218 39 solutions instructor manual chapter 4 acceleration

8/13/2019 218 39 Solutions Instructor Manual Chapter 4 Acceleration

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Chapter 4

Acceleration

4.1 The position vector of a point is defined by the equation ( )3 4 3 10t t= +R i j whereRisin mm and tis in seconds. Find the acceleration of the point at t = 2 s.

( ) ( )3 4 3 10t t t= +R i j

( ) ( )2 4t t= R i ( ) 2t t= R i ( ) ( ) 2 2 s 2 2 4 mm/s= = R i i Ans.

4.2 Find the acceleration at t = 3 s of a point which moves according to the equation( ) ( )2 3 3 = 6 3t t t +R i j . The units are meters and seconds.

( ) ( ) ( )2 3 3 = 6 3t t t t +R i j

( ) ( )2 2

= 2 2t t t t +R i j

( ) ( ) = 2 2t t t +R i j ( ) ( ) ( ) 2 3 s = 2 3 2 3 6 m/s + = +R i j i j Ans.

4.3 The path of a point is described by the equation 2 /10= ( 4) j tt e +R where R is inmillimeters and tis in seconds. For t= 20 s, find the unit tangent vector for the path, thenormal and tangential components of the points absolute acceleration, and the radius ofcurvature of the path.

( ) 2 /10= ( 4) j tt t e +R

( ) /10 2 /10= 2 ( 4)10

j t j tjt te t e +R

( )2

/10 /10 2 /10= 2 ( 4)5 5 100

j t j t j tj t j tt e e t e

+

R

Noticing, at t= 20 s, that /10 2 1.0j t je e = = , we find that

( ) 220 s = (20 4) 404 mm+ =R


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( ) ( ) 220 s = 2 20 (20 4) 40.00 126.92 133.07 72.5 mm/s10

jj

+ = = R

( )2

2 220 2020 s = 2 (20 4) 37.873 25.133 mm/s5 5 100

j jj

+ =

R

From the direction of the velocity we find the unit tangent and unit normal vectors 1 72.5 cos 72.5 sin 72.5 0.30058 0.95376= = + = i j i j Ans.

sin 72.5 cos 72.5 0.95376 0.30058= = = k i j i j

From these, the components of the points absolute acceleration are

( )( )2 2 0.95376 0.30058 37.873 25.133 mm/s 43.676 mm/snA = = =R i j i ji i Ans.

( )( )2 2 0.30058 0.95376 37.873 25.133 mm/s 12.586 mm/stA = = =R i j i ji i Ans.Then, from Eq. (4.2) or Eq. (4.14), the radius of curvature is

( )2 2

2

133.07 mm/s

405.4 mm43.676 mm/snA= = =

R

Ans.

Where the negative sign indicates that the point is in the negative direction from the

center of curvature of the points path.

4.4 The motion of a point is described by the equations 34 cosx t t= and ( )3 6 sin 2y t t= where x and y are in meters and t is in seconds. Find the acceleration of the point at

1.40 st= .

( ) ( )3 3 4 cos 6 sin 2t t t t t = + R i j

( ) ( ) ( )

3 3 3 2 3 4 cos 12 sin 2 sin 2 3 cos 2t t t t t t + t t = +

R i j

( ) ( )2 3 2 5 3 2 2 2 48 sin 36 cos 1 2 3 sin 2 cos 2t t t t t t t t +2 t t = + R i j

( ) 2 1.40 s 1 112.620 19.753 1 112.796 1.0 ft/s= = R i j Ans.

4.5 Link 2 in the figure has an angular velocity of 2 120 rad/s ccw = and an angularacceleration of 4800 rad/ 2s ccw at the instant shown. Determine the absolute

acceleration of pointA.

( )

2 2 2

2 2

2

2 2

2

120 0.500 4 800 0.500

n t

A O AO AO

AO AO

= + +

= +

= +

A A A A

R k R

i k i

2 2 7 200 2 400 m/s 7 589 161.6 m/sA

= + = A i j Ans.


