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Unformatted text preview: Solutions to Chapter 2 Exercise Problems Problem 2.1 In the mechanism shown below, link 2 is rotating CCW at the rate of 2 rad/s (constant). In the position shown, link 2 is horizontal and link 4 is vertical. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vC4, ω3, and ω4. b) Determine aC4, α3, and α4. Link lengths: AB = 75 mm, CD = 100 mm ω2 A 2 B 4 D 250 mm 3 C 50 mm Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2 v B2 = v A2 + v B2 / A2  48  v A2 = 0 Therefore, vC3 + v B3 /C3 = v A2 + v B2 / A2 Now, (1) ω 2 = 2 rad / s CCW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (2 rad / s)(75 mm) =150 mm / s vC3 / B3 = ω 3 × rC / B (⊥ to rC / B ) vC4 / D4 = ω 4 × rC / D (⊥ to rC / D ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, vC3 / B3 = 156 mm / s vC4 / D4 = vC4 = 43 mm / s Now, ω3 = vC3 / B3 156 = = .86 rad / s rC / B 182 From the directions given in the position and velocity polygons ω 3 = .86 rad / s CW
Also, ω4 = vC4 / D4 43 = = .43 rad / s rC / D 100 From the directions given in the position and velocity polygons ω 4 = .43 rad / s CW
Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 2 2 ⋅ 75 = 300 mm / s2 B B  49  in the direction of  rB2 / A2
t aB2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = .862 ⋅182 =134.6 mm / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = .432 ⋅100 = 18.5 mm / s 2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aC3 / B3 = 19.22 mm / s 2 t aC4 / D4 = 434.70 mm / s 2 Then,
t aC3 / B3 67, 600 = = 27, 900 rad / s2 α3 = 2.42 rC / B t aC4 / D4 434.70 = = 4.347 rad / s2 100 rC / D α4 = To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly counterclockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly counterclockwise. From the acceleration polygon, aC4 = 435 mm / s 2  50  Problem 2.2 In the mechanism shown below, link 2 is rotating CCW at the rate of 500 rad/s (constant). In the position shown, link 2 is vertical. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vC4, ω3, and ω4. b) Determine aC4, α3, and α4. Link lengths: AB = 1.2 in, BC = 2.42 in, CD = 2 in Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2 v B2 = v A2 + v B2 / A2 v A2 = 0 Therefore,  51  vC3 + v B3 /C3 = v A2 + v B2 / A2 Now, (1) ω 2 = 500 rad / s CCW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (500 rad / s)(1.2 in) = 600 in / s vC3 / B3 = ω 3 × rC / B (⊥ to rC / B ) vC4 / D4 = ω 4 × rC / D (⊥ to rC / D ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, vC3 / B3 = 523.5 in / s vC4 / D4 = vC4 = 858in / s Now, ω3 = vC3 / B3 523.5 = = 216.3 rad / s 2.42 rC / B From the directions given in the position and velocity polygons ω 3 = 216.3 rad / s CCW
Also, ω4 = vC4 / D4 858 = = 429 rad / s 2 rC / D From the directions given in the position and velocity polygons ω 4 = 429 rad / s CCW
Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 500 2 ⋅1.2 = 300000 in / s 2 B B in the direction of  rB2 / A2
t aB2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 )  52  r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 216.32 ⋅ 2.42 = 113, 000 in / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 4292 ⋅ 2 = 368, 000 in / s 2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aC3 / B3 = 67561in / s 2 t aC4 / D4 = 151437 in / s 2 Then,
t aC3 / B3 67561 = = 27, 900 rad / s2 α3 = 2.42 rC / B t aC4 / D4 151437 = = 75, 700 rad / s2 α4 = 2 rC / D To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly clockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly clockwise. From the acceleration polygon, aC4 = 398, 000 in / s 2  53  Problem 2.3 In the mechanism shown below, link 2 is rotating CW at the rate of 10 rad/s (constant). In the position shown, link 4 is vertical. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vC4, ω3, and ω4. b) Determine aC4, α3, and α4. Link lengths: AB = 100 mm, BC = 260 mm, CD = 180 mm C 3 B ω2 2 A 250 mm D 4 Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2 v B2 = v A2 + v B2 / A2 v A2 = 0 Therefore, vC3 + v B3 /C3 = v A2 + v B2 / A2  54 (1) Now, ω 2 =10 rad / s CW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (10 rad / s)(100 mm) =1000 mm / s vC3 / B3 = ω 3 × rC / B (⊥ to rC / B ) vC4 / D4 = ω 4 × rC / D (⊥ to rC / D ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, vC3 / B3 = 31.3 mm / s vC4 / D4 = vC4 = 990 mm / s Now, ω3 = vC3 / B3 31.3 = = .12 rad / s 260 rC / B From the directions given in the position and velocity polygons ω 3 = .12 rad / s CCW
Also, ω4 = vC4 / D4 990 = = 5.5 rad / s 180 rC / D From the directions given in the position and velocity polygons ω 4 = 5.5 rad / s CW
Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 10 2 ⋅100 =10, 000 mm / s 2 B B in the direction of  rB2 / A2
