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Given Information: Gear A (80 mm diameter) has a constant rotational speed of 145 of 145 rpm clockwise.
Required: What is the rotational speed and direction of gears of gears B (200 mm diameter) and C (120mm diameter)?
Solution: Consider the contact point 1 where gears A and B meet
80 58 145 200 Similarly, consider the contact point 2 where gears B and C meet
96.67 58 200 120 Answer:
96.58 6 7
Problem 2.2
Given Information: The disk shown has a radius of 4 of 4 inches [101.6 mm] and rotates about the fixed point O at a constant rate of 75 of 75 rpm clockwise.
Required: Determine the magnitude and direction of the of the velocities of points of points A and B.
Solution: The velocity of disk of disk at point A is
2 75 75 60 ̂ ̂ ̂ ̂ 0 0 2 75 60 ̂ ̂ ̂ ̂ 0 0 75 30 ̂ ̂ 4̂3̂ 75 3023.56̂31.42̂ ⁄ The velocity оf disk f disk at point B is
75 30 ̂ 70° ̂ 70° ̂ 470°̂ 3 4 470° 70° ̂ 53.09̂ 10.76̂ ⁄ Answer:
39.27 ; 53.16°↗ 54.17 ; 11.46° ↘
Problem 2.3
Given Information: The disk shown has a radius of 4 of 4 inches [101.6 mm] and rotates about the fixed point O at a constant rate of 75 of 75 rpm clockwise.
Required: Determine the magnitude and direction of the of the normal and tangential accelerations of point of point A.
Solution:
, ⁄
The normal acceleration of point A is
, where
is the
unit vector that indicates the position of point of point A w.r.t the fixed point O. This unit vector is equal to
⁄⁄ 4̂ 533̂ 0.88̂ 0.66̂ 23.56̂31.42̂ ⁄ ⇒ 39.27⁄ . , ⁄ 0.8̂ 0.6̂ 246.7̂185.06̂ ⁄ Moreover, from problem 2.2, . Thus, the normal acceleration of point of point A is
In addition, the tangential acceleration of point of point A is equal to zero because the disk rotates about point with constant angular velocity. That is, the angular acceleration of the of the disk is equal to zero. Answer:
О
, 308.4 ⁄ @0 36.88° ↘ ,
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Given Information: At the instant shown, the disk shown has an angular velocity of 10 of 10 2 rad/s clockwise and is accelerating at a rate of 1.4 of 1.4 rad/s .
Required: Determine the velocity and acceleration of point of point B.
Solution: The acceleration of point of point B w.r.t fixed point
О is
, ,
where is the unit vector that indicates the position of point of point B w.r.t the fixed point О, and is equal to
3 70°° ̂ 470°6.̂ 9 703 370 0.199̂ 0.98̂ Furthermore, the velocity of point of point B w.r.t the fixed point
О is
equal to
10 470°̂ 3 370 3 70°° ̂ 67.7̂ 13.7̂ ⁄ ⇒ 68.98 ⁄ ⁄ Substituting the speed and unit vector of point of point B w.r.t the fixed point
О into
the acceleration equation
1.4 1.37̂ 6.76̂ 68.6.9980⁄ 0.199̂0.98̂ 9.464̂ 1.918̂ 137.23̂ 675.8̂ ⁄ 127.77 77̂ 677.72̂ ⁄ 68.98 ⁄ @11. 4 6° ↘ 689.7⁄ @ 1185.3° Answer:
Problem 2.5
Given Information: At the instant shown the 5 inch radius gear A rotates counterclockwise at a rate of 8 of 8 rad/s and is decelerating at 2 a rate of 2.3 of 2.3 rad/s . Gear A drives the 8.4 inch radius gear B. The 5.5 inch radius pulley C is on the same shaft as gear B. A belt around pulley C drives pulley D, which has a 9 inch radius.
Required: What is the rotational velocity and acceleration of pulley of pulley D?