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4.6 Link 2 is rotating clockwise as shown in the figure. Find its angular velocity andacceleration and the acceleration of its midpoint C.

n t

B A BA BA= + +A A A A

Construct the acceleration polygon.2178.38 m/s

356.8 rad/s cw0.5m

n

BA

BA

A

R= = = Ans.

Note the ambiguous sign of this square root. The sense of cannot be determined from

the accelerations, but here is found from the problem statement.2251.03 m/s 102.1 rad/s cw

0.5m

t

BA

BA

A

R= = = Ans.

292.76 44.0 m/sC= A Ans.

4.7 For the data given in the figure, find the velocity and acceleration of pointsBand C.

B A BA= +V V V

( )( )24 rad/s 0.4 m 9.6 m/sBA BAV R= = =

Construct the velocity polygon.

3.6 270 m/sB = V Ans.

2.51 12.1 m/sC= V Ans.n t


( ) ( )22 224 rad/s 0.4 m 230.4 m/snBA BAA R= = =

( ) ( )2 2160 rad/s 0.4 m 64 m/stBA BAA R= = = Construct the acceleration polygon.

2118.53 165 m/sB = A Ans.263 240.3 m/sC= A Ans.


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4.8 For the straight-line mechanism shown in the figure, 2 20 rad/s cw = and2

2 140 rad/s cw = . Determine the velocity and acceleration of pointBand the angular

acceleration of link 3.

2 2A O AO= +V V V

( )( )2 22

20 rad/s 0.100 m 2.000 m/sAO AOV R= = =

C A CA= +V V V Construct the velocity polygon of link 3.

3.864 270 m/sB

= V Ans.

2 2 2

n t

A O AO AO= + +A A A A

( ) ( )2 2

22 2

2 20 rad/s 0.100 m 40.000 m/sn

AO AOA R= = =

( ) ( )2 2

2 2

2 140 rad/s 0.100 m 14.000 m/st

AO AOA R= = =

n t

C A CA CA= + +A A A A

( ) ( )22 22.000 m/s 0.100 m 40.000 m/sn

CA CA CAA V R= = =

Construct the acceleration polygon of link 3.26.340 90 m/sB = A Ans.

2

23

14.000 m/s 140 rad/s cw0.100 m

t

BA

BA

AR

= = = Ans.

4.9 In the figure, the slider 4 is moving to the left with a constant velocity of 2.0 m/s. Find theangular velocity and acceleration of link 2.

2 2A C AC O AO= + = +V V V V V . Construct the velocity polygon of link 3.

2

2

2

3.864 m/s38.64 rad/s cw

0.100 m

AO

AO

V

R = = = Ans.


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4 4 4BO BO

n t n t

B A BA BA O= + + = + +A A A A A A A

( ) ( )22 2/ 358.5 mm/s / 250 mm 514.09 mm/snBA BA BAA V R= = = (Ignore compared to other

parts.)

( ) ( )4 4 4

22 2/ 4620 mm/s / 300 mm 71148 mm/snBO BO BOA V R= = = 2

2

3

140817.5 m/s563.3 rad/s ccw

250 mm

t

BA

BA

A

R = = = Ans.

4

4

22

4

37102.5 in/s123.7 rad/s ccw

300 mm

t

BO

BO

A

R = = = Ans.

4.12 For Problem 3.10, find the acceleration of point Cand the angular accelerations of links 3and 4.

Assume 2 = constant. Therefore 2 0 = .

2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 60 rad/s 0.150 m 540.0 m/sn

AO AOA R= = =

4 4 4BO BO

n t n t


( ) ( )22 2/ 13.02 m/s / 0.300 m 565.1 m/snBA BA BAA V R= = =

( ) ( )4 4 4

22 2/ 11.36 m/s / 0.300 m 430.1 m/snBO BO BOA V R= = =

Construct the acceleration polygon of link 3.2209.6 10.6 m/sC= A Ans.

2

23 136.5 m/s 455.0 rad/s ccw

0.300 m

t

BA

BA

AR

= = = Ans.

4

4

22

4

46.04 m/s153.5 rad/s cw

0.300 m

t

BO

BO

A

R = = = Ans.


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4.13 For Problem 3.11, find the acceleration of point Cand the angular accelerations of links 3and 4.