t aB2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = .12 2 ⋅ 260 = 3.744 mm / s2  55  in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 5.52 ⋅180 = 5, 445 mm / s 2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aC3 / B3 = 4784 mm / s 2 t aC4 / D4 = 1778 mm / s 2 Then, α3 = α4 = t aC3 / B3 4785 = = 18.4 rad / s2 260 rC / B t aC4 / D4 1778 = = 9.88 rad / s2 180 rC / D To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly counterclockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly counterclockwise. From the acceleration polygon, aC4 = 5, 700 mm / s 2  56  Problem 2.4 In the mechanism shown below, link 2 is rotating CW at the rate of 4 rad/s (constant). In the position shown, θ is 53˚. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vC4, ω3, and ω4. b) Determine aC4, α3, and α4. Link lengths: AB = 100 mm, BC = 160 mm, CD = 200 mm A 2 ω2 C B 160 mm 220 mm D 3 4 θ  57  Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2 v B2 = v A2 + v B2 / A2 v A2 = 0 Therefore, vC3 + v B3 /C3 = v A2 + v B2 / A2 Now, (1) ω 2 = 4 rad / s CW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (4 rad / s)(100 mm) = 400 mm / s vC3 / B3 = ω 3 × rC / B (⊥ to rC / B ) vC4 / D4 = ω 4 × rC / D (⊥ to rC / D ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, vC3 / B3 = 500 mm / s vC4 / D4 = vC4 = 300 mm / s Now, ω3 = vC3 / B3 500 = = 3.125 rad / s rC / B 160 From the directions given in the position and velocity polygons ω 3 = 3.125 rad / s CCW
Also, ω4 = vC4 / D4 300 = = 1.5 rad / s 200 rC / D From the directions given in the position and velocity polygons ω 4 =1.5 rad / s CCW
Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3  58  t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 4 2 ⋅100 =1600 mm / s 2 B B in the direction of  rB2 / A2
t aB2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 3.1252 ⋅160 = 1560 mm / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 1.52 ⋅ 200 = 450 mm / s 2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aC3 / B3 = 618.5 mm / s 2 t aC4 / D4 = 3, 220 mm / s 2 Then, α3 = α4 = t aC3 / B3 618.5 = = 3.87 rad / s2 160 rC / B t aC4 / D4 3220 = = 16.1rad / s2 200 rC / D To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly counterclockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly counterclockwise. From the acceleration polygon, aC4 = 3250 mm / s 2  59  Problem 2.5 In the mechanism shown below, link 2 is rotating CCW at the rate of 4 rad/s (constant). In the position shown, link 2 is horizontal. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vC4, ω3, and ω4. b) Determine aC4, α3, and α4. Link lengths: AB = 1.25 in, BC = 2.5 in, CD = 2.5 in D 1.0 in A 4 0.75 in 2 B ω2 3 C Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2  60  v B2 = v A2 + v B2 / A2 v A2 = 0 Therefore, vC3 + v B3 /C3 = v A2 + v B2 / A2 Now, (1) ω 2 = 4 rad / s CCW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (4 rad / s)(1.25in) = 5in / s vC3 / B3 = ω 3 × rC / B (⊥ to rC / B ) vC4 / D4 = ω 4 × rC / D (⊥ to rC / D ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, vC3 / B3 = 6.25in / s vC4 / D4 = vC4 = 3.75in / s Now, ω3 = vC3 / B3 6.25 = = 2.5 rad / s 2.5 rC / B From the directions given in the position and velocity polygons ω 3 = 2.5 rad / s CCW
Also, ω4 = vC4 / D4 3.75 = =1.5 rad / s 2.5 rC / D From the directions given in the position and velocity polygons ω 4 =1.5 rad / s CW
Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now,  61  a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 4 2 ⋅1.25 = 20 in / s 2 B B in the direction of  rB2 / A2
t aB2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 2.52 ⋅ 2.5 = 15.6 in / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 1.52 ⋅ 2.5 = 5.6 in / s 2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aC3 / B3 = 4.69 in / s 2 t aC4 / D4 = 4.69 in / s 2 Then, α3 = t aC3 / B3 4.69 = = 1.87 rad / s2 2.5 rC / B t aC4 / D4 4.69 = = 1.87 rad / s2 2.5 rC / D α4 = To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly counterclockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly clockwise. From the acceleration polygon, aC4 = 7.32 in / s 2  62  Problem 2.6 In the mechanism shown below, link 2 is rotating CW at the rate of 100 rad/s (constant). In the position shown, link 2 is horizontal. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vC4 and ω3 b) Determine aC4 and α3 Link lengths: AB = 60 mm, BC = 200 mm A 2 120 mm C 4 3 B ω2 Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2 v B2 = v A2 + v B2 / A2 v A2 = 0 Therefore, vC3 + v B3 /C3 = v A2 + v B2 / A2  63 (1) Now, ω 2 =100 rad / s CW
v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) = (100 rad / s)(60 mm) = 6000 mm / s vC3 / B3 = ω 3 × rC / B (⊥ to rC / B ) vC4 / D4 → parallel to the ground. Solve Eq. (1) graphically with a velocity polygon. From the polygon, vC3 / B3 = 7, 500 mm / s vC4 / D4 = vC4 = 4500 mm / s Now, ω3 = vC3 / B3 7500 = = 37.5 rad / s 200 rC / B From the directions given in the position and velocity polygons ω 3 = .12 rad / s CW
Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 100 2 ⋅ 60 = 600, 000 mm / s 2 B B in the direction of  rB2 / A2
