Solution manual vector mechanics engineers dynamics 8th beer chapter 10

Consider a virtual rotation δφ of link AB... Consider a virtual rotation δφ of link AB... Equation 1: WW... Since cosθ = is not a solution of the equation, we can divide all terms by 0 k

δφ = δθ

3G

x = x δxG =3δxD4

x = x δxH =4δxD5

I D

x = x δxI =5δxD(a) Virtual Work: δU =0: F xEδ E −FSPδxI = 0

δ =δ =δ = δVirtual Work:

Have y A = 2 cos ; l θ δy A = −2 sinl θ δθ

Chapter 10, Solution 10

Virtual Work:

2 cosA

3 sinF

δ = − θ δθVirtual Work: δU = 0: Q xδ A +P yδ F = 0

(2 cos ) ( 3 sin ) 0

Q l θ δθ + P − l θ δθ =

3 tan2

Q = P θ

Thus: P(−3 sinl θ δθ)+Q l( cosθ δθ)= 0

3 sin− Pl θ +Qlcosθ = 0

3 sin

3 tancos

P

θ

Chapter 10, Solution 12

A

x = a+b θ δx A = −(a+b)sinθ δθsin

G

y =a θ δy G =acosθ δθVirtual Work:

The reactions at A and B are perpendicular to the displacements of A and B hence do no work

sin cos4

(35 kN) sin 2 1.10365 kN

9 16sin

θ

θ =+

ABC: sin y B = a θ ⇒ δy B = acosθδθ

y = a θ ⇒δy = a θδθ

CDE: Note that as ABC rotates counterclockwise, CDE rotates clockwise

while it moves to the left

Virtual Work: δU = 0: cosMδθ −(P β δ) x D −(Psinβ δ) y D = 0

( cos )( sin ) ( sin )(3 cos ) 0

(c) For P directed , β =180° Equation (1): M = Pl(3sin180 cos° θ −cos180 sin° θ)

sin

M = Pl θ

Analysis of the geometry:

Virtual Work: δU = 0: −P xδ C −Mδθ = 0

sin

0cos

For the given conditions:P =1.0 kip =1000 lb, AB =2.5 in., and BC =10 in.:

(a) When 30 : sin 2.5sin30 , 7.181

From the analysis of Problem 10.19, sin( )

Chapter 10, Solution 21

Consider a virtual rotation δφ of link AB

Disregarding the second-order rotation of link BC,

C C

Consider a virtual rotation δφ of link AB

Disregarding the second-order rotation of link BC,

C C

(75 lb) 2 60 lb( ) sin

cos2

θθ

=

 

 

 sin 0.625cos

θ = °

36.4

θ = °
(Additional solutions discarded as not applicable are θ = ±180 )°

From the solution to Problem 10.16

2

2cos 1 sinsin

=

2

sincos

y =l θ δy = −l θ δθVirtual Work:

=

2

sincos

y =l θ δy = −l θ δθVirtual Work:

or θ = 39.1°

Then P l (sinθ −cos tanθ φ δθ) −kl3−(2cosφ +cosθ) (l sinθ +cos tanθ φ δθ) = 0

or 3 (2cos cos ) sin cos tan

sin cos tan

First note: y D =(250 mm sin) θ δy D =(250 mm cos) θ δθ

Virtual Work:

0:δU= (250 N)δy D −F SPδx A= 0

(250 N 250 mm cos)( ) θ δθ −(1500 N cos)( θ −cos 45°)(600 mm sin) θ δθ = 0

or 5−72 tanθ(cosθ −cos 45° = ) 0

Solving numerically

15.03 and 36.9

or (150 lb 15 in cos)( ) θ δθ +(375 lb 1 cos)( − θ)  −(30in sin) θ δθ=0

or 2250cosθ −11250 1 cos( − θ)sinθ = 0

or (1 cos− θ)tanθ =0.200

Solving numerically, θ =40.22°

40.2

θ = ° W

Chapter 10, Solution 35

s =rθ

s r

δ = δθSpring is unstretched at θ = ° 0

SP

F =ks=k rθsin

C

x =l θcos

θθ

=

θ =1.054 rad 60.39

60.4

θ = ° W

=A

y lcos

1 2sin 45 cos 45 cos 2 cos 45

 ° + 

= −

Now, with P =150 lb, l =30 in., and k =40 lb/in.