2 2 2AO AOn t

A O= + +A A A A

( ) ( )

2 2

2

2

2 248 rad/s 200 mm 18 432.0 in/s

n

AO AOA R=

= =

4 4 4BO BO

n t n t


( ) ( )22 2/ 267.5 mm/s / 800 mm 89.45 mm/snBA BA BAA V R= = = (Ignore compared to other

parts.)

( ) ( )4 4 4

22 2/ 9365 mm/s / 400 mm 219.26 m/snBO BO BOA V R= = =


22

3

1,393,250 mm/s1 741.6 rad/s ccw

800 mm

t

BA

BA

A

R = = = Ans.

4

4

22

4

1,222,300 mm/s3 055.8 rad/s ccw

400 mm

t

BO

BO

A

R = = = Ans.

4.14 Using the data of Problem 3.13, solve for the accelerations of points C and D and theangular acceleration of link 4.


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2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 1.0 rad/s 0.300 m 0.300 m/sn

AO AOA R= = =

4 4 4BO BO

n t n t


( ) ( )22 2/ 0.263 5 m/s / 0.150 m 0.462 9 m/snBA BA BAA V R= = =

( ) ( )4 4 4

22 2/ 0.227 0 m/s / 0.300 m 0.171 8 m/snBO BO BOA V R= = =

Construct the acceleration polygon of link 3.20.515 4 119.8 m/sC= A Ans.

20.492 5 21.8 m/sD = A Ans.

4

4

22

4

0.221 2 m/s0.737 4 rad/s cw

0.300 m

t

BO

BO

A

R = = = Ans.

4.15 For Problem 3.14, find the acceleration of point Cand the angular acceleration of link 4.


2 2 2AO AO

n tA O= + +A A A A

( ) ( )2 2

22 2

2 60.0 rad/s 150 mm 540 m/sn

AO AOA R= = =

4 4 4BO BO

n t n t


( ) ( )22 2/ 4235 mm/s / 150 mm 119.57 m/snBA BA BAA V R= = =

( ) ( )4 4 4



4

4

2

24 373.75 m/s 1 495 rad/s ccw

0.25 m

t

BO

BO

A

R = = = Ans.


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4.16 Solve Problem 3.16 for the acceleration of point Cand the angular acceleration of link 4.


2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 30.0 rad/s 75 mm 67.5 m/sn

AO AOA R= = =

4 4 4BO BO

n t n t



( ) ( )4 4 4

22 2/ 0.0 mm/s / 125 mm 0.0 m/snBO BO BOA V R= = =


4

4

22

4

108 m/s720 rad/s ccw

0.150 m

t

BO

BO

A

R = = = Ans.

4.17 For Problem 3.17, find the acceleration of pointBand the angular accelerations of links 3and 6.


2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 10.0 rad/s 63 mm 6300 mm/sn

AO AOA R= = =

( ) ( )22 2/ 468.75 mm/s / 250 mm 878.9 mm/snBA BA BAA V R= = =

n t



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25022.5 0 mm/sB = A Ans.2

2

3

4372.5 mm/s17.49 rad/s ccw

250 mm

t

BA

BA

A

R = = = Ans.

Construct the acceleration polygon of link 3, orCB CB

n t n t

C A CA CA B= + + = + +A A A A A A A

6 6 6DO DO

n t n t D C DC DC O= + + = + +A A A A A A A

( ) ( )22 2/ 15 mm/s / 100 mm 2.25 mm/snDC DC DCA V R= = = (Ignore compared to other

parts.)

( ) ( )6 6 6

22 2/ 604.5 mm/s / 150 mm 2436.13 mm/snDO DO DOA V R= = =

6

6

22

6

1621.954 mm/s10.81 rad/s cw

150 mm

t

DO

DO

A

R = = = Ans.

4.18 For the data of Problem 3.18, what angular acceleration must be given to link 2 for theposition shown to make the angular acceleration of link 4 zero?