t aB2 / A2 = 0 since link 2 rotates at a constant speed ( α 2 = 0 ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 37.52 ⋅ 200 = 281, 000 mm / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) a C4 / D4 = a C4 → parallel to ground Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,  64  t aC3 / B3 = 211, 000 mm / s 2 a C4 / D4 = a C4 = 248, 000 mm / s 2 Then, α3 = t aC3 / B3 211, 000 = = 1060 rad / s2 200 rC / B To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly counterclockwise. From the acceleration polygon, aC4 = 248, 000 mm / s 2 Problem 2.7 In the mechanism shown below, link 4 is moving to the left at the rate of 4 ft/s (constant). Write the appropriate vector equations, solve them using vector polygons, and a) Determine ω3 and ω4. b) Determine α3 and α4. Link lengths: AB = 10 ft, BC = 20 ft. B 3 2 120˚ A 8.5 ft vC 4 C 4  65  Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2 v B2 = v A2 + v B2 / A2 v A2 = 0 Therefore, vC3 + v B3 /C3 = v A2 + v B2 / A2 Now, vC4 = 4 ft / s parallel to the ground v B3 /C3 = ω 3 × rB/C (⊥ to rB/C ) v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 /C3 = 2.3 ft / s v B2 / A2 = 2.3 ft / s or (1) ω3 = v B3 /C3 2.3 = = .115 rad / s 20 rB/C From the directions given in the position and velocity polygons  66  ω 3 = .115 rad / s CW
Also, ω2 = v B2 / A2 2.3 = = .23 rad / s 10 rB/ A From the directions given in the position and velocity polygons ω 2 = .23 rad / s CCW ω 4 = 0 rad / s since it does not rotate
Acceleration Analysis: aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = .232 ⋅10 = .529 ft / s 2 B B in the direction of  rB2 / A2
t t aB2 / A2 = α 2 × rB/ A ⇒ aB2 / A2 = α 2 ⋅ rB/ A (⊥ to rB/ A ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = .1152 ⋅ 20 = .264 ft / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) a C4 / D4 = 0 link 4 is moving at a constant velocity Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aC3 / B3 = 0.045 ft / s 2 a tB2 / A2 = 0.017 ft / s 2 Then,
t aC3 / B3 0.45 = = .023 rad / s2 α3 = 20 rC / B t aB2 / A2 0.017 = = .0017 rad / s2 α2 = 10 rB/ A  67  To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly clockwise. To determine the direction of α2, determine the direction that rB/ A must be rotated to be parallel to t aB2 / A2 . This direction is clearly counterclockwise. Problem 2.8 In the mechanism shown below, link 4 is moving to the right at the rate of 20 in/s (constant). Write the appropriate vector equations, solve them using vector polygons, and a) Determine ω3 and ω4. b) Determine α3 and α4. Link lengths: AB = 5 in, BC = 5 in. A 2 7 in C 4 3 vC 4 45˚ B  68  Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2 v B2 = v A2 + v B2 / A2 v A2 = 0 Therefore, vC3 + v B3 /C3 = v A2 + v B2 / A2 Now, vC4 = 20 in / s parallel to the ground v B3 /C3 = ω 3 × rB/C (⊥ to rB/C ) v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 /C3 = 14.1in / s v B2 / A2 = 14.1in / s or (1) ω3 = v B3 /C3 14.1 = = 2.82 rad / s 5 rB/C From the directions given in the position and velocity polygons ω 3 = 2.82 rad / s CCW
Also, ω2 = v B2 / A2 14.1 = = 2.82 rad / s 5 rB/ A From the directions given in the position and velocity polygons ω 2 = 2.82 rad / s CCW ω 4 = 0 rad / s since it doesn’t rotate
Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3  69  t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 2.82 2 ⋅ 5 = 39.8in / s 2 B B in the direction of  rB2 / A2
t t aB2 / A2 = α 2 × rB/ A ⇒ aB2 / A2 = α 2 ⋅ rB/ A (⊥ to rB/ A ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 2.82 2 ⋅ 5 = 39.8 in / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) a C4 / D4 = 0 link 4 is moving at a constant velocity Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aC3 / B3 = 38.8 in / s 2 a tB2 / A2 = 38.8 in / s 2 Then, α3 = t aC3 / B3 38.8 = = 7.76 rad / s2 5 rC / B t aB2 / A2 38.8 = = 7.76 rad / s2 5 rB/ A α2 = α 4 = 0 (link 4 isnot rotating)
To determine the direction of α3, determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly counterclockwise. To determine the direction of α 2 , determine the direction that rB/ A must be rotated to be parallel to t aB2 / A2 . This direction is clearly clockwise.  70  Problem 2.9 In the mechanism shown below, link 4 is moving to the left at the rate of 0.6 ft/s (constant). Write the appropriate vector equations, solve them using vector polygons, and determine the velocity and acceleration of point A3. Link lengths: AB = 5 in, BC = 5 in. A 3 B 135˚ 2 4 C vC 4 Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2 v B3 = v A3 + v B3 / A3 Therefore, vC3 + v B3 /C3 = v A3 + v B3 / A3  71 (1) Now, vC4 = .6 ft / s parallel to the ground v B3 /C3 = ω 3 × rB/C (⊥ to rB/C ) v B3 / A3 = ω 3 × rB/ A (⊥ to rB/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 /C3 = .85 ft / s or ω3 = v B3 /C3 = .85 = 2.04 rad / s (5 /12) rB/C From the directions given in the position and velocity polygons ω 3 = 2.04 rad / s CW
Now, v B3 / A3 = ω 3 × rB/ A (⊥ to rB/ A ) = (2.04)(5 /12) = .85 ft / s Using velocity image, v A3 =1.34 ft / s Acceleration Analysis: aC4 = aC3 = 0 aB3 = aB2 = aB3 /C3 = a r B3 /C3 + a t B3 /C3 Now, a r 3 /C3 = ω 3 × (ω 3 × rB/C ) ⇒ a r 3 /C3 = ω 3 2 ⋅ rB/C = 2.04 2 ⋅ (5 /12) = 1.73 ft / s 2 B B in the direction of  rB3 /C3
