m 1 m 1.764 mAB = + + =

( ) ( ) ( )0.6 m 1.3 m 1 m1.764 m

AB

AB AB AB

AB TT T

AB = = = + + T i j k

JJJG

( )0.3436 0.7444 0.5726AB ABT= + +T i j k

and ( ) ( ) ( )0.7 m 1.4 m 1 mAC = + i j kJJJG

( ) ( ) ( )2 2 2

0.7 m 1.4 m 1 m 1.8574 mAC = + + =

( ) ( ) ( )0.7 m 1.4 m 1 m1.764 mAC

AC AC AC

AC TT T

AC = = = + T i j k

JJJG

( )0.3769 0.7537 0.5384AC ACT= + T i j k

Then: 0: 0AB AC = + + + =F N T T W

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PROBLEM 2.121 CONTINUED

With ( )( )200 kg 9.81 m/s 1962 N,W = = and equating the factors ofi, j,and kto zero, we obtain the linear algebraic equations:

: 0.3436 0.3769 0AB ACT T + =

i (1)

: 0.7444 0.7537 0.8944 1962 0AB AC

T T N+ + =j (2)

: 0.5726 0.5384 0.4472 0AB AC

T T N + =k (3)

Using conventional methods for solving Linear Algebraic Equations(elimination, MATLAB or Maple, for example), we obtain

1311 NN =

551 NAB

T =

503 NACT =

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PROBLEM 2.122

Solve Problem 2.121 assuming that a third worker is exerting a force

(180 N)= P i on the counterweight.

Problem 2.121: Using two ropes and a roller chute, two workers are

unloading a 200-kg cast-iron counterweight from a truck. Knowing that at

the instant shown the counterweight is kept from moving and that the

positions of points A, B, and C are, respectively, A(0, 0.5 m, 1 m),

B(0.6 m, 0.8 m, 0), and C(0.7 m, 0.9 m, 0), and assuming that no friction

exists between the counterweight and the chute, determine the tension in

each rope. (Hint: Since there is no friction, the force exerted by the chuteon the counterweight must be perpendicular to the chute.)

SOLUTION

From the geometry of the chute:

( ) ( )2 0.8944 0.44725

NN= + = +N j k j k

As in Problem 2.11, for example, the force in each rope can be written as

the product of the magnitude of the force and the unit vector along the

cable. Thus, with

( ) ( ) ( )0.6 m 1.3 m 1 mAB = + +i j kJJJG

( ) ( ) ( )2 2 2

0.6 m 1.3 m 1 m 1.764 mAB = + + =

( ) ( ) ( )0.6 m 1.3 m 1 m1.764 m

AB

AB AB AB

AB TT T

AB = = = + + T i j k

JJJG

( )0.3436 0.7444 0.5726AB ABT= + +T i j k

and ( ) ( ) ( )0.7 m 1.4 m 1 mAC = + i j kJJJG

( ) ( ) ( )2 2 2

0.7 m 1.4 m 1 m 1.8574 mAC = + + =

( ) ( ) ( )0.7 m 1.4 m 1 m1.764 mAC

AC AC AC

AC TT T

AC = = = + T i j k

JJJG

( )0.3769 0.7537 0.5384AC ACT= + T i j k

Then: 0: 0AB AC = + + + + =F N T T P W

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PROBLEM 2.122 CONTINUED

Where ( )180 N= P i

and ( )( )2200 kg 9.81 m/s = W j

( )1962 N= j

Equating the factors ofi, j, and kto zero, we obtain the linear equations:

: 0.3436 0.3769 180 0AB AC

T T + =i

: 0.8944 0.7444 0.7537 1962 0AB AC

N T T+ + =j

: 0.4472 0.5726 0.5384 0AB AC

N T T =k

Using conventional methods for solving Linear Algebraic Equations

(elimination, MATLAB or Maple, for example), we obtain

1302 NN =

306 NAB

T =

756 NAC

T =

138

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PROBLEM 2.123

A piece of machinery of weight Wis temporarily supported by cablesAB,

AC, and ADE. Cable ADE is attached to the ring at A, passes over thepulley atD and back through the ring, and is attached to the support at E.