θθ

 ° + 

From geometry: y A =lsinθ

0:δU = PδS −F SPδy C = 0

(480 N 75 mm)( )δθ −(300 N tan) θ(375 mm sec) 2θ δθ =0

or 3.125 tan secθ 2θ = 1

Have siny A =l θ cosδy A=l θ δθ

θ

2.2222 radcos

θ

θ =

Have dAC = 212+7.22 =22.2 in.

7.2tan

21

φ = or φ =18.9246° Now α θ φ= +

By Law of Cosines:

2 22.22 102 2 22.2 10 cosAB

α

cyl =99.3 lb

Have dAC = 212 +7.22 =22.2 in.

7.2tan

21

φ = or φ =18.9246° Now α θ φ= +

By Law of Cosines:

2 22.22 102 2 22.2 10 cosAB

α

589.5791 1 cos− α =592.84 444 cos− α2

cos α −0.75307cosα +0.0055309 0= Solving α =41.785° and α =89.575°

18.9246

θ α φ α= − = − °

For Bar ABC:

2

C

y a

For the given position of member CD, is isosceles.∆CDE

Thus l D =a and l C =2 cos55a °

Then (2 cos55a ° −acos55°)δy C =(a−2 cos 55a 2 °)δy D

=

°Thus for given data

Equation (1):

WW

Link BC: xB =lcosθ

sinB

δ = − θ δθ

sinC

y =l θcosC

δ = θ δθ

2B

θ µ

=

−Substituting N into relationship for Mmax:

s

s s s

s

PP

For the linkage:

δ = θ δθ

3 sinF

δ = − θ δθ Virtual Work:

Recall Figure 8.9a Draw force triangle

Q W= θ φ+tan so that tan

θ φ δ

=

θη

A = − or Ay = 60 N

To determine A consider a horizontal displacement x, δxA:

Virtual Work: δU =0: A xxδ A = or 0, A = x 0

To determine M consider a counterclockwise rotation A, δθA:

Note that δyB =600δθA δyC =900δθA

D = −

or Dy =1.260 kN

From the solution of Problem 10.41

δ = −

or δdAB =1.451 in shorter( )

Chapter 10, Solution 56

2370 lbBD

Apply vertical load P at C:

FG

F = P (T) Virtual Work:

Remove member FG and replace it with forces F FG and −F FG at pins F

and G, respectively Denoting the virtual displacements of F and G as

Virtual Work:

Remove member FG and replace it with forces F FG and −F FG at pins F

and G, respectively Since P and δx C have the same direction, and since

Chapter 10, Solution 62

lb12.5in

lb12.5in.

Pkl

θθ

 ° + 

Then withP=150 lb, l =30 in and k = 40 lb/in.

θθ

 ° + 

2ks SP Py A

21

x = φ =  θPotential Energy:

d V Wl d

2

d V Wl d

θ

θ

CD

W d

2

CD

W d

( )

2.5 0

= WL > ∴ θ = 270 , Stable°

For equilibrium 0: 1.5sin1.5θ sinθ 0

dVdSolutions: One solution, by inspection, is θ =0,and a second angle less than 180° can be found numerically:

= > ∴ 137.8 ,θ = ° Stable

1.06896 rad 61.247

61.2

θ = °
Stability

Have ( 0), 4 in., 0 20 rad

V = ky +Wy

2 0

π

B AB

y =l θ 18 in.l AB =Potential Energy:

21

V = ky +Wy

2 0

12.5 1 lb/in  x−20 in. +12.5 1 lb/in −x +40x−279 in + 10 lb  x−20 in.=0

2 4.24

=

or x =14.6579 in and x=25.342 in

Deflection of spring = s, where s = l2+ y2 − l

∠ = ∠ =  ° − ° + 

452

θ = −

and cos 0.2942812

θ = −

34.2

θ = ° and θ =145.8° W

34.2 :θ = ° 2 2( )

2

10.2940 0.9558 0.8993 0 Stable2

d V

ka

(b) For Stable equilibrium:

Since cosθ = is not a solution of the equation, we can divide all terms by 0 kl2cosθ and write

2cos cos tan mg

(b) For the given data

Solving by trial and error: θ =51.96 ,° 52.0θ = ° W

Stability: Differentiating Eq (1):