t t aB3 /C3 = α 3 × rB/C ⇒ aB3 /C3 = α 3 ⋅ rB/C (⊥ to rB/C ) (2) a C4 / D4 = 0 link 4 is moving at a constant velocity Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aB3 /C3 = 1.73 ft / s 2 Then,  72  α3 = t aB3 /C3 = 1.73 = 4.15 rad / s2 (5 /12) rB/C To determine the direction of α3, determine the direction that rB/C must be rotated to be parallel to t aB3 /C3 . This direction is clearly clockwise. Using acceleration image, a A3 = 4.93 ft / s 2 Problem 2.10 In the mechanism shown below, link 4 moves to the right with a constant velocity of 75 ft/s. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vB2, vG3, ω2, and ω3. b) Determine aB2, aG3, α2, and α3. Link lengths: AB = 4.8 in , BC = 16.0 in , BG = 6.0 in B G 2 A 42˚ 3 C 4 Position Analysis: Draw the linkage to scale.  73  B G 2 42˚ A ov a 1, a 2 25 ft/sec c 3, c4 3 AB = 4.8" BC = 16.0" BG = 6.0" AC = 19.33" C Velocity Polygon g3 b2, b3 Velocity Analysis: v B3 = vC3 + v B3 /C3 v B3 = v B2  74  v B2 = v A2 + v B2 / A2 v A2 = 0 Therefore, vC3 + v B3 /C3 = v A2 + v B2 / A2 Now, vC3 = 75 ft / s in the direction of rC / A v B3 /C3 = ω 3 × rB/C (⊥ to rB/C ) v B2 / A2 = ω 2 × rB/ A (⊥ to rB/ A ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 /C3 = 69.4 ft / s or (1) ω3 = v B3 /C3 = 69.4 = 52 rad / s rB/C 16(1 /12) From the directions given in the position and velocity polygons ω 3 = 52 rad / s CCW
Also, ω2 = v B2 / A2 = 91.5 = 228 rad / s 4.8(1 /12) rB/ A From the directions given in the position and velocity polygons ω 2 = 228 rad / s CW
To compute the velocity of G3, vG3 = v B3 + vG3 / B3 = v B3 + ω 3 × rG3 / B3 Using the values computed previously ω 3 × rG3 / B3 = 52(6.0) = 312 in / s
and from the directions given in the velocity and position diagrams ω 3 × rG3 / B3 = 312 in / s ⊥ rG3 / B3
Now draw vG3 on the velocity diagram vG3 = 79.0 ft / s in the direction shown.  75  Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 2282 ⋅ (4.8 /12) = 20, 900 ft / s 2 B B in the direction of  rB2 / A2
t t aB2 / A2 = α 2 × rB/ A ⇒ aB2 / A2 = α 2 ⋅ rB/ A (⊥ to rB/ A ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 52 2 ⋅ (16 /12) = 3605 ft / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) a C4 / D4 = 0 link 4 is moving at a constant velocity Solve Eq. (2) graphically with an acceleration polygon. From the acceleration polygon,
t aC3 / B3 = 28, 700 ft / s 2 a tB2 / A2 = 20, 000 ft / s 2 Then, α3 = t aC3 / B3 28, 700 = = 21, 500 rad / s2 (16 /12) rC / B t aB2 / A2 = 20, 000 = 50, 000 rad / s2 (4.8 /12) rB/ A α2 = To determine the direction of α 3 , determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly clockwise. To determine the direction of α 2 , determine the direction that rB/ A must be rotated to be parallel to t aB2 /C2 . This direction is clearly counterclockwise. From the acceleration polygon, aB2 = 28, 900 ft / s 2  76  To compute the acceleration of G3, use acceleration image. From the acceleration polygon, aG3 = 18, 000 ft / s 2 Problem 2.11 For the fourbar linkage, assume that ω2 = 50 rad/s CW and α2 = 1600 rad/s2 CW. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vB2, vC3, vE3, ω3, and ω4. b) Determine aB2, aC3, aE3,α3, and α4. E 3 C B 2 120˚ A Position Analysis Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B and then C. Then locate E. Velocity Analysis: v B3 = v B2 = v B2 / A2 vC3 = vC4 = vC4 / D4 = v B3 + vC3 / B3 Now, v B2 / A2 = ω 2 × rB/ A ⇒ v B/ A = ω 2 ⋅ rB/ A = 50 ⋅1.75 = 87.5 in / s (⊥ to rB/ A ) vC4 / D4 = ω 4 × rC / D ⇒ vC4 / D4 = ω 4 ⋅ rC / D (⊥ to rC / D ) vC3 / B3 = ω 3 × rC / B ⇒ vC3 / B3 = ω 3 ⋅ rC / B (⊥ to rC / B ) (1) AB = 1.75" AD = 3.55" CD = 2.75" BC = 5.15" BE = 2.5" EC = 4.0" 4 D  77  E C B Velocity Scale 50 in/s b3 e3 o' d'4 o c3 A 3 4 2 D r aC 3 / B3 r aC
4 /D 4 r aB 2 / A2 atC 3 / B3 b'3 e'3 t aB 2 / A 2 at /D C4 4 Acceleration Scale 2000 in/s 2 Solve Eq. (1) graphically with a velocity polygon and locate the velocity of point E3 by image. From the polygon, vC3 / B3 = 65.2 in / s vC4 / D4 = 92.6 in / s and vE3 = 107.8 in / s in the direction shown. Now c' 3 ω3 = vC3 / B3 65.2 = = 12.7 rad / s 5.15 rC / B  78  and ω4 = vC4 / D4 92.6 = = 33.7 rad / s 2.75 rC / D To determine the direction of ω 3 , determine the direction that rC / B must be rotated to be parallel to vC3 / B3 . This direction is clearly clockwise. To determine the direction of ω 4 , determine the direction that rC / D must be rotated to be parallel to vC4 / D4 . This direction is clearly clockwise. Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 50 2 ⋅1.75 = 4375 in / s2 B B in the direction of  rB2 / A2
t t aB2 / A2 = α 2 × rB/ A ⇒ aB2 / A2 = α 2 rB/ A = 1600 ⋅1.75 = 2800 in / s2 (⊥ to rB/ A ) r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 12.72 ⋅ 5.15 = 830.6 in / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 33.72 ⋅ 2.75 = 3123 in / sec 2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with an acceleration polygon and determine the acceleration of point E3 by image. From the acceleration polygon,