4 4 4BO BO

n t

B O= + +A A A A

( ) ( )4 4 4


2 2 2AO AO

n t n t

A B AB AB O= + + = + +A A A A A A A

( ) ( )22 2/ 5.004 m/s / 0.425 m 58.93 m/snAB AB ABA V R= = =

( ) ( )2 2 2

22 2/ 5.600 m/s / 0.35 m 89.60 m/snAO AO AOA V R= = =

2

2

22

2

18.54 m/s52.98 rad/s ccw

0.350 m

t

AO

AO

A

R = = = Ans.


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4.19 For the data of Problem 3.19, what angular acceleration must be given to link 2 for theangular acceleration of link 4 to be 100 rad/ 2s cw at the instant shown?

4 4 4BO BO

n t

B O= + +A A A A

( ) ( )4 4 4


( )( )4 4

2 2

4 100 rad/s 200 mm 20 m/st

BO BOA R= = =

2 2 2AO AO

n t n t

A B AB AB O= + + = + +A A A A A A A

( ) ( )22 2/ 646.75 mm/s / 200 mm 2091.43 mm/snAB AB ABA V R= = = (Ignorable compared to other

parts.)

( ) ( )2 2 2

22 2/ 4500 mm/s / 125 mm 162 m/snAO AO AOA V R= = =

2

2

22

2

522575 mm/s4 180.7 rad/s ccw

125 mm

t

AO

AO

A

R = = = Ans.

4.20 Solve Problem 3.20 for the acceleration of point Cand the angular acceleration of link 3.


2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 8.0 rad/s 0.150 m 9.600 m/sn

AO AOA R= = =

4 4 4BO BO

n t n t


( ) ( )22 2/ 3.783 9 m/s / 0.250 m 57.270 m/snBA BA BAA V R= = =

( ) ( )4 4 4

22 2/ 3.483 7 m/s / 0.250 m 48.544 m/snBO BO BOA V R= = =


22

3

15.647 m/s62.59 rad/s cw

0.250 m

t

BA

BA

A

R = = = Ans.


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4.21 For Problem 3.21, find the acceleration of point Cand the angular acceleration of link 3.

Assume 2 = constant.

Therefore 2 0 = .

2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 56.0 rad/s 0.150 m 470.4 m/sn

AO AOA R= = =

4 4 4BO BO

n t n t


( ) ( )22 2/ 6.966 m/s / 0.250 m 194.09 m/sn

BA BA BAA V R= = =

( ) ( )4 4 4



22

3

18.52 m/s74.08 rad/s cw

0.250 m

t

BA

BA

A

R = = = Ans.

4.22 Find the accelerations of pointsBandDof Problem 3.22.


2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 42.0 rad/s 50 mm 88.2 m/sn

AO AOA R= = =


284.68 0 in/sB = A Ans.


n t n t

C A CA CA B= + + = + +A A A A A A A n t

D C DC DC= + +A A A A

( ) ( )22 2/ 1541.5 mm/s / 200 mm 11.88 m/snDC DC DCA V R= = =

n t



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240.4 90 m/sD = A Ans.

4.23 Find the accelerations of pointsBandDof Problem 3.23.


2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 209.4 rad/s 50 mm 2.19 m/sn

AO AOA R= = =

( ) ( )22 2/ 5468.75 mm/s / 150 mm 199.38 m/snBA BA BAA V R= = =

2732.17 240 m/sB = A Ans.


n t n t

C A CA CA B= + + = + +A A A A A A A n t

D C DC DC= + +A A A A

( ) ( )22 2/ 6725 mm/s / 125 mm 361.8 m/snDC DC DCA V R= = =

21.209 120 m/sD = A Ans.

4.24 to 4.30 The nomenclature for this group of problems is shown in the figure, and thedimensions and data are given in the accompanying table. For each problem,

determine 3 4 3 4 3 4, , , , , and . The angular velocity 2 is constant for each

problem, and a negative sign is used to indicate the clockwise direction. The dimensionsare given in millimeters.

This group of problems was solved on a programmable calculator. The position solutionvalues were found from Eqs. (2.25) through (2.32). The velocity values were found fromEqs. (3.22). The acceleration values were found from Eqs. (4.31) and (4.32).