Solution: Let point 1 be intersection of gears of gears A and B. The speed of gear of gear A at point 1 is equal to the speed of gear of gear B at the same point
r
where represents the radius of each of each pulley. Moreover, pulley C rotates with the same speed as gear B. Substituting this speed and the radii values yields
8.54 8⁄ 4.762⁄
Furthermore, because a belt connects pulleys C and D, the pulleys have the same speed as the belt at all points of contact. of contact. Thus
5. 5 9 4.762⁄ 2.91 ⁄
and by inspection, pulley D rotates in the clockwise direction. Similarly, the magnitudes of the of the tangential accelerations of gears of gears A and B at contact point 1 are equal
, , 2.3 ⁄ 8.54 1.37⁄
Also, because pulley C is attached to gear B, they both have the same angular acceleration. Similarly, pulleys C and D have the same angular acceleration because the belt connects them
1.37⁄ Note that the negative sign means pulley D is decelerating. Answer:
1.2.37 937 1 ⁄⁄
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Given Information: The hexagon shown is rotating about point O at a speed of 5.6 of 5.6 2 rad/s clockwise and accelerating at a rate of 2.7 of 2.7 rad/s .
Required: Determine the point along the perimeter that will have the highest velocity magnitude. Calculate the velocity and acceleration at that point.
Solution: Let point Q be the point along the perimeter of the hexagon that has the highest speed. Because the speed of any point along the perimeter of the of the hexagon is directly proportional to its angular speed and the distance to that point from pivot point О, point Q will be located at the maximum distance from point O. That is, at either of the opposite vertices of the hexagon
2180° 120° 22180° 6 2 6 Applying the law of cosines of cosines to find the distance between opposite vertices,
120° 4 120° 4 120° 90° 90° 6.928 To find the distance from point
О to
point Q
⁄ 2 6.928 2 7.211 ⁄ ⁄ ⁄ ̂ ̂ 73.89° ⁄ 7.211 73.89° 9° ̂ 73.89° 9° ̂ 2 6.93̂
To calculate the velocity of point of point Q, the position vector
where
and the velocity of point of point Q is Q is
. As a result,
is required
⁄ 5.6 7.21 cos cos73.89° 9°̂ sin sin73.89° 9° ̂ 38.80̂11.21̂ , , ⁄ cos cos sin ⁄ / cos73.89° sin73.89° ../ 2. 2.7 2̂ 6.93̂ 44.21̂222.9̂ 40.4 / @ 16.16.11° ↘ 227.2 / @ @ 78.7°7° ↙
To find the acceleration of point of point Q
Answer:
Problem 2.7
Given Information: In the mechanism shown the 8 inch diameter gear A turns at a constant speed of 1100 of 1100 rpm clockwise and drives the 20 inch diameter gear B. Attached to gear B is a 15 inch diameter disk.
Required: Determine the angular velocity of the of the connecting rod CD and the velocity of slider of slider D for the instant shown.
Solution: First, determine the angular speed of gear of gear B
1100 20 8 440
Then, determine the velocity of point of point C
2 440 440 60 ̂ ̂ 7. 5 cos cos 25° 25° sin sin 25° 25° 313.2̂ 146.1 ̂ / sin 7.5 cos3025° 10 34.05°
From trigonometry,
can be found as follows
Then, determine the velocity of the of the slider at point D:
coŝ sin̂ 30 cos34.05°̂̂ sin 34.05° ̂ 313.2̂146. 1313. ̂ /216. 8 ̂ 146. 1 24.86 Because the slider moves in the horizontal direction, the vertical component of the of the velocity is zero
0 146.1 24.86 5.88 / 313. 313.2 16. 16 .8 313.2 16.8 5.88 412 / Answer:
5.412 88 / / / / @180 @ 180°°
Problem 2.8
Given Information: Continue Example Problem 2‐4 to determine the rotation speed of link of link 4 [BO4].
Solution: From problem 2‐4
cos 24
94.71
v B
4
v B r 4
vB
94.71
103.7
cos 24
103.7
in s
in s
7in Answer:
4
14.8
rad s
CW
Problem 2.9
Given Information: The slider‐crank mechanism shown in has a constant crank (link OA) rotation speed of 850 of 850 rpm clockwise.
Required: Determine the velocity and acceleration of the of the piston (point B) at the instant shown.
