Problem 6-62A carton turnover machine is shown below.

16 in

15 in

7.3 in

7 in

600

2 in3 in

Problem 6-62At the instant shown, the 15 in link is driven clockwise at 5 rad/sec, and decelerating at

40 rad/sec2. Determine the instantaneous torque required to operate the 15 in arm, and the forces at the two lower bearings. The two long links closely resemble slender members, made from steel, with a width of 1 in and a thickness of

0.25 in. The carrier and carton weigh 26 lb and its mass moment of inertia, relative to an axis through its center, is 2.75 lb in s2.

Kinematic Diagram

2

A

B

C

D

G2

G4

G3 3

4

310

7.3”

16”

15”5.83”

600

3.6

220

7

56.30

2

3

Velocity Analysis

VB = 75 in/s

VB = 2 rAB = (5 rad/s)(15 in) = 75 in/s

VC/B

VC

VC = VB +> VC/B

to CB

to CDVG3/BVG3/C

VG3

to G3C

to G3B

VG3 = VB +> VG3/B

= VC +> VG3/C

Velocity Results

• VC = 40.4 in/s 680

• VC/B = 49.8 in/s

• VX/B = 25.6 in/s 56.30

• VX/C = 41.5 in/s 310

• VG3 = 77.7 in/s 490

• VG2 = 37.5 in/s 300

• VG4 = 20.2 in/s 680

Acceleration Analysis

• anC = (40.4)2/16 = 102.0 in/s2 22.00

• anG3/C = (41.5)2/5.83 = 295.4 in/s2 59.00

• anG3/B = (25.6)2/3.6 = 182.0 in/s2 33.70

• atB = (15)(40) = 600.0 in/s2 30.00

• anC/B = (49.8)2/7 = 354.3 in/s2

• anB = (75)2/15 = 375.0 in/s2 60.00

Acceleration Analysis

anC+>at

C= anB+>at

B+> anC/B+>at

C/B

to CB to CD

anC

anB

atB

anC/B

atC

atC/B

aB

aC

Acceleration Analysis• at

C = 250.8 in/s2 68.00

• =(250.8)/16 =15.68 rad/s2 ccw

• =(332.7)/7 =47.53 rad/s2

ccw

• aC = 270.7 in/s2

• atC/B = 332.7 in/s2

• aG4 = 135.4 in/s2

• aB = 707.5 in/s2 620

• aG2 = 353.8 in/s2

620

anC

anB

atB

anC/B

atC

atC/B

anG3/B an

G3/C

atG3/B at

G3/C

to G3C

to G3BaG3

• aG3 = 672 in/s2 830

aG3= aB+>anG3/B+>at

G3/B

= aC+> anG3/C+>at

G3/C

Inertia Forces: Link 2

15 in600

• aG2 = 353.8 in/s2

620

• Vol2 = (16 in)(1 in)(.25 in) = 4 in3

• W2 = (.283 lb/in3)(4 in3) = 1.13 lb

• FiG2 = (1.13 lb)(353.8 in/s2)/(386.4 in/s2)

= 1.03 lbs 620

FiG2

Inertia Forces: Link 2

15 in600

FiG2

• IG2 = ¼ [(1.13 lb)/(386.4 in/s2)](16 in)2

= 0.19 lb in s2

• = 40 rad/s2 ccw

• TiG2 = (0.19 lb in s2)(40 rad/s2)

= 7.5 lb in cw

TiG2

• Assume slender rod

Inertia Forces: Link 3

• aG3 = 672 in/s2 830

• =47.53 rad/s2ccw

• FiG3 = (26 lb)(672 in/s2)/(386.4 in/s2)

= 45.2 lbs 830

FiG3

• TiG3 = (2.75 in lb s2)(47.53 rad/s2)

= 130.7 in lbs cw

TiG3

• Vol4 = (17 in)(1 in)(.25 in) = 4.25 in3

• W4 = (.283 lb/in3)(4.25 in3) = 1.20 lb

Inertia Forces: Link 4

16 in220

• aG4 = 135.4 in/s2

• FiG4 = (1.20 lb)(135.4 in/s2)/(386.4 in/s2)

= 0.42 lb

FiG4

16 in220

Inertia Forces: Link 4

FiG4

• IG4 = ¼ [(1.20 lb)/(386.4 in/s2)](17 in)2

= 0.22 lb in s2

• TiG4 = (0.22 lb in s2)(15.68 rad/s2)

= 3.4 lb in cw

TiG4

• Assume slender rod

• = 15.68 rad/s2

ccw

Free Body Diagrams

Cx

Cy

Cy

Cx

Dy

Dx

Ay

Ax

Bx

ByBy

Bx

FiG3

TiG3

FiG2

TiG4

FiG2

TiG2

W3

W4

W2

TA

FBD: Link 2

7.5 600

Ay

Ax

By

Bx

FiG2

TiG2

620

15 in

Fx = 0

Ax - Bx + FiG2 cos62 = 0

Fy = 0

Ay - By +FiG2 sin62 - W2 = 0

W2

A = 0

Bx[15(sin60)] + By[15(cos60)] + W2[7.5(cos60)] - TiG2 - TA

- [FiG2(sin62)][7.5(cos60)] – [Fi

G2(cos62)][7.5(sin60)] = 0

TA

FBD: Link 3

Bx

Cx

FiG3

3 in

7 in

2 in

Cy

By

TiG3

W3

830

Fx = 0

Bx + Cx + FiG3 cos83 = 0

Fy = 0

By + Cy + FiG3 sin83 – W3 = 0

c = 0

W3(3) - Bx(7) - (FiG3 sin83)(3) - (Fi

G3 cos83)(5) - TiG3 = 0

FBD: Link 4

16 in

220

Dy

Dx

Cy

Cx FiG4

TiG4

W4

8

Fx = 0

Dx - Cx = 0

Fy = 0

Dy - Cy + FiG4 - W4 = 0

D = 0

Cx[16(sin22)] + Cy[16(cos22)] + W4[8(cos22)]

- FiG4[8(cos22)] - Ti

G4 = 0

Finally, Solving the Nine Simultaneous Equations:

• Ax = -23.3 lbs• Ay = -11.5 lbs• Bx = -22.8 lbs• By = -11.7 lbs

• Cx = 17.3 lbs• Cy = -7.2 lbs• Dx = 17.3 lbs• Dy = -6.4 lbs

• TA = -394.1 in lbs