Prob.3 , deg 4 , deg 3, rad/s 4 , rad/s

2

3, rad/s 2

4 , rad/s

4.24 105.29 159.60 0.809 3 0.525 0 0.230 28 0.008 09

4.25 171.01 195.54 70.452 7 47.566 8 3 196.657 49 3 330.841 37

4.26 45.57 91.15 -4.000 0 -4.000 0 -1.120 22 54.890 98

4.27 28.32 55.88 -0.632 6 -2.155 7 7.822 40 6.704 18

4.28 24.17 63.73 -1.516 7 1.712 9 41.414 93 74.975 93

4.29 38.42 155.60 -6.855 2 -1.234 5 62.500 44 -96.514 21

4.30 73.16 138.51 -0.505 1 7.275 3 -206.384 28 -94.122 01

n t



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4.31 Crank 2 of the system shown in the figure has a speed of 60 rev/min ccw. Find thevelocity and acceleration of pointBand the angular velocity and acceleration of link 4.

2

rev rad 1 min60 2

min rev 60 s

6.283 rad/s

=

=

2 2 2 2 4 2 / 4A O A O A A= + = +V V V V V

( ) ( )2 2 2 22

6.283 rad/s 175 mm

1.099 m/s

A O A OV R= =

=


4 4

4 4

4

910.75 mm/p

138.57 mm

A O

A O

V

R = =

4 6.572 rad/s cw = Ans.

4 4B O BO= +V V V

( )( )4 44

6.572 rad/s 700 mmBO BOV R= =

4.6 19.1 m/sB = V Ans.

Since we know the path ofA2on link 4, we write

2 2 2 2 2 2 4 2 4 2 2/ 4 / 4

n t c n t

A O A O A O A A A A A= + + = + + +A A A A A A ; 4 4 4 4 4 4n t

A O A O A O= + +A A

( ) ( )2 2 2 2

22 2

2 6.283 rad/s 175 mm 6.9 m/sn

A O A OA R= = =

( )( )2 4 2

2

/ 42 2 6.572 rad/s 616 mm/s 8.096 19.1 m/sc

A A A= = = 4A V

( )2

2

2

22

/ 4

/ 4

/ 4

616 mm/s0

An

A

A

VA

= = =

Construct the acceleration polygon.

4 4

4 4

22

4

11969.5 mm/s86.38 rad/s ccw

138.57 mm

t

A O

A O

A

R = = = Ans.

Construct the acceleration image of link 4.267.57 187.5 m/sB = A Ans.

4.32 The mechanism shown in the figure is a marine steering gear called Rapsons slide. 2O Bis the tiller, andACis the actuating rod. If the velocity ofACis 3 m/min to the left, findthe angular acceleration of the tiller.


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( )( )43 m/min

50 mm/s60 s/min

AV = =

2 2 2 2 4 2 / 4A O A O A A= + = +V V V V V


Assume 4AV = constant. Therefore 4 0AA = .


2 2 2 2 2 2

n t

A O A O A O= + +A A ; and

4 2 4 2 4 4/ 2 / 2

c n t

A A A A A A= + + +A A A A A

( )2 2

2 2

2 2

22

243.3 mm/s 0.9 mm/s2078.5 mm

A On

A O

A O

VA

R= = =

( ) ( )4 2 4

2

2 / 22 2 0.020 83 rad/s 25 mm/s 1.042 30.0 mm/sc

A A A= = = A V

( )4

4

4

22

/ 2

/ 2

/ 2

25 mm/s0

An

A

A

VA

= = =


2 2

2 2

22

2

1.042 mm/s0.000 50 rad/s cw

2078.5 mm

t

A O

A O

A

R = = = Ans.

4.33 Determine the acceleration of link 4 of Problem 3.26.



2 2 2 2 2 2 4 2 4 2 2/ 4 / 4

n t c n t

A O A O A O A A A A A= + + = + + +A A A A A A

( ) ( )2 2 2 2

22 2

2 36.0 rad/s 0.250 m 324.0 m/sn

A O A OA R= = =

( ) ( )4 2 24 / 4

2 2 0.0 rad/s 6.588 m/s 0cA A A= = =A V ;( )

2

4

2

22

/ 4

/ 2

/ 4

6.588 in/s0

An

A

A

VA

= = =


4

2290.5 180.0 m/sA = A Ans.