Solution: First, find the velocity of point of point A
cos ̂ sisin ̂ cos 68° sin 2 3 sin68° 850 288.0160/
where
, and
Substituting the speed of point of point A to obtain its velocity
̂̂ sin ̂ 288. 0 1107. / /9̂cos cos 68° 68° sin 68° 68° 267.04 /
Then, find the velocity of point of point B
/ / / cos ̂ sin sin̂ 68° ̂ ̂ sinsin68° ̂ 107.9 ̂267. 0 4107. ̂ 9/ 7.4 2̂ 267. 26 87.043 cos 3 Because the piston is constrained to move in the vertical direction only
0 107.9 7. 7.42 14.54 / Substituting the angular speed into the velocity of the of the piston equation
267.04 3 14.544 ̂ 310.7̂ / Then, find the acceleration of the of the piston using the relative accelerations method
/ , , /, /, / / /
89.01/ 89.01/ 3̂ ̂ 14.54̂ / 68° sin 68° 14.5 4 / 8 cos s in 68°̂ ̂ sin68° ̂ ̂ 23134. 8 87.cos 42 1569 3
Because the piston is constrained to move in the vertical direction only, the acceleration in the horizontal direction is zero
As a result, the acceleration is
0 23134.8 7.42 3117.9 / ̂ 1569 3 3 3117. 9 7784.7̂/
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7784. 310 / / @270 @ 270° ° 7/ @ 90°
Problem 2.10
Given Information: The slider‐crank mechanism is decelerating. At the instant shown the crank is rotating counterclockwise at a speed of 640 of 640 rpm. It is decelerating at a constant rate and comes to a complete stop in 1.2 minutes.
Required: Determine the velocity and acceleration of the of the piston at the instant shown.
Solution: First, find the velocity of point of point A at the given instant
cos ̂ sin in ̂ cos 68° sin 2 3 sin68° 640 60 216.9 /
where
, and
Substituting the speed of point of point A to obtain its velocity
̂ ̂ 216. 9 / cos 68° sin 68° 81.23 ̂ 201.1 ̂ /
Then, find the velocity of point of point B
/ / / cos ̂ sin in ̂ ̂ ̂ 8 68° 81.23 ̂201. 181. ̂ /23 7.42 ̂ 201. 1 cos cos 68° sin sin 68° 68° 3 ̂ Because the piston is constrained to move in the vertical direction only
0 81.23 7. 7.42 10.95 / Substituting the angular speed into the velocity of the of the piston equation
201. 201.1 3 3 10.95 ̂ 234̂ / The angular acceleration of arm of arm OA is required to find the acceleration of the of the piston. Assuming that the mechanism is decelerating uniformly, the angular acceleration of arm of arm OA can be calculated as
2 0 640 Δ 1.2 60 0.931 / And the acceleration of the of the piston is
/ , , /, /, / / / / 3̂ 67.02 / 0.931 / 0.931/ ̂ 67.02/ 3 10. 9 5 / 10. 95 / 8 ̂ cos68° ̂̂ sin68° ̂ 68° ̂ 13115. 8 4 7.cos cos42 68° 68°̂ sin sin892. 68° 892.2 3 Because the piston is constrained to move in the vertical direction only, the acceleration in the horizontal direction is zero
As a result, the acceleration is
0 13115.4 7.42 1767.6 / ̂ 892. 2 3 1767. 6 4411̂/
Answer:
4411 234 / @ 90° / @ 90°
Problem 2.11
Given Information: The spool shown is unwinding such that point O moves straight down. At the instant shown the spool has an angular velocity of 2 of 2 rad/s and an angular acceleration of 2 3.5 rad/s (both counterclockwise).
Required: Find the acceleration of point of point O and point A.
Solution: First, find the acceleration of the of the center of the of the spool using the relative accelerations method
/ /, /, / / ̂ / / / / / 2̂ 2 / 2 ̂ 2 / / 2 / 3.5/ 2̂ 7 7 / Then, find the acceleration of point of point A noting that the center of the of the spool moves vertically only
/ /, /, / / / / / 3̂ 7̂ / 2/ / 2 2/ 3.5 /̂ ̂3̂ 10.55 19 / 7 / @ 90° 21.71 / @ 61. 61.1° ↗ Answer:
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Given Information: The disk shown rotates about point O at a constant speed of 45 of 45 rpm. Connecting rod AB is attached to a rack that turns the 6 inch radius gear.
Required: What is the rotational speed and rotational acceleration of the of the gear at the instant shown?
