Solution manual vector mechanics engineers dynamics 8th beer chapter 04

Express all forces in terms of rectangular components: continued... Express all forces in terms of rectangular components: continued... Express all forces in terms of rectangular compone

(52.426 in.) 80 lb 14.7413 in.( ) 80 lb 4.9183 in.( ) 0

14.9896 lbP

∴ = or P =14.99 lb
(b) From Equation (2)

14.9896 lb 2 80 lb− +2B = 0

72.505 lbB

∴ = or B =72.5 lb

Free-Body Diagram: a1 =(20 in sin) α −( )8 in cosα

(49.027 in.) 80 lb 11.6577 in.( ) 80 lb 6.7274 in.( ) 0

a=
(b) The corresponding value of Ay is

For the loading to be safe, cables must not be slack and tension must not exceed 12 kN

Thus, making 0≤ TB ≤ 12 kN in (1), we have

Note that W = mgis the weight of the crate in the free-body diagram, and that

A = ≤ which satisfies the constraint on Ay

For the largest allowable value of E y:

Free-Body Diagram:

Calculate lengths of vectors BD and CD:

ft8.23ft)0.21()2.11( 2 + 2 =

ft2.11ft24lb221ft8.23

ft2.11ft24lb161:







+

−

=

150.000 lbCD

(a) Equilibrium for ABCD:

430 N

∴A = 50.2 ,° B= 275 N
(c) Substituting α =30° into (1), (2), and (3) and solving for A and B:

70.753

y x

AA

219 N

∴A = 71.2 ,° B=141.5 N 60°

Equations of equilibrium:

(82.5N m) sin (0.3m) cos (0.5m) 0:

141.506 N, x 70.753 N, y 122.548 N

N506.141)548.122()753.70

70.753

y x

AA

N5.141

=

∴A 60.0 ,° B=141.5 N 60°

N

From free-body diagram of lever BCD

CC

C= x2 + y2 = 1.9 2+ 1.2 2 =2.2472Since Cmax = 500 N,

Free-Body Diagram:

(a)

(0.9in.) 0cos

in

2.4:

258 N

∴A = 22.8°
From For (b)

1 187.504tan tan 20.226

237.50

y x

AA

253 N

∴A = 20.2°

From free-body diagram for (a):

B = −0: 100 N 0

tan tan 28.072

187.5

y x

BB

213 N

∴ =B 28.1°
From free-body diagram or (b):

Equilibrium for lever:

(a) ΣMC =0: TADcos 27.603 8 in.°( ) (− 60 lb 12 in cos20) ( ) ° = 0

95.435 lbAD

and tan 1 tan 1124.348 60.456

70.478

y x

CC

142.9 lb

∴ =C 60.5°

A =Thus: A= Ax2+ Ay2 = (586.80) (2 + 269.19)2 = 645.60 N

and tan 1 tan 1269.19 24.643

586.80

y x

AA

646 N

∴A = 24.6°

409 N

∴A = 22.4°

y x

CC

216 N

∴ =C 33.7°

36.4 lb

∴A = 81.0°

Free-Body Diagram:

Geometry:

Distance BC= (7.2)2+(3)2 =7.8 in

Distance CD= (10.8)2+(3)2 =11.2089 in

Equilibrium for bracket:

0: (10 lb 9 in.)( ) 7.2 (4 in.) 3 (9 in.) 10.8 (4 in.)

+  =101.014 lb

and tan 1 tan 175.887 86.919

4.0853

y x

AA

76.0 lb

∴A = 86.9°

252 N

=

PB

tancos

Equilibrium for bracket:

Equilibrium for bracket:

Equilibrium for rod:

Free-Body Diagram:

Equilibrium for rod:

(a) ΣME =0: ( )6 lb cos60°( ) (dOE − Tcosθ) ( )dOE = 0

3 lbcos

Equilibrium for bracket:

F = − or FCD =126.138 N0: 126.138 N 741.82 N 0

(a) ΣME =0: P y( ) (E − Qsinβ)(yE − yD) (− Qcosβ)(2.4 in.)= 0

(3.44 in.) ( )2 lb sin12.6804 1.24 in.( ) ( )2 lb cos12.6804 (2.4 in.) 0

With M = 0 and Ti= T0= 12 lb

0:

xF

Σ = C −x 12 lb = 0

Cx= 12 lb 0:

y

F

Σ = C −y 12 lb 0=

CY= 12 lb Thus: C = 2 2

C +C = (12)2 +(12)2 = 16.9706 lb

or C = 16.97 lb 45°
0:

F

Σ = Cy – 8 lb = 0

Cy= 8 lb Thus: C = 2 2

CC

∴ C = 17.89 26.6°
0:

C

M

Σ = MC – (16 lb)[(1.8 – 1) in.] + (8 lb)[(2 + 1 –2.4) in.] – 8 lb in.⋅ = 0

MC = 16.00 lb in.⋅ or MC = 16.00 lb in.⋅

Σ = MA – (2 lb)(8 in.) – (1 lb)(12 in.) + (1.2 lb)(16 in.) = 0

= 8.8 lb in

Set MA=20 lb in.⋅ clockwise to find Fmax:

(ME )2 = (1.5 T − 127.44) kN m⋅ (a) The maximum absolute value of ME is obtained when (ME )1 = − (ME ) and

1.5 T − 23.76 kN = − (1.5 T – 127.44 kN)

T = 50.400 kN or T = 50.4 kN
(b) For this value of T:

ME= 1.5(50.400) kN m 23.76 kN m⋅ − ⋅

Geometry:

Distance BD = (1.8)2+(4)2 = 4.3863 m

Note also that: W = mg = (160 kg)(9.81 m/s2) = 1569.60 N

With MA= 360 N m⋅ clockwise: (i.e corresponding to Tmax )

0:

AM

Σ = 360 N m⋅ – [(540 N) cos 15o](5.6 m) +

min

1.84.3863 T

For both parts (a) and (b)

2

2(3 N)tan

Using W = mg, and h = tan

= °

or θ =141.1°

φ

⋅

= =Since the torsion spring is unstretched when θ = 0:

(70 mm)φ =(35 mm)θ

φ = 1θ

2Therefore:

2 2

Pl θ −kl θ θ =

cos sin 0

P θ −kl θ = tan P

tan kl tan (2) 63.435kl

 

1 Three non-concurrent, non-parallel reactions

(b) Reactions determined by dynamics

Note that the wheel is a three-force body, and let point D be the intersection of the three forces

With a = 75mm, it follows from the force triangle that

Free-Body Diagram:

Note that the wheel is a three-force body, and let point D be the intersection of the three forces

From the force triangle it follows that

( )2 2

57.735 mm3

From (1) it follows that A will decrease as a increases Therefore the value of a

calculated is a lower limit:

57.7 mm

Equilibrium: force triangle

Using the law of sines on the force triangle:

sin120 sin 60sp sin

FB

φφ

11.05 lb

B =

Free-Body Diagram:

Note that the bent rod is a three-force body D is the point where the lines of action of the three

forces intersect

(a) The requirement B = means that the force triangle must be isosceles Therefore C θ φ=

Which leads to the force triangle shown

From the geometry it follows that

1tanθ = 2

26.565

(b) From the force triangle:

2 sinB θ =P, or with sin 1

P

=

C 26.6°

The bracket is a three-force body and A is the intersection of the lines of action of the three forces

1 6tan 26.56512

 

 From the force triangle:

(75 lb)cot(75 lb)cot 26.565150.000 lb

Let E be the intersection of the lines of action of the three

forces acting on the bracket

Solving for A and C:

194.452 lb253.10 lb

AC

=

=

or A =194.5 lb

or C =253 lb 77.9°

Let C be the intersection of the lines of action of the three

forces acting on the girder

6000 lbsin30°A sin16.1021°B sin133.898°

From force triangle:

1 2600 Ntan 83.636

l

(900 mm sin50)

76.894 mmtan 83.636

°

or l = 76.9 mm

θ = 13 °

Also:

9.2 1.8 m 2.8 m2

AG = −  =

9.2 m cos 4.6 m cos2

BD=  θ = θ

CD CG GD= +

sinsin

AG

BG θθ

9.2sinsin 2 θθ

2.8

continued

227.28 N

BD = θ = θ = but also that

1.751.75tan m

9.2 m cos 4.6 m cos2

This implies:

1.754.6cos 1.75tan

Then sin31.722

a

° =9.2 −5.8717 m

Free-Body Diagram:

Let D be the intersection of the lines of action

of the three forces acting on the tool

From the free-body diagram:

20 lbsin sin 20

Let E be the intersection of the lines of action

of the three forces acting on the tool

From the free-body diagram, using law of sines:

Free-Body Diagram:

Based on the roller having impending motion to the left, the only contact between the roller and floor will be at the edge of the tile

First note W =mg = (20 kg 9.81 m/s) ( 2) =196.2 NFrom the geometry of the three forces acting on the roller

= ° Applying the law of sines to the force triangle,

∴ =

or P =128.0 N 30°

Based on the roller having impending motion to the right, the only contact between the roller and floor will be at the edge of the tile

First note W =mg =(20 kg 9.81 m/s) ( 2)

196.2 N

=From the geometry of the three forces acting on the roller

= ° Applying the law of sines to the force triangle,

∴ =

or P =77.5 N 30°

Free-Body Diagram:

Note that the clamp is a three-force body D is the intersection

of the lines of action of the three forces

From the free-body diagram it follows that:

(4.2 in.) tan 78 19.7594 in

sin12B sinA sin(90

Note that the hoist AD is a three-force body

E is the intersection between the lines of action of the three forces

acting on the hoist

From the free-body diagram:

(48 in.)cos30 41.5692 in

(48 in.)sin 30 24 in

tan 75 (41.5692 in.)tan75155.1384 in

AE AD

BE AD

xy

1 16 in 1 139.1384

41.569273.36588

BE AD

yx

260 lbsin sin sin15

260 lbsin 1.63412 sin 163.366 sin 15

Free-Body Diagram:

Note that the member is a three-force body In the free-body diagram, D is the intersection between the lines

of action of the three forces

(a) From the force triangle:

110 N 3

4

TT

Note that the member is a three-force body In the free-body diagram, E is the intersection between the lines

of action of the three forces

From the free-body diagram:

1 15tan 61.928

12

 

 From the force triangle:

Free-Body Diagram:

Note that the peavey is a three-force body

In the free-body diagram, D is the intersection of the lines

of action of the three forces acting on the peavey

From the force triangle, using the law of sines:

sin sin sin135

80 lbsin 40.236 sin 4.7636 sin135

or A =87.6 lb 85.2°

Note that the peavey is a three-force body

In the free-body diagram, D is the intersection of the lines

of action of the three forces acting on the peavey

(46.9282 in tan 30) 27.0940 in

sin sin sin120

80 lbsin51.424 sin8.5755 sin120

C and A:

Free-Body Diagram:

From the free-body diagram, the member AB is a force body Let D be the intersection of the lines of action of the three forces acting on AB Then, using triangle BCD:

three-(250 mm tan 60) 433.01 mm

Also:

(300 433.01) mm733.01 mm

FDAF

From the force triangle

180 30 18.8324131.168

α = ° − ° − °

Using the law of sines

330 Nsin30 sin18.8324 sin131.168

From the free-body diagram it follows that

9.6 8sin20tan

8cos20

°42.397

b) C =142.935 lb C=142.935 lb 60.5°

sin sin 28.955

θα

(a) Note that the rod is a three-force body Using the law of cosines on triangle ABC:

2 2 2 2 2 cos

R = R +L − R L θ

2 2cos

4

R LRL

Note that the wheel is a two-force body and therefore the force at C is directed along CA and perpendicular to the incline

The wheelbarrow is a three-force body Let D be the intersection of the lines of action of the three forces acting

on the wheelbarrow Then, using the triangle DEG

120 lbsin18 sin 50.667 sin111.333

39.809 lb,

B = C =9.644 lb(a) Noting that the force on each handle is B/2:

1 19.90 lb

2B = 39.3°
(b) Reaction at C:

99.6 lb

=

C 72.0°

Free-Body Diagram:

From the free-body diagram:

(40 in tan) 32 in 8 in. (23 in tan18)

(a) 1 24.2 lb

2B= 58.0°
(b) C =83.0 lb 72.0°

diagram, E is the intersection between the lines of action

of the three forces

Using triangle ACF in the free-body diagram:

tanCF

y =d θ

2tan tan

β =where

(a) As shown in the free-body diagram of the slender rod AB, the three forces intersect at C From the geometry of the forces

tan CB

BC

xy

β = where

1 sin2

CB

x = L θ

and yBC = Lcosθ

1 tan tan

β

°

sin aL

 

Note that the athlete is a three-force body From the free-body diagram

1

2tan tan

From triangle BCD:

1.25 2cos

La

θ

=1.6 cos

θ

−

= +

− or 3

The forces acting on the three-force member intersect at D

(a) From triangle ACO

tan tan 18.4349

rr

   

    or θ =18.43°
(b) From triangle DCG tan r

1

tan DO

AG

yx

  where yDO =( )DO cosθ = ( )4 cos18.4349r °

3.4947r

= and xAG =( )2 cosr θ = ( )2 cos18.4349r °

1.89737r

=

1 3.4947 tan 63.435

1.89737

rr

∴ =   = ° where 90° +(α θ− ) = 90° + 45° =135.00° Applying the law of sines to the force triangle,

or 7 in 7 in 2 in 530 lb 2 in 192 lb 3 in 530 lb

3 in 96 lb 3 in 265 lb 6 in 265 lb 0

or −(7 in.)Cyk+(7 in.)Czj+ 2 in 530 lb( )( )k+ 2 in 192 lb( )( ) (j− 3 in 530 lb)( )i

(3 in 265 lb)( ) (3 in 96 lb)( ) (6 in 96 lb)( ) 0

Solving the equation one component at a time:

From i component: 1.2 sin 60Cz ° −(105.948 N cos60) ° = or 0, Cz =50.974 N

366 N/m

k =elongation of spring x =( )yE θ=180° −( )yE θ= °0

0:

x

F

Σ = Dx = 00:

y

F

Σ = C +y 52.338 N 40.26 N 0,− = or C = −y 12.078 N0:

z

F

Σ = Cz +150 N 150 N 180 N+ − =0, or Cz = −120.000 NTherefore:

Σ = + + − + = or C = −z 109.308 NTherefore:

N = W =

continued

617.67 N,

B

N =and from (1)

C

N =Therefore the forces on the blocks are:

continued

0.10.7 m

S

S

WW

z

=

−: 0.3N +2.4N +0.6N −1.2W − x WS S = 0

S

S

WW

x

=

For ( )WS min, (1) and (2) imply that xS, should be chosen as small as possible and that zS should be chosen as

large as possible with the constraint that

(1.1 m−xS) (= zS −0.7 m)

or xS + zS =1.8 m

The smallest xS and the largest zS that satisfy this condition are

0.6 mS

x =
1.2 m

S

z =
The corresponding value of WS is:

0.1 1667.7 N

333.54 N1.1 m 0.6 m

S

−Therefore the smallest mass of the bucket of sand is

333.54 N

34.000kg9.81 m/s

S

Free-Body Diagram:

First note WAB =(5 lb/ft 2 ft)( ) =10 lb

(5 lb/ft 4 ft)( ) 20 lbBC

3.18 lbC

T =
21.8 lbD

a = −

=Results: (a)

0: 3 in 1.5 in 1.4 in 0.44 lb

1.5 in 2.8 in 0.53 lb 1.5 in 3.2 in 0

or (3 in.)Cyi −(1.5 in 0.44 lb)( ) (i + 1.4 in 0.44 lb)( )k −(1.5 in 0.53 lb)( )i

(2.8 in 0.53 lb)( ) (1.5 in.)Ay (3.2 in.)Ay 0

0: 3.2 in 1.5 in 3.2 in 1.5 in.