t aC3 / B3 = 1563 in / s2 t aC4 / D4 = 4881 in / s2 Then,  79  α3 = α4 = t aC3 / B3 1563 = = 303 rad / s2 5.15 rC / B t aC4 / D4 4881 = = 1775 rad / s2 2.75 rC / D To determine the direction of α 3 , determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly clockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly clockwise. Also aE3 = 5958 in / s2 Problem 2.12 Resolve Problem 2.11 if ω2 = 50 rad/s CCW and α 2 = 0 . Position Analysis Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B and then C. Then locate E. Velocity Analysis: The velocity analysis is similar to that in Problem 2.18. v B3 = v B2 = v B2 / A2 vC3 = vC4 = vC4 / D4 = v B3 + vC3 / B3 Now, v B2 / A2 = ω 2 × rB/ A ⇒ v B/ A = ω 2 ⋅ rB/ A = 50 ⋅1.75 = 87.5 in / s (⊥ to rB/ A ) vC4 / D4 = ω 4 × rC / D ⇒ vC4 / D4 = ω 4 ⋅ rC / D (⊥ to rC / D ) vC3 / B3 = ω 3 × rC / B ⇒ vC3 / B3 = ω 3 ⋅ rC / B (⊥ to rC / B ) Solve Eq. (1) graphically with a velocity polygon and locate the velocity of point E3 by image. From the polygon, vC4 / D4 = 103.1 in / s and vE3 = 116 in / s  80 (1) in the direction shown. Now ω3 =
and vC3 / B3 88.8 = = 17.2 rad / s 5.15 rC / B ω4 = vC4 / D4 103.1 = = 37.5 rad / s 2.75 rC / D To determine the direction of ω 3 , determine the direction that rC / B must be rotated to be parallel to vC3 / B3 . This direction is clearly counterclockwise. To determine the direction of ω 4 , determine the direction that rC / D must be rotated to be parallel to vC4 / D4 . This direction is clearly counterclockwise.  81  E C B 3 4 Velocity Scale 50 in/s c3 A 2 D o' d'4 o g3 b3
r aB 2/ A 2 b'3
r aC 3 / B3 Acceleration Scale 1000 in/s 2
r aC 4 / D 4 t aC
3 / B3 e'3 at C 4 /D 4 c' 3  82  Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 50 2 ⋅1.75 = 4375 in / s2 B B in the direction of  rB2 / A2
t t aB2 / A2 = α 2 × rB/ A ⇒ aB2 / A2 = α 2 rB/ A = 0 ⋅1.75 = 0 r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 17.24 2 ⋅ 5.15 = 1530 in / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 37.492 ⋅ 2.75 = 3865 in / s2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with an acceleration polygon and determine the acceleration of point E3 by image. From the acceleration polygon,
t aC3 / B3 = 2751 in / s2 t aC4 / D4 = 1405 in / s2 Then, α3 = t aC3 / B3 2751 = = 534 rad / s 2 5.15 rC / B at α 4 = C4 / D4 = 1405 = 511 rad / s2 2.75 rC / D To determine the direction of α 3 , determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly clockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly clockwise.  83  Also aE3 = 2784 in / s2 Problem 2.13 In the mechanism shown below, link 2 is rotating CW at the rate of 180 rad/s. Write the appropriate vector equations, solve them using vector polygons, and a) Determine vB2, vC3, vE3, ω3, and ω4. b) Determine aB2, aC3, aE3, α3, and α4. Link lengths: AB = 4.6 in, BC = 12.0 in, AD = 15.2 in, CD = 9.2 in, EB = 8.0 in, CE = 5.48 in. C Y 3 4 E 2 D A 120˚ X B Position Analysis: Draw the linkage to scale. Velocity Analysis: v B2 = v B2 / A2 = ω 2 × rB2 / A2 ⇒ v B2 = ω 2 rB2 / A2 = 180(4.6) = 828 in / s v B3 = v B2 vC3 = v B3 + vC3 / B3 vC3 = vC4 = vD4 + vC4 / D4 and vD4 = 0  84  C AD = 15.2" DC = 9.2" BC = 12.0" AB = 4.6" EC = 5.48" EB = 8.0" 3 4 E B 2 D A b 2, b 3 120˚ Velocity Polygon 400 in/sec ov a1 , a2 c 3, c 4 e3 Therefore, vC4 / D4 = v B3 + vC3 / B3 Now, v B3 = 828 ft / s (⊥ to rB/ A) vC3 / B3 = ω 3 × rC / B (⊥ to rC / B ) vC4 / D4 = ω 4 × rC / D (⊥ to rC / D ) Solve Eq. (1) graphically with a velocity polygon. The velocity directions can be gotten directly from the polygon. The magnitudes are given by: vC3 / B3 = 583 in / s ⇒ ω 3 = vC3 / B3 583 = = 48.6 rad / s 12 rC / B (1) From the directions given in the position and velocity polygons ω 3 = 48.6 rad / s CCW
Also,  85  vC4 / D4 = 475 rad / s ⇒ ω 4 = vC4 / D4 475 = = 51.6 rad / s 9.2 rC / D From the directions given in the position and velocity polygons ω 4 = 51.6 rad / s CW
To compute the velocity of E3, vE3 = v B3 + vE3 / B3 = vC3 + vE3 /C3 Because two points in the same link are involved in the relative velocity terms vE3 / B3 = ω 3 × rE / B (⊥ to rE / B ) and vE3 /C3 = ω 3 × rE /C (⊥ to rE /C ) Equation (2) can now be solved to give vE3 = 695 in / s Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (1) (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 180 2 ⋅ 4.6 = 149, 000 in / s2 B B in the direction of  rB2 / A2
t t aB2 / A2 = α 2 × rB/ A ⇒ aB2 / A2 = α 2 rB/ A = 0 ⋅ 4.6 = 0 r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 48.992 ⋅12 = 28,800 in / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 50.4 2 ⋅ 9.2 = 23,370 in / s2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D )  86  Solve Eq. (2) graphically with an acceleration polygon and determine the acceleration of point E3 by image. From the acceleration polygon,
t aC3 / B3 = 96,880 in / s2 t aC4 / D4 = 9785 in / s2 Then, α3 = α4 = t aC3 / B3 96876 = = 8073 rad / s 2 12 rC / B t aC4 / D4 9785.5 = = 1063.6 rad / s2 9.2 rC / D To determine the direction of α 3 , determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly clockwise.  87  To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly clockwise. Also aE3 = 123, 700 in / s2 and aC3 = 149, 780 in / s2 Problem 2.14 The accelerations of points A and B in the coupler below are as given. Determine the acceleration of the center of mass G and the angular acceleration of the body. Draw the vector representing aG from G. B  63˚ G aA = 7000 in/s 2 a B = 7000 in/s 2
AG = 1.5" BG = 1.5" AB = 2.8" aA 50˚ 22˚ A Acceleration Analysis: Draw the accelerations of points A and B on an acceleration polygon. Then locate the accleration of point G by image.