Solution:
sin40° sin 24 12.37°8
First, find the angle
using the law of sines of sines
Next, find the velocity of the of the linkage point B
/ / / 45 260 / 45 / 8cos40°̂ sin sin40° ̂ ̂ ̂ 24 cos 12. 3 7° sin s in 12. 3 7° 24.23 5.14 ̂ 28.9 23.44 ̂
Because the rack is constrained to move in the horizontal direction only
and
0 28.9 23.44 1.23 / 1.23̂ 24. 2 330. 5.5714̂/
Thus, the rotational speed of the of the gear at the instant shown is
30.567/ 5.095 /
Similarly, to find the angular acceleration of the of the gear, determine the acceleration of linkage of linkage point B
/ , /, /, / / / 4.71/ / 4. 4.71/ 8cos40°̂ ̂ sin40° ̂̂ 1.23/ 1. 2 3 / 24 cos 12. 3 7° sin 12. 3 7° ̂ ̂ 24 cos 12. 3 7° sin s in 12. 3 7° ̂ 171.84 5.14 106.38 23.44 ̂ 0 106.38 23.4 4 4.54 / 171. 171 148. .84 5. 5 5.̂/ 144.544̂
Because the rack is constrained to move in the horizontal direction only
and
Finally, determine the angular acceleration of the of the gear noticing that
, 148.65 / 24.75/ 5/ Answer:
24. 5.09575 / / /
Problem 2.13
Given Information: The linkage mechanism shown is driven by link O2A, which rotates at a constant speed of 90 of 90 rpm clockwise.
Required: Determine the rotational velocity and acceleration of link of link O4B in the position shown when the drive link is vertical.
Solution: First, find the velocity of point of point A
/ 2 2̂ 90 90 18.85 ̂60/ Then, find the internal angles of the of the mechanism at the given instant
6 2 2√ 1010 Using the law of sines: of sines:
and
sin sin180° 2 4 37.2√ 8101°0 180° 2 22 104.5° tan 6 18.43° 3 7° 56.13° 180° 19.37°
Then, find the velocity of linkage of linkage AB at point B
/ / ̂ ̂ 4 18.85̂ / cos cos 19. 3 7° 7° sin sin 19. 3 7° 7° 18.85 1.33 ̂ 3.8 ̂ Eq. (1)
Moreover, the velocity of point of point B can be found from linkage BO4
/ / / 4 cos cos56.13° 3° ̂sin sin56.13° 3° ̂ 3.32/ ̂ 2.23 23/̂ Eq. (2)
Setting Eq. (1) equal to Eq. (2) in their respective directions
and
18.8585 1. 1.33̂ 3.32/̂ 1.33 3.32/ 18.85 3.8 ̂ 2.23/ ̂ 3.8 2.23/ 0 / 4.6 / 2.71 /
Solving for the system of equations of equations yields
To find the angular acceleration of linkage of linkage BO4, first find the acceleration of point of point B
/ , /, /, / / / 9.425 / 18.85 ̂/ 2.706/ 2.706/ ̂ cos19.37°̂ sin19.37° ̂̂ 4 4 4 cos cos 19. 3 7° 7° sin sin 19. 3 7° 7° ̂ 27.63 1.33 187.41 3.77 ̂
Eq. (3) Similarly, find an equation for the acceleration of point of point B from linkage BO4
/, /, / / / / / ̂ sin ̂ / 4.6/ 4. 6 / 4 4 cos cos 56. 1 3° 3° sin 56. 1 3° 3° cos 4 4 47. cos2 56.3.3123° 3°̂ sin sin̂ 56.70. 13° 3°282 8̂ 2.233 ̂ / / Eq. (4)
Setting Eq. (3) equal to Eq. (4) in their respective directions
and
27.63 1.33̂ 47.2 3.32 / ̂ 1.33 3.32/ 74.83 187.41 3.77 ̂ 70.28 2.23/ ̂ 3.77 2.23/ 117.13 / 22.25 / / 4.6 / / / 22.25 /
Solving for the system of equations of equations yields
Answer:
Problem 2.14
Given Information: The linkage mechanism shown in Figure 2.42 is driven by link O2A, which rotates at a constant speed of 90 of 90 rpm clockwise.
Required: Determine the velocity magnitude of the of the midpoint of link of link AB.
Solution: First, find the velocity of the of the midpoint of link of link AB
/ /
where
and
were found in problem 2.13. As a result
and
/ / ̂ 18.85 ̂/ 2. 7 1 / / 2 cos cos 19. 3 7° 7° sin sin 19. 3 7° 7° 17.05̂5.125̂/ 17.055 5.125 25/ 17.80 / / 17.80 / Answer:
Problem 2.15
Given Information: For the linkage shown the drive link O2A rotates at 5 rad/s clockwise and is accelerating at a rate of 1.2 of 1.2 rad/s2.
Required: Determine the rotational velocity and acceleration of link of link O4B in the position shown.
