Solution manual vector mechanics engineers dynamics 8th beer chapter 02

See Problem 2.101 for the figure and the analysis leading to the linear algebraic Equations 1, 2, and 3... See Problem 2.105 for the figure and the analysis leading to the linear algebra

(a) Parallelogram law:

(b) Triangle rule:

10.5 kN

R=22.5

α = °

10.5 kN

=

(a) Parallelogram law:

Using the triangle rule and the Law of Sines

sinβ =0.5303332.028

N200sinbα

F

276 N

bb

Using the triangle rule and the Law of Sines

sinα =0.4242625.104

Using the triangle rule and the Law of Cosines,

135

β = ° Then:

72.236

(a) α =72.2°!

(b) R=1.391 kN!

By trigonometry: Law of Sines

Using the Law of Sines

Using the Law of Sines: 60 N 80 N

Using the triangle rule and the Law of Sines

Using the triangle rule and the Law of Sines

sinα = sin 35°sinα =0.6555240.959

We observe that force P is minimum when α = °90 Then:

69.829 N

=

(a) T BC =4.88 N 6.00°!

Using the force triangle and the Laws of Cosines and Sines

Using the Law of Cosines and the Law of Sines,

56.6 lb

=

R 85.8° !

Using the force triangle and the Laws of Cosines and Sines

20 kNsin sin110

Using the force triangle and the Laws of Cosines and Sines

20 kNsin sin110

Using the force triangle and the Laws of Cosines and Sines

0.803 kN

y

5 kips: Fx = (5 kips cos 40) °

or Fx =3.83 kips

(5 kips sin 40)y

Determine the following distances:

We compute the following distances:

( ) ( ) ( ) ( ) ( ) ( )

°

(a)

cos 40

x P

P=

°

30 lbcos 40

P =

°

100 Ntan 75

x

P =

°

or P x =26.8 N"

P =

°

260 lbtan 50

=

°

218.16 lb

From the solution to Problem 2.21:

From the solution to Problem 2.22:

From the solution to Problem 2.24:

256 lb

α =

−6.6839

α = − °

258 lb

=

Cable BC Force:

( ) 84

145 lb 105 lb116

x

( ) 80

145 lb 100 lb116

x

100 lb 80 lb5

x

156 lb 60 lb13

21

α =

140tan 62.321

Thus: R = 45.2 lb 62.3°! !

(a) Since R is to be horizontal, R y = 0

13 1−sin α = 7 sinα +9

169 1−sin α = 49 sin α + 126 sinα +81

( )218 sin2α +( )126 sinα −88=0Solving by quadratic formula: sinα =0.40899

x F

(a) Since R is to be vertical, R x = 0

Then, R x = Σ =F x 0

(600 N cos) α +(300 N cos) (α + ° −35 ) (700 N cos) α =0

Expanding: 3 cos cos 35( α ° −sin sin 35α ° −) cosα =0

Then:

1cos 35

3tan

Selecting the x axis along aa′,we write

Free-Body Diagram Force Triangle

Solving Equations (1) and (2) simultaneously:

(a) T AB =38.6 lb!

(b) T AC =44.3 lb!

Free-Body Diagram:

(a) Σ =F x 0: −T ABcos 30° + Rcos 65° =0

cos 30cos 65 AB

°0:

°

or R=830 N!

From Similar Triangles we have:

BC

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

(a) For T AB to be a minimum

Note: In problems of this type, P may be directed along one of the cables, with T =Tmax in that cable and

0

T = in the other, or P may be directed in such a way that T is maximum in both cables The second

possibility is investigated first

Note: Refer to Note in Problem 2.60

T must be perpendicular to F AC to be as small as possible

Free-Body Diagram: C Force Triangle is a Right Triangle

Free-Body Diagram: B (a) Have: TBD +FAB +TBC =0

where magnitude and direction of T are known, and the direction BD

of F is known AB

Then, in a force triangle:

170.5 lb

=

Free-Body Diagram of Pulley

687 N

Free-Body Diagram of Pulley and

687 N

Free-Body Diagram: Pulley C

(b) ΣFy =0: TACB(sin 30° +sin50° +) (800 N sin50) ° −Q = 0

(2303.5 N sin 30)( ° +sin50° +) (800 N sin 50) ° −Q =0

Free-Body Diagram: Pulley C

First replace 30 lb forces by their resultant Q:

(a) F x =(950 lb sin 50 cos 40) ° °

950 lb

θy = −

467.78 lbcos

950 lb

θz =

(a) F x = −(810 lb cos 45 sin 25) ° °

810 lb

θy = −

or θy =135.0°!