t For the angular acceleration of the body, resolve the acceleration aB/ A in terms of components along and perpendicular to rB/ A . The tangential component is perpendicular to rB/ A . t t aB/ A = α × rB/ A ⇒ aB/ A = α rB/ A aB and at α = r B/ A = 3141 = 1122 rad / s2 2.8 B/ A  88  B a' G aG o' A g' Acceleration Scale 2000 in/s 2 at A B/ b' To determine the direction of α , determine the direction that rB/ A must be rotated to be parallel to t aB/ A . This direction is clearly clockwise. Also aG = 6980 in / s2 in the direction shown.  89  Problem 2.15 Crank 2 of the pushlink mechanism shown in the figure is driven at a constant angular velocity ω2 = 60 rad/s (CW). Find the velocity and acceleration of point F and the angular velocity and acceleration of links 3 and 4. C Y AB = 15 cm BC = 29.5 cm CD = 30.1 cm AD = 7.5 cm BE = 14.75 cm EF = 7.5 cm 3 E F 4 A 30˚ 2 B Position Analysis: D X Draw the linkage to scale. First located the pivots A and D. Next locate B, then C, then E, then F. Velocity Analysis: vA2 = vB3 = vB2 / A2 =ω 2·rB2 / A2 vC3 = vB3 + vC3 / B3 vC3 = vC4 = vC4 / D4 =ω 4·rC / D Now, vB3 = 9 m / s (^ to rB / A) vB2 = ω 2 rB2 / A2 = 60 (0.15) = 9 m / s (1) vC3 / B3 =ω 3·rC / B (^ to rC / B) vC4 =ω 4·rC / D (^ to rC / D) Solve Eq. (1) graphically with a velocity polygon. The velocity directions can be gotten directly from the polygon. The magnitudes are given by: vC3 / B3 = 12.82 m / s ω 3 = Using velocity image, vF3 = 4.94 m / s  90 vC3 / B3 12.82 = = 43.45 rad / s rC / B 0.295 in the direction shown. C 5 cm 3 E Acceleration Polygon 100 m/s 2 a r 2 /A2 B f '3
r aC 3 /B 3 4 F A D b'2 , b'3 B 30° 2 Velocity Polygon 2 m/s b2 , b3 c 3 , c4 o' ar 4 /D4 C c' 3 , c'4 atC3 /B3
t aC4 /D 4 a2 o From the directions given in the position and velocity polygons ω 3 = 43.45 rad / s CW
Also, vC4 / D4 = 11.39 m / s ω4 = vC4 / D4 11.39 = = 37.84 rad / s rC / D 0.301 From the directions given in the position and velocity polygons ω 4 = 37.84 rad / s CW
Acceleration Analysis: aB2 = aB3 = ar B2 / A2 =ω 2· ( 2·rB / A) ω  91  aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3 atC4 / D4 + arC4 / D4 = ar B2 / A2 + at B2 / A2 + arC3 / B3 + atC3 / B3 Now, arC4 / D4 = ω 4 2 rC / D = 37.842 0.301 = 430.99 m / s2 in the direction opposite to rC/ D ) arC3 / B3 = ω 3 2 rC / B = 43.452 0.295 = 556.93 m / s2 in the direction opposite to rC/ B ) ar B2 / A2 = ω 2 2 rB / A = 602 0.15 = 540 m / s2 in the direction opposite to rB/ A ) atC3 / B3 = α 3·rC / B α 3 = atC4 / D4 = α 4·rC / D α 4 = atC3 / B3 (^ to rC / B) rC / B atC4 / D4 (^ to rC / D) rC / D (2) Solve Eq. (2) graphically with an acceleration polygon. The acceleration directions can be gotten directly from the polygon. The magnitudes are given by: α3 =
Also, atC3 / B3 142.79 = = 484 rad / s2 CW 0.295 rC / B α4 = atC4 / D4 41.01 = = 136 rad / s2 CCW 0.301 rC / D Using acceleration polygon, aF3 = 256 m / s2 in the direction shown. Problem 2.16 For the straightline mechanism shown in the figure, ω2 = 20 rad/s (CW) and α2 = 140 rad/s2 (CW). Determine the velocity and acceleration of point B and the angular acceleration of link 3. B DA = 2.0" AC = 2.0" AB = 2.0" 2 15
o D A 3 C 4  92  Velocity Analysis: vA2 = vA2 / D2 = vA3 =ω 2·rA2 / D2 vC3 = vC4 = v A3 + vC3 / A3 Now, v A3 = ω 2 rA2 / D2 = 20 ⋅ 2 = 40 in / s (⊥ to rA2 / D2 ) vC3 in horizontal direction vC3 / A3 = ω 3 × rC3 / A3 ⇒ vC3 / A3 = ω 3 rC3 / A3 (⊥ to rC3 / A3 ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B3 = 77.3 in / s Also, vC3 / A3 = 40 in / s or (1) ω3 =
Also, vC3 / A3 40 = = 20 rad / s CCW 2 rC3 / A3 vC3 = 20.7 in / s Acceleration Analysis: aC3 = aC4 = a A3 + aC3 / A3
t r r aC3 = a A2 / D2 + a t 2 / D2 + aC3 / A3 + aC3 / A3 A (2) Now, aC3 in horizontal direction a r 2 / D2 = ω 2 × (ω 2 × rA2 / D2 ) ⇒ a r 2 / D2 = ω 2 2 ⋅ rA/ D = 20 2 ⋅ 2 = 800 in / s2 A A in the direction opposite to rA/ D a t 2 / D2 = α 2 × rA2 / D2 ⇒ a t 2 / D2 = α 2 ⋅ rA/ D = 140 ⋅ 2 = 280 in / s2 (⊥ to rA/ D ) A A