Solution: First, find the velocity and acceleration of point of point A
/ 2̂ / 5 10 ̂/ , , / / ̂ 1.2 / 2̂̂ 5 5 / 10 / 2.4 50̂ / Then, find the internal angles of the of the mechanism at the given instant
6 2 2√ 1010 Using the law of sines: of sines:
and
Then, find the velocity of point of point B
sin sin in180° 2 2 4 37.2√ 8101°0 180° 2 22 104.5° tan 6 18.43° 180° 19.37° 56.13°
/ / 10 ̂/ 4 cos19. 37° 10̂sin1.319.337° ̂̂ 3.8 ̂
Eq. (1)
Moreover, the velocity of point of point B can be found from linkage BO4
/ / / 4 cos cos56.13° 3° ̂sin sin56.13° 3° ̂ 3.32/ ̂ 2.23 23/ ̂ Eq. (2)
Setting Eq. (1) equal to Eq. (2) in their respective directions
and
10 1.33̂ 3.32/ ̂ 1.33 3.32/ 10 3.8 ̂ 2.23/ ̂ 3.8 2.23/ 0
Solving for the system of equations of equations yields
/ 2.44 / 1.44 /
To find the angular acceleration of linkage of linkage BO4, first find the acceleration of point of point B
/ , , /, /, , , / / 2.4̂ 50 50̂ / 2.̂ 706/ ̂ 2. 7 06 / 4 cos 19. 3 7° sin 19. 3 7° 4 cos 25. 19.327°3 1. ̂3sin319.3̂ 7° 59. ̂ 714 3.77 ̂ Eq. (3)
Similarly, find an expression for the acceleration of point of point B from linkage BO4
/, /, / / / / / 4.6 / 4.6/ ̂ 4cos56.13°̂ sin56.13° ̂̂ / 4 4 cos cos 56. 1 3° 3° sin sin 56. 1 3° 3° 47.2 3.32/̂ 70.2828 2.23/ ̂ Eq. (4)
Setting Eq. (3) equal to Eq. (4) in their respective directions
and
25.23 1.33̂ 47.2 3.32 / ̂ 1.33 3.32/ 72.43 59.714 3.77 ̂ 70.28 2.23/ ̂ 3.77 2.23/ 10.57 12. 7 / / 16.73 / / 2.44 / / 16.73 /
Solving for the system of equations of equations yields
Answer:
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Given Information: The linkage mechanism shown is driven by link O2A. At the instant shown the driving link is rotating at 8 radians per second clockwise and accelerating at 2.5 radians per second squared.
Required: Determine the rotational velocity and acceleration of links of links AB and O4B in the position shown.
Solution: First, find the velocity of point of point A at the instant shown
̂ sin ̂ / 8 / 2 cos 50° s in 50° 12.26 ̂ 10.28̂ / Then, find the internal angles of the of the linkage mechanism at the instant shown. Applying the law of cosines of cosines to triangle O2AO4:
2 6 4.22962 66 cos cos50° 50° Applying the law of sines of sines to triangle O2AO4:
6 2 sin4.96 sin50°2 sin sin50° 18°sin Applying the law of cosines of cosines to triangle BAO4:
4 3.5 4.96 53.2213.°5 4.4.96 cos cos 35.1° 4 sin sin
Applying the law of sines of sines to triangle BAO4:
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4 4.96 sin53. 1°82.6sin° 62. 3180° 180 ° ° 90° 27.7
and
Then, determine the velocity of point of point B
/ / / coŝsin̂ 12.26 ̂ 10.28̂ / ̂ 4 4cos cos62.3°°̂ ̂sin sin62.3°° ̂ 12.26 3.55 10.28 1.83 Eq. (1)
The velocity of point of point B can also be calculated from linkage BO4 as follows
/ / cos ̂ sin ̂ 3.5 ̂ cos35.1° ̂̂ sin35.1° ̂ 2.02 2.86 86 Eq. (2)
Setting Eq. (1) equal to Eq. (2) in their respecitve directions
12.26 3.55̂ 2.02 ̂ 3.55 2.02 12.26 10.28 1.83 ̂ 2.86 ̂ 1.83 2.86 10.28
and
Solving for the system of equations of equations yields
Furthermore, substituting
1.4.02335 / / into Eq. (2) to determine the velocity of point of point B yields
2.17786 ̂ 4./25 ̂ 2. 028.54.9925̂ ̂ 12.
Next, find the acceleration of point of point A
, , / 12.26 ̂ 10.̂ 28̂ /̂ 2.5 / 8/ 2 cos 2 cos 50° 50° sin sin 50° 50° 78.41̂ 101.3̂ / Then, find an expression for the acceleration of point of point B