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4.34 For Problem 3.27, find the acceleration of pointE.


2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 2

22 2

2 72.0 rad/s 38 mm 196.99 m/sn

AO AOA R= = =

4 4 4BO BO

n t n t


( ) ( )2

2 2/ 3780 mm/s / 263 mm 54.33 m/snBA BA BAA V R= = =

( ) ( )4 4 4


Construct the acceleration polygon of link 3.Since we know the path of C3on link 6, we write

3 6 3 6 3 3/ 6 / 6

c n t

C C C C C C = + + +A A A A A and 6 6 6 6 6 6n t

C O C O C O= + +A A

( ) ( )3 6 3

2

6 / 62 2 9.673 rad/s 1253.75 mm/s 24.255 m/sc

C C C= = =A V ;

( )3

3

3

22

/ 6

/ 6

/ 6

1253.75 mm/s0

Cn

C

C

VA

= = =

( )6 66 6

6 6

2221076.75 mm/s 10.41 m/s

111.325 mm

C On

C O

C O

VA

R= = =

Construct the acceleration polygon of link 6.2180.82 252.8 m/sE= A Ans.

4.35 Find the acceleration of pointBand the angular acceleration of link 4 of Problem 3.24.Assume 2 = constant. Therefore 2 0 = .

2 2 2AO AO

n t

A O= + +A A A A

( ) ( )2 222 2

2 24.0 rad/s 200 mm 115.2 m/sn

AO AOA R= = =


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Since we know the path of P3on link 4, we write

3 3 3 3 3 3 4 3 4 3 3/ 4 / 4

n t c n t

P A P A P A P P P P P= + + = + + +A A A A A A

( ) ( )3 3 3 3 3 3

22 2/ 4045 mm/s / 656.75 mm 24.91 m/snP A P A P AA V R= = =

( )( )3 4 3

2

/ 42 2 6.159 rad/s 2592.5 mm/s 31.934 102.4 m/sc

P P P= = = 4A V

( )3

3

3

22

/ 4

/ 4

/ 4

2592.5 mm/s0

Pn

P

P

VA

= = =

Construct the acceleration polygon of link 3.2123.765 25.7 m/sB = A Ans.

Since links 3 and 4 remain perpendicular,

3 3

3 3

22

4 3

30070 mm/s45.78 rad/s ccw

656.75 mm

t

P A

P A

A

R = = = = Ans.

4.36 For Problem 3.25, find the acceleration of pointBand the angular acceleration of link 3.


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Assume A =V constant. Therefore 0A =A .

Since we know the path of P3on link 4, we write

3 3 3 3 3 3 4 3 4 3 3/ 4 / 4

n t c n t

P A P A P A P P P P P= + + = + + +A A A A A A

( ) ( )3 3 3 3 3 3

22 2/ 68.375 mm/s / 225 mm 20.78 mm/snP A P A P AA V R= = =

( ) ( )3 4 3 2/ 42 2 0.3039 rad/s 307.75 mm/s 187 102.8 mm/scP P P= = = 4A V

( )3

3

3

22

/ 4

/ 4

/ 4

307.75 mm/s0

Pn

P

P

VA

= = =

Construct the acceleration polygon of link 3.2505.75 102.1 mm/sB = A Ans.

3 3

3 3

22

3

280.5 mm/s1.247 rad/s cw

225 mm

t

P A

P A

A

R = = = Ans.

4.37 Assuming that both links 2 and 3 of Problem 3.28 are rotating at constant speed, find theacceleration of point 4P .

Since 2 = constant and 3 = constant, 2 3 0 = = .

Since we know the paths of P4on link 2 and on link 3, we write

2 2 2 2 2 2P O P O

n t

P O= + +A A A A ; 3 3 3 3 3 3P O P On t

P O= + +A A A A

( ) ( )2 2 2 2

22 2

2 30.0 rad/s 0.054 9 m 49.410 m/sn

P O P OA R= = = .