r r a C3 / A3 = ω 3 × (ω 3 × rC3 / A3 ) ⇒ a C3 / A3 = ω 3 2 ⋅ rC3 / A3 = 20 2 ⋅ 2 = 800 in / s2 in the direction opposite to rC3 / A3  93  t t aC3 / A3 = α 3 × rC3 / A3 ⇒ aC3 / A3 = α 3 ⋅ rC3 / A3 (⊥ to rC3 / A3 ) Solve Eq. (2) graphically with a acceleration polygon. From the polygon, aB3 = 955 in / s2 aC3 / A3 = 280 in / s2 Also, α3 = t aC3 / A3 280 = = 140 rad / s2 CCW 2 rC3 / A3 B D 2 A 3 C 4 o v C3 c3 , c 4 Acceleration Polygon 400 in/s 2 c' 3
t a C3 / A 3 v A3 vC 3 / A 3 o' a C3 a A2 / D 2 a A2/D 2 a C 3 / A3
r t r v B3 a2 , a3 '' a2 , a3 a B3 v B3 / A3 Velocity Polygon 20 in/s b3 ' b3  94  Problem 2.17 For the data given in the figure below, find the velocity and acceleration of points B and C. Assume vA = 20 ft/s, aA = 400 ft/s2, ω2 = 24 rad/s (CW), and α 2 = 160 rad/s2 (CCW). C vA 90˚ A 15 o aA Position Analysis Draw the link to scale α2 AB = 4.05" AC = 2.5" BC = 2.0" B ω2 Velocity Polygon 5 ft/sec Acceleration Polygon o' 80 ft/sec 2 b'2 a tB2 /A 2 a'2
r a B2 /A 2 a2 c2 b2 c' 2 o Velocity Analysis: v B2 = v A2 + v B2 / A2 Now,  95 (1) v A2 = 20 ft / sec in the positive vertical direction v B2 / A2 = ω 2 × rB/ A ⇒ v B2 / A2 = ω 2 ⋅ rB/ A = 24 ⋅ 4.05 = 97.2in / sec v B2 / A2 = 8.1 ft / sec(⊥ to rB/ A ) in the positive vertical direction Solve Eq. (1) graphically with a velocity polygon. From the polygon, v B2 = 11.9 ft / sec Also, from the velocity polygon, vC2 = 15.55 ft / sec in the direction shown Acceleration Analysis:
t r aB2 = a A2 + aB2 / A2 = a A2 + aB2 / A2 + aB2 / A2 (2) Now, a A2 = 400 ft / sec 2 in the given direction
t aB2 / A2 = α 2 ⋅ rB/ A = 160 ⋅ 4.05 = 648 in / sec 2 = 54 ft / sec 2 r aB2 / A2 = ω 2 2 ⋅ rB/ A = 24 2 ⋅ 4.05 = 2332 in / sec 2 = 194 ft / sec 2 Solve Eq. (2) graphically with an acceleration polygon. From the polygon, aB2 = 198.64 ft / sec 2 in the direction shown. Determine the acceleration of point C by image. From the acceleration image, aC2 = 289.4 ft / sec 2 in the direction shown.  96  Problem 2.18 In the mechanism shown below, link 2 is turning CCW at the rate of 10 rad/s (constant). Draw the velocity and acceleration polygons for the mechanism, and determine aG3 and α4. C 3 G 4 B 2 A 90˚ ω 2 , α2 D AB = 1.0" BC = 2.0" BG = 1.0" CD = 3.0" AD = 3.0" Position Analysis Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B and then C. Then locate G. Velocity Analysis: v B3 = v B2 = v B2 / A2 vC3 = vC4 = vC4 / D4 = v B3 + vC3 / B3 Now, v B2 / A2 = ω 2 × rB/ A ⇒ v B/ A = ω 2 ⋅ rB/ A = 10 ⋅ 2 = 20 in / s (⊥ to rB/ A ) vC4 / D4 = ω 4 × rC / D ⇒ vC4 / D4 = ω 4 ⋅ rC / D (⊥ to rC / D ) vC3 / B3 = ω 3 × rC / B ⇒ vC3 / B3 = ω 3 ⋅ rC / B (⊥ to rC / B )  97 (1) Solve Eq. (1) graphically with a velocity polygon. From the polygon, vC3 / B3 = 14.4 in / s vC4 / D4 = 13.7 in / s in the direction shown. Now ω3 =
and vC3 / B3 11.4 = = 5.7 rad / s 2 rC / B ω4 = vC4 / D4 13.7 = = 4.57 rad / s 3 rC / D To determine the direction of ω 3 , determine the direction that rC / B must be rotated to be parallel to vC3 / B3 . This direction is clearly clockwise. To determine the direction of ω 4 , determine the direction that rC / D must be rotated to be parallel to vC4 / D4 . This direction is clearly counterclockwise. The velocity of point G3 vG3 = v B3 + vG3 / B3 = ω 3 × rG/ B = ω 3 ⋅ rG/ B = 5.7 ⋅1 = 5.7 in / s Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now,  98  a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 10 2 ⋅1 = 100 in / s2 B B in the direction of  rB2 / A2
t t aB2 / A2 = α 2 × rB/ A ⇒ aB2 / A2 = α 2 rB/ A = 0 r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 5.72 ⋅ 2 = 64.98 in / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 4.572 ⋅ 3 = 62.66 in / sec 2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically for the accelerations.  99  From the acceleration polygon,
t aC3 / B3 = 38 in / s2 t aC4 / D4 = 128 in / s2 Then, α3 = α4 = t aC3 / B3 38 = = 19 rad / s2 2 rC / B t aC4 / D4 128 = = 42.67 rad / s2 3 rC / D To determine the direction of α 3 , determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly counterclockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly clockwise. Determine the acceleration of point G3