2cos cos cos 2 mgsin

2 cos51.96 cos103.92 0.4905sin 51.96

d V kl

21.2431kl 0

dθ = (P Q+ ) (sin θ β− )=Psin cosθ β

or (P Q+ )(sin cosθ β −cos sinθ β)=Psin cosθ β

Substracting P(sin cosθ β) from each member yields

(P Q)cos sinθ β Qsin cosθ β 0

dθ = (P Q+ ) (sin θ β− )=Psin cosθ β

or (P Q+ )(sin cosθ β −cos sinθ β)=Psin cosθ β

Substracting P(sin cosθ β) from each member yields

(P Q)cos sinθ β Qsin cosθ β 0

First note, by Law of Cosines:

First note that cable tension is uniform throughout,

41

49

Equilibrium condition:

10:

−

= −Solving numerically, θ = 0.9580 rad =54.9° θ =54.9°

mgkr

Have xC = dsin θ yB =hcosθ

Consider a small clockwise rotation θ of the plate about its center

a

d =   +a

 

 52

and

21

When 0, dV 0

d

θ

θ

= = for all values of P

For stable equilibrium when θ =0,require

a a

2 0

d V

dθ = 3

2 3

5sin 2 0 unstable4

44cos 2 2cos cos 2

9

≤ < W

From geometry:

sin 2 sinC

x = −a θ = − a φFor small values of θ φ,

2

θ = φ1

or =

2

cos 3 cosA

s x= = −a θPotential Energy:

7

P< ka

or 1

10

P < ka

1 0

0

dV

dθ = 2

2 0

d V

dθ = 3

= − < ⇒

a c a

φ

θ

=+ −sincos

a

a c a

θθ

=+ −sin

1 c cosa

θθ

=+ −

Differentiating both sides with respect to θ:

For θ φ= =0, and recalling Eq (1),

2 2

d V

dθ and

4 4

d V

dθ for ( )

2 2

1 m c m

ab

= In practice, however we shall want to keep m1 below this value

Chapter 10, Solution 96

For small values of θ and :φ aθ =bφ

ab

Stability Conditions for stability (see page 583)

2 2

20: 0

From the analysis of Problem 10.97 with

cos 2θ sin sinθ θ 2 cosθθ

sin sinθ θ cos 2θ cosθθ

Apply Equations 10.24

10: condition satisfied

Vθ

∂ =

∂

20: condition satisfied

Vθ

cos sin sin 2 cos

θ

∂2

θ

∂

When θ1=θ2= 0

10

Vθ

∂ =

∂2

2 2 1

10: Condition satisfied

Vθ

∂ =

∂

20: Condition satisfied

Vθ

∂ =

∂2

∂ <

∂

Thus, for stable equilibrium when θ1=θ2= 0:

0≤ <P 0.21922kawith k =2 kN/m anda= 350 mm

0≤ <P 0.21922 2000 N/m 0.35 m

or 0≤P< 153.5 N

Chapter 10, Solution 102

First note, by the Law of Cosines

2 = 3 ft + 2 ft −2 3 ft 2 ft cosθDB

Also yA = 4.5cosθThen δyA = −4.5sinθδθVirtual Work

θ

DBF

or FDB =6 13 12cos− θ

17.895 kips

=DBF

17.90 kips

DB =

Given:

3.6 in

AB

l =1.6 in

BC

l =1.2 in

CD

l =1.6 in

DE

l =1.6 in

EF

l =4.8 in

49

θ

−

=

or θ =25.0° W

12

Chapter 10, Solution 108

Assuming

Ay

Apply vertical load P at D

Virtual Work:

We remove member BF and replace it with forces F and BF −F at pins BF

F and B, respectively Denoting the virtual displacements of points B and

F as δr and B δr respectively, and noting that P and F, δuuurD have the same direction, we have

δ = +

41.7 mmD

We remove member BF and replace it with forces F and BF −F at pins BF

F and B, respectively Denoting the virtual displacements of points B and

F as δr and B δr respectively, and noting that P and F, δuuurD have the same direction, we have

By inspection, one solution is cosθ =0 or θ =90.0°

Solving numerically: θ =0.38338 rad = 9.6883 and° θ = 0.59053 rad = 33.8351°

Stability

2 2