Knowing that W = 320 lb, determine the tension in each cable. (Hint:

The tension is the same in all portions of cableADE

.)

SOLUTION

The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along

the cable. That is, with

( ) ( ) ( )9 ft 8 ft 12 ftAB = + i j kJJJG

( ) ( ) ( )2 2 2

9 ft 8 ft 12 ft 17 ftAB = + + =

( ) ( ) ( )9 ft 8 ft 12 ft17 ft

AB

AB AB AB

AB TT T

AB = = = + T i j k

JJJG

( )0.5294 0.4706 0.7059AB ABT= + T i j k

and

( ) ( ) ( )0 8 ft 6 ftAC = + +i j kJJJG

( ) ( ) ( )2 2 2

0 ft 8 ft 6 ft 10 ftAC = + + =

( ) ( ) ( )0 ft 8 ft 6 ft10 ft

AC

AC AC AC

AC TT T

AC = = = + + T i j k

JJJG

( )0.8 0.6AC ACT= +T j k

and

( ) ( ) ( )4 ft 8 ft 1 ftAD = + i j kJJJG

( ) ( ) ( )2 2 2

4 ft 8 ft 1 ft 9 ftAD = + + =

( ) ( ) ( )4 ft 8 ft 1 ft9 ft

ADE

AD AD ADE

AD TT T

AD = = = + T i j k

JJJG

( )0.4444 0.8889 0.1111AD ADET= + T i j k

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PROBLEM 2.123 CONTINUED

Finally,

( ) ( ) ( )8 ft 8 ft 4 ftAE = + +i j kJJJG

( ) ( ) ( )2 2 28 ft 8 ft 4 ft 12 ftAE = + + =

( ) ( ) ( )8 ft 8 ft 4 ft12 ft

ADE

AE AE ADE

AE TT T

AE = = = + + T i j k

JJJG

( )0.6667 0.6667 0.3333AE ADET= + +T i j k

With the weight of the machinery, ,W= W j atA, we have:

0: 2 0AB AC AD

W = + + =F T T T j

Equating the factors of , , andi j k to zero, we obtain the following linear algebraic equations:

( )0.5294 2 0.4444 0.6667 0AB ADE ADET T T + = (1)

( )0.4706 0.8 2 0.8889 0.6667 0AB AC ADE ADET T T T W + + + = (2)

( )0.7059 0.6 2 0.1111 0.3333 0AB AC ADE ADET T T T + + = (3)

Knowing that 320 lb,W = we can solve Equations (1), (2) and (3) using conventional methods for solving

Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for example) to obtain

46.5 lbAB

T =

34.2 lbAC

T =

110.8 lbADE

T =

140

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PROBLEM 2.124

A piece of machinery of weight Wis temporarily supported by cablesAB,

AC, and ADE. Cable ADE is attached to the ring at A, passes over thepulley atD and back through the ring, and is attached to the support at E.

Knowing that the tension in cable AB is 68 lb, determine (a) the tension

inAC

, (b

) the tension inADE

, (c) the weight

W. (

Hint:The tension is thesame in all portions of cableADE.)

SOLUTION

See Problem 2.123 for the analysis leading to the linear algebraic Equations (1), (2), and (3), below:

( )0.5294 2 0.4444 0.6667 0AB ADE ADET T T + = (1)

( )0.4706 0.8 2 0.8889 0.6667 0AB AC ADE ADET T T T W + + + = (2)

( )0.7059 0.6 2 0.1111 0.3333 0AB AC ADE ADET T T T + + = (3)

Knowing that the tension in cable AB is 68 lb, we can solve Equations (1), (2) and (3) using conventional

methods for solving Linear Algebraic Equations (elimination, matrix methods via MATLAB or Maple, for

example) to obtain

(a) 50.0 lbAC

T =

(b) 162.0 lbAET =

(c) 468 lbW =

141