1.8 in 0.44 lb 1.8 in 1.4 in cos 1.4 in sin 0.53 lb 0

( 1 ) (0.44 lb 0.53 lb 1.8 in 0.53 lb) 1.4 (3.2 in sin) (1.4 in cos)

2.1333sin cos cos

cos cos sin sin

AA

T = W
(b) Using TC in (1):

W L+W L− W L Wz+ =

 1

1 2 33

C

x = − L

11

2 3

C

z = − L

0.450 m

continued

R =
43.4 N

C

(a) Ay =17.36 lb

(a) Ay =13.50 lb
(b) By = 0
(c) Cy =13.50 lb

Express all forces in terms of rectangular components:

continued

Express the forces in terms of their rectangular components:

( ) (2 )2

27 57.752720

A = −Therefore:

(4.00 kips) (2.73 kips)

First express tensions in terms of rectangular components:

(73.9 N) (878 N) (127.9 N)

First express tensions in terms of rectangular components:

Express all forces in terms of rectangular components:

continued

Express all forces in terms of rectangular components:

continued

Express all forces in terms of rectangular components:

Express all forces in terms of rectangular components:

or −(26 in.)Ayk+(26 in.)Azj+(13 in 75 lb)( )k +(16 in 75 lb cos50)( ) °i

(26 in 0.94521)( FCD) (26 in 0.32646)( FCD) (7 in 0.94521)( FCD) 0

Express tension in terms of rectangular components:

26.231 lbx

0.08 97.119 N 00.33

0.3 232.99 N 00.38

Express tension, weight in terms of rectangular components:

Express tension, weight in terms of rectangular components:

B =Therefore:

Express forces, weight in terms of rectangular components:

( )( ) ( )( )

: 0.74278FCE 3 ft − 300 lb 1.5 ft =0

k

201.94 lbCE

Express forces, weight in terms of rectangular components:

(0.600 kN)

z = −

Notice that the forces in the belts can be equivalently moved to the center of the pulley because their net moment about this point is zero

(a) ΣFx =0: 3 lb+( )3 lb cos30° − = T 0

5.5981 lb

T =

or T =5.60 lb
(b) ΣFy =0: Dy = 0

Express the tension in terms of its rectangular components:

(b) ΣFx =0: Ex −44 lb 0, or= Ex = 44.000 lb

80: 74.8 lb 0, or 35.200 lb

Express forces and moments in terms of rectangular components:

Express tension, weight in terms of rectangular components:

B = −

270: 68.67 N 39.422 N 0

Express tensions, load in terms of rectangular components:

13 TCF − =i

650.00 lbCF

650.00 lbBE

= 1920 lb − 300 lb

Express tensions, load in terms of rectangular components:

or 60 0 0 in 60 0 0 in 60 0 30 in 60 0 15 lb in.

= 6.14 kips − 1.260 kips

Free-Body Diagram: First note

0.65 m 0.2 m 0.44 m0.65 0.2 0.44 m

    0.108642TDI 0.0196721TFG 21.6

187.816 N 0.45 0.2 0.360.61

: −cCy +MA =0i

36 lb ft 36.0 lb

1 ft

A

y MCc

Β = −

20.0 lb 36.0 lb 16.00 lb

B = −A −C = − = − Therefore:

: −cCy +MA =0i

6.3 N m 35.0 N0.180 m

A

y MCc

Β = −

12.50 N 35.0 N 22.5 N

B = −A −C = − = − Therefore:

Express the forces in terms of rectangular components:

x =Then,

A = −

160: 29.43 N 3.6787 N 7.2425 N 0

(a) TBC =7.24 N
(b) A = −(4.48 N) (i + 20.2 N) (j− 1.379 Ν)k

(3.68 N + 2.76 N) ( )

B =

perpendicular to AB and CD Letting λλλλE= direction cosines for force E,

/ /

B A E

E 5.3333 lb

∴ = and E=5.3333 lb 0.6 0.8( i+ j )

or E=(3.20 lb) (i + 4.27 lb)j
(b) From j-coefficient −18 5.3333 lb( )+32B= 0

B 3.00 lb

∴ =