( ) ( )3 3 3 3

22 2

3 20.0 rad/s 0.102 6 m 41.039 m/sn

P O P OA R= = =

4 2 4 2 4 4 3 4 3 4 4/ 2 / 2 /3 /3

c n t c n t

P P P P P P P P P P P= + + + = + + +A A A A A A A A A

( )( )4 2 4

2

2 / 22 2 30.0 rad/s 1.683 m/s 100.98 30.0 m/sc

P P P= = = A V

( ) ( )4 3 4

2

3 /32 2 20.0 rad/s 1.155 m/s 46.22 225.0 m/sc

P P P= = = A V

( )4

4

4

22

/ 2

/ 2

/ 2

1.683 m/s0

Pn

P

P

VA

= = =

;

( )4

4

4

22

/ 2

/ 2

/ 2

1.155 m/s0

Pn

P

P

VA

= = =


4

2125.73 66.6 m/sP = A Ans.


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4.38 Solve Problem 3.32, for the accelerations of pointsAandB.

Assume

2 = constant. Therefore 2 0 = .

Since we know the path of F2on link 3, we write

2 2 2 2 2 2 3 2 3 2 2/ 3 /3

n t c n t

F E F E F E F F F F F = + + = + + +A A A A A A and 3 3 3 3 3 3n t

F G F G F G= + +A A A A

( ) ( )2 2 2 2

22 2

2 25.0 rad/s 25 mm 15.625 m/sn

F E F E A R= = =

( )3 3

3 3

3 3

22

2102.75 mm/s 69.46 m/s152 mm

F Gn

F G

F G

VA

R= = =

( )( )2 3 2

2

3 /32 2 0.676 rad/s 616.5 mm/s 833.5 80.5 mm/sc

F F F= = = A V

( )2

2

2

22

/3

/ 3

/ 3

616.5 mm/s0

Fn

F

F

VA

= = =

Construct the acceleration polygon of link 3.n t

A C AC AC= + +

A A A A and

n t

B D BD BD= + +

A A A A ( ) ( )

22 2/ 195.175 mm/s / 150 mm 253.95 mm/snAC AC ACA V R= = = 24230 0 mm/sA = A Ans.

( ) ( )22 2/ 190.55 mm/s / 150 mm 242.1 mm/snBD BD BDA V R= = =

220.152 0 m/sB = A Ans.

4.39 For Problem 3.33, determine the acceleration of point 4C and the angular acceleration oflink 3 if crank 2 is given an angular acceleration of 2 rad/ 2s ccw.

n t


( ) ( )22 2

2 10.0 rad/s 75 mm 7500 mm/sn

BA BAA R= = =

( ) ( )2 22 2.0 rad/s 75 mm 150 mm/stBA BAA R= = =

4 4 /1

r n t

D D B DB DB= = + +A A A A A

( ) ( )22 2/ 750 mm/s / 25 mm 225 m/snDB DB DBA V R= = =


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Draw the acceleration image of link 4.

4

215 91.1 m/sC = A Ans.

3 4 3 / 4

r n t

C C C A CA CA= + = + +A A A A A A

( ) ( )22 2/ 1500 mm/s / 50 mm 45 m/snCA CA CAA V R= = =

Draw the acceleration image of link 3.

3 3

3 3

22

3


50 mm

t

C A

C A

A

R = = = Ans.

4.40 Determine the angular accelerations of links 3 and 4 of Problem 3.30.


Since we know the path ofD4on link 2, we write

2 2 2 2 2 2

n t

D A D A D A= + +A A A A


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( ) ( )2 2 2 2

22 2

2 15.0 rad/s 31 mm 6975 mm/sn

D A D AA R= = = .

4 2 4 2 4 4 4 4 4 4 4/ 2 / 2

c n t n t

D D D D D D E D E D E= + + + = + +A A A A A A A A

( ) ( )4 2 4

2

2 / 22 2 15.0 rad/s 1235.25 mm/s 37.05 80.8 m/sc

D D D= = = A V

( ) ( )4 4 4

22 2

/ 2 / 2 / 2

1235.25 mm/s 62.5 mm 24.413 m/snD D D

A V = = =

( ) ( )4 4 4 4 4 4

22 2381 mm/s 88 mm 1.65 m/snD E D E D EA V R= = =

Draw the acceleration images of links 2 and 4.