519.09 lbcos

810 lb

θz =

or θz = 50.1°!

(a) F x =(900 N cos 30 cos 25) ° °

900 N

θy =

or θy =60.0°!

329.40 Ncos

900 N

θz = −

or θz =111.5°!

(a) F x = −(1900 N sin 20 sin 70) ° °

1900 N

z

θ =

orθz =83.3°!

(a) F x =(180 lb cos 35 sin 20) ° °

180 lb

z

θ =

θ = °!

(a) F x =(180 lb cos 30 cos 25) ° °

180 lb

z

θ =

orθz =68.5°!

(a) F x = −(220 N cos 60 cos35) ° °

(a) F x =180 N With F x =Fcos 60 cos 35° °

225 N

x x

F F

225 N

y y

F F

225 N

z z

F F

θ = = −

152.7

z

θ = °!

1950 N

x x F F

1950 N

y y F F

1950 N

z z F F

θ = = −

157.4

z

θ = °!

(a) We have ( )2 ( )2 ( )2

cosθx + cosθy + cosθz =1

cosθy = −1 cosθx − cosθz

Since 0F y < we must have cosθy <0

F F

θ

=

50 lb0.67597

−73.968 lb

(a) We have ( )2 ( )2 ( )2

cosθx + cosθy + cosθz =1

or cosθz = −1 cosθx − cosθy

Since 0F z < we must have cosθz <0

F F

θ

−

−39.025 lb

(a) We have F y =Fcosθy

θ =

80 Ncos

θ =

224.47 Ncos

(a) Have F x =Fcosθx

(36 in.) (42 in.) (36 in.)

(36 in.) (45 in.) (48 in.)

(4 kips cos30 sin 20) [ sin30 cos30 cos 20 ]

RR

z

θ = °

(6 kips cos30 sin 20) [ sin30 cos30 cos 20 ]

2.0144 kips

y y

RR

z

θ = °

1121.80 N

y

275.4 Ncos

1121.80 N

z

1121.80 N

x

559.2 Ncos

1121.80 N

y

θ = θy =60.1°!

30.6 Ncos

1121.80 N

z

θ = − θz =91.6°!

Have TAB =(760 lb sin50 cos40)( ° ° −i cos50° +j sin50 sin 40° °k )

( cos 45 sin 25 sin 45 cos 45 cos 25 )

0cos

2038.1 lb

x

1543.81 lbcos

2038.1 lb

y

1330.65 lbcos

2038.1 lb

z

Have TAB =TAB(sin 50 cos40° ° −i cos50° +j sin50 sin 40° °k )

(980 lb)( cos 45 sin 25 sin 45 cos45 cos 25 )

1338.35 lb

y

873.78 lbcos

1338.35 lb

z

Cable AB: uuurAB= −( ) (4 m i − 20 m) ( )j+ 5 m k

AC AC

AC T

See Problem 2.101 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)

See Problem 2.103 for the analysis leading to the linear algebraic Equations (1), (2), and (3)

2037

AB

or T AB =240 NAnd from Equation (3)

See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and

The force in each cable can be written as the product of the magnitude of the force and the unit vector along the cable That is, with

( 0.158730 0.79365 0.58730 )

continued

In Equations (1), (2) and (3), set T AB =3.6 kN,and, using conventional methods for solving Linear Algebraic Equations (MATLAB or Maple, for example), we obtain:

Based on the results of Problem 2.107, particularly Equations (1), (2) and (3), we substitute T AC = 2.6 kN and solve the three resulting linear equations using conventional tools for solving Linear Algebraic Equations (MATLAB or Maple, for example), to obtain

AB AB

AD AD

See Problem 2.109 for the analysis leading to the linear algebraic Equations (1), (2), and (3) shown below

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 BE is identical to the unit vector along the generator AB

65

AB = BE = ° +i j− °kλλλλ λλλλ

Then, isolating the factors of i, j, and k, we obtain three algebraic equations:

WithP = and the tension in cord 0 BE =0.2 lb:

See Problem 2.111 for the Figure and the analysis leading to the linear algebraic Equations (1), (2), and (3)

With W =1.6 lb, the range of values of P for which the cord CF is taut can found by solving Equations (1),

(2), and (3) for the tension T CF as a function of P and requiring it to be positive ( 0).>

Solving (1), (2), and (3) with unknown P, using conventional methods in Linear Algebra (elimination, matrix

methods or iteration – with MATLAB or Maple, for example), we obtain:

From the solutions of 2.107 and 2.108:

From the solutions to Problems 2.111 and 2.112, have

Applying the method of elimination to obtain a desired result:

Multiplying (2 )′ by sin 45°and adding the result to (3):

(sin 45 sin 30 ) (sin 45 sin15 ) 0.2 65 sin 45

Multiplying (2 )′ by sin 30°and subtracting (3) from the result:

(sin 30 sin 45 ) (sin 30 sin15 ) 0.2 65 sin 30

Refer to the solution of problem 2.119 and the resulting linear algebraic Equations (1), (2), (3) Include force

Note: BE shares the same unit vector as AB

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

And writing ΣF =y 0 gives:

1.92535 81.9 lb −0.84327 17 lb −W =0

or W =143.4 lb

See Problem 2.123 for the analysis leading to the linear algebraic Equations (3) and (4) below:

AE =T AEλAE

T

(2.4 m) (2.4 m) (1.2 m)3.6 m

See Problem 2.125 for the analysis leading to the linear algebraic Equations ( ) ( )1 , 2 ,′ ′ and ( )3′ below:

Free-Body Diagrams of collars For both Problems 2.127 and 2.128:

Where y and z are in units of meters, m

From the F.B Diagram of collar A:

Setting the k coefficient to zero gives:

From the analysis of Problem 2.127, particularly the results:

∴ =and

Using the triangle rule and the Law of Sines

sinα = sin30°sinα =0.71428

We compute the following distances:

Note that the force exerted by BD on the pole is directed along BD, and the component of P along AC

Free-Body Diagram Force Triangle

Law of Sines:

5 kNsin115 sin 5 sin 60

Free-Body Diagram: Pulley A ΣFx =0: 2 sin 25P ° − Pcosα =0

(a) F x = Fsin 30 sin 50° ° = 220.6 N (Given)

220.6 N

575.95 Nsin30 sin50

F F

67.5

x

θ = °! cos 30 498.79 N

F F

30.0

y

θ = °! sin 30 cos 50

F F

108.7

z

(a) F z = Fcosθz =(600 lb cos136.8) °

F F

F F

(500 lb) [ cos 30 sin15 sin 30 cos 30 cos15 ]

R R

65.0

x

θ = °! 635.67 lb

757.98 lb

y y

R R

33.0

y

θ = °! 261.04 lb

757.98 lb

z z

R R

69.9

z

θ = °!

The forces applied at A are:

Equilibrium Condition with W = −Wj

Substituting TAB =3 kNin Equations (1), (2) and (3) and solving the resulting set of equations, using

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

(32 in.) (48 in.) (36 in.)

(25 in.) (48 in.) (36 in.)

In Equations (1), (2) and (3), set T AD =120 lb,and, using conventional methods for solving Linear Algebraic

Equations (MATLAB or Maple, for example), we obtain:

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

(0.31169 0.93506 0.16883 )

At A: Σ =F 0: TAB +TAC + +P Q+W =0Noting that T AB =T AC because of the ring A, we equate the factors of