t r aG3 = aB3 + aG3 / B3 = aB2 + a G3 / B3 + aG3 / B3 t t aG3 / B3 = α 3 × rG/ B ⇒ aG3 / B3 = α 3 ⋅ rG/ B = 19 ⋅1 = 19 in / sec 2 r r a G3 / B3 = ω 3 × (ω 3 × rG/ B ) ⇒ a G3 / B3 = ω 3 2 ⋅ rG/ B = 5.72 ⋅1 = 32.49 in / sec 2 From the acceleration polygon, aG3 = 116 in / s2 Problem 2.19 If ω2 = 100 rad/s CCW (constant) find the velocity and acceleration of point E. A ω2 2 B 3 C 70˚ E 4 AB = 1.0" BC = 1.75" CD = 2.0" DE = 0.8" AD = 3.0" D  100  Position Analysis Draw the linkage to scale. Start by locating the relative positions of A and D. Next locate B and then C. Then locate E. Velocity Analysis: v B3 = v B2 = v B2 / A2 vC3 = vC4 = vC4 / D4 = v B3 + vC3 / B3 Now, v B2 / A2 = ω 2 × rB/ A ⇒ v B/ A = ω 2 ⋅ rB/ A = 100 ⋅1 = 100 in / s (⊥ to rB/ A ) vC4 / D4 = ω 4 × rC / D ⇒ vC4 / D4 = ω 4 ⋅ rC / D (⊥ to rC / D ) vC3 / B3 = ω 3 × rC / B ⇒ vC3 / B3 = ω 3 ⋅ rC / B (⊥ to rC / B ) Solve Eq. (1) graphically with a velocity polygon. (1) From the polygon, vC3 / B3 = 77.5 in / s vC4 / D4 = 71 in / s in the direction shown.  101  Now ω3 =
and vC3 / B3 77.5 = = 44.29 rad / s rC / B 1.75 ω4 = vC4 / D4 71 = = 35.5 rad / s 2 rC / D To determine the direction of ω 3 , determine the direction that rC / B must be rotated to be parallel to vC3 / B3 . This direction is clearly clockwise. To determine the direction of ω 4 , determine the direction that rC / D must be rotated to be parallel to vC4 / D4 . This direction is clearly counterclockwise. The velocity of point E3 vE4 = vD4 + vE4 / D4 = ω 4 × rD/ E = ω 4 ⋅ rD/ E = 35.5 ⋅ 0.8 = 28.4 in / s Acceleration Analysis: aB3 = aB2 = aB2 / A2 aC3 = aC4 = aC4 / D4 = aB3 + aC3 / B3
t t t r r r a C4 / D4 + aC4 / D4 = aB2 / A2 + aB2 / A2 + aC3 / B3 + aC3 / B3 (2) Now, a r 2 / A2 = ω 2 × (ω 2 × rB/ A ) ⇒ a r 2 / A2 = ω 2 2 ⋅ rB/ A = 100 2 ⋅1 = 10, 000 in / s2 B B in the direction of  rB2 / A2
t t aB2 / A2 = α 2 × rB/ A ⇒ aB2 / A2 = α 2 rB/ A = 0 r r a C3 / B3 = ω 3 × (ω 3 × rC / B ) ⇒ a C3 / B3 = ω 3 2 ⋅ rC / B = 44.292 ⋅1.75 = 3432.8 in / s2 in the direction of  rC / B
t t aC3 / B3 = α 3 × rC / B ⇒ aC3 / B3 = α 3 ⋅ rC / B (⊥ to rC / B ) r r a C4 / D4 = ω 4 × (ω 4 × rC / D ) ⇒ a C4 / D4 = ω 4 2 ⋅ rC / D = 35.52 ⋅ 2 = 2520.5 in / sec 2 in the direction of  rC / D
t t aC4 / D4 = α 4 × rC / D ⇒ aC4 / D4 = α 4 ⋅ rC / D (⊥ to rC / D ) Solve Eq. (2) graphically with acceleration.  102  From the acceleration polygon,
t aC3 / B3 = 3500 in / s2 t aC4 / D4 = 10, 900 in / s2 Then, α3 = α4 = t aC3 / B3 3500 = = 2000 rad / s2 1.75 rC / B t aC4 / D4 10, 900 = = 5450 rad / s2 2 rC / D To determine the direction of α 3 , determine the direction that rC / B must be rotated to be parallel to t aC3 / B3 . This direction is clearly counterclockwise. To determine the direction of α 4 , determine the direction that rC / D must be rotated to be parallel to t aC4 / D4 . This direction is clearly clockwise. Determine the acceleration of point E4
t aE4 = aD4 + aE4 / D4 = a r 4 / D4 + aE4 / D4 E  103  t t aE4 / D4 = α 4 × rE / D ⇒ aE4 / D4 = α 4 ⋅ rE / D = 5450 ⋅ 0.8 = 4360 in / sec 2 a r 4 / D4 = ω 4 × (ω 4 × rE / D ) ⇒ a r 4 / D4 = ω 4 2 ⋅ rE / D = 35.52 ⋅ 0.8 = 1008.2 in / sec 2 E E From the acceleration polygon, aE4 = 4600 in / s2 Problem 2.20 Draw the velocity polygon to determine the velocity of link 6. Points A, C, and E have the same vertical coordinate. 2
1 ω2 = B rad 6s 3 AB = 1.80" BC = 1.95" CD = 0.75" DE = 2.10" 4 E C 5 D 6 A 50˚ Velocity Analysis: v B3 = v B2 = v B2 / A2 vC4 = vC3 = v B3 + vC3 / B3 v D5 = vD3 vE5 = vE6 = vD5 + vE5 / D5 Now, v B2 / A2 = ω 2 × rB2 / A2 ⇒ v B2 / A2 = ω 2 ⋅ rB2 / A2 = 6 ⋅1.8 = 10.8 in / s (⊥ to rB2 / A2 ) vC3 is in the vertical direction. Then, vC3 / B3 = ω 3 × rC3 / B3 ⇒ vC3 / B3 = ω 3 ⋅ rC3 / B3 (⊥ to rC3 / B3 ) Solve Eq. (1) graphically with a velocity polygon. From the polygon, using velocity image, vD3 = 18.7 in / s (2) (1)  104  B 2 3 A C E 4 D e5 o 5 6 b3 Velocity Polygon c3 d3 Now, vE5 in horizontal direction vE5 / D5 = ω 5 × rE / D ⇒ vE5 / D5 = ω 5 ⋅ rE / D (⊥ to rE / D ) Solve Eq. (2) graphically with a velocity polygon. From the polygon, using velocity image, vE5 = vE6 = 8.0 in / s d5 10 in/sec  105  ...
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 Winter '11
 Shudong

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