4 4

4 4

22

4


88 mm

t

D E

D E

A

R = = = Ans.

3 2 3 3 3 3 3 3/ 2

r n t

C C C D C D C D= + = + +A A A A A A

( ) ( )3 3 3 3 3 3

22 21047.7 mm/s 13 mm 84.45 m/snC D C D C DA V R= = =

3 3

3 3

22

3

5792.5 mm/s445.6 rad/s ccw

13 mm

t

C D

C D

A

R

= = = Ans.

4.41 For Problem 3.31, determine the acceleration of point Gand the angular accelerations oflinks 5 and 6.


BA BA

n t

B A= + +A A A A

( ) ( )22 2

2 10 rad/s 25 mm 2.5 m/sn

BA BAA R= = = n t n t

C B CB CB D CD CD= + + = + +A A A A A A A

( ) ( )22 2/ 333.25 mm/s / 100 mm 1.11 m/snCB CB CBA V R= = =

( ) ( )22 2/ 166.5 mm/s / 50 mm 0.56 m/snCD CD CDA V R= = =

Construct the acceleration image of link 3.Since we know the path ofE3on link 6, we write

3 6 3 6 3 3/ 6 / 6

c n t

E E E E E E= + + +A A A A A and 6 6 6 6 6 6n t

E H E H E H= + +A A

( ) ( )3 6 3

2

6 / 62 2 3.774 rad/s 272.25 mm/s 2.05 104.5 m/sc

E E E= = = A V ;


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62

( )3 3 3

22

/ 6 / 6 / 6/ 272.25 mm/s / 0n

E E EA V = = =

( )6 6 6 6 6 6

22 2121.5 mm/s 32.25 mm 457.74 mm/snE H E H E HA V R= = =

Construct the acceleration image of link 6.28785 64.5 mm/sG = A Ans.

6 6

6 6

22

6

3547.5 mm/s110 rad/s cw

32.25 mm

tE H

E H

A

R = = = Ans.

5 6 5 5 5 5 5 5/ 6

r n t

F F F E F E F E = + = + +A A A A A A

( ) ( )5 5 5 5 5 5

22 2319.5 mm/s 13 mm 7.85 m/snF E F E F E A V R= = =

5 5

5 5

2

5

0 in/s0

13 mm

t

F E

F E

A

R = = = Ans.

4.42 Find the inflection circle for motion of the coupler of the double-slider mechanism shownin the figure. Select several points on the centrode normal and find their conjugatepoints. Plot portions of the paths of these points to demonstrate for yourself that theconjugates are indeed the centers of curvature.


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4.43 Find the inflection circle for motion of the coupler relative to the frame of the linkageshown in the figure. Find the center of curvature of the coupler curve of point C andgenerate a portion of the path of Cto verify your findings.

Since point C is on the inflection circle; its center of curvature is at infinity; and its point

path is a straight line in the vicinity of the position shown.


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4.44 For the motion of the coupler relative to the frame, find the inflection circle, the centrodenormal, the centrode tangent, and the centers of curvature of points C and D of thelinkage of Problem 3.13. Choose points on the coupler coincident with the instant centerand inflection pole and plot their paths.


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4.45 On 450- by 600-mm paper, draw the linkage shown in the figure in full size, placing A at 150 mm from the lower edge and 175 mm from the right edge. Better utilization of thepaper is obtained by tilting the frame through about 15 as shown.

(a) Find the inflection circle.(b) Draw the cubic of stationary curvature.(c) Choose a coupler point C coincident with the cubic and plot a portion of its

coupler curve in the vicinity of the cubic.(d) Find the conjugate point C . Draw a circle through C with center at C and

compare this circle with the actual path of C.(e)Find Balls point. Locate a point Don the coupler at Balls point and plot a portion

of its path. Compare the result with a straight line.

Drawn with a precise CAD system above, the circle around center C matches the couplercurve near C within better than visual comparison can detect for the 30 of crankrotation shown. Similarly, Balls point D follows an almost perfect straight line over thesame range as shown.

218 39 solutions instructor manual chapter 4 acceleration

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