# ch04 - Solutions to Chapter 4 Exercise Problems Problem 4.1...

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- 1 - Solutions to Chapter 4 Exercise Problems Problem 4.1 Locate all of the instant centers in the mechanism shown below. B 1 2 3 A 30˚ 63˚ AB = 1.5" 4 Position Analysis Draw the linkage to scale. Start by locating point B. Then draw the line on which B must slide and draw a horizontal line on which link 4 must slide. This will locate all of the ponts and directions required for the analysis. Instant Center Locations Locate the obvious instant centers (I 12 ,I 23 34 14 ). Then I 24 is found in a straight forward manner using the procedures given in Section 2.15. To locate I 13 , note that it must lie on the line AB. It also lies on the line through I 14 and I 43 . However, both points are at infinity and, the line through the two ponits lies at infinity. Therefore, the line through AB intersects this line at infinity meaning that I 13 must be at infinity in the direction indicated.

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- 2 - B 1 2 3 A 4 I 12 I 13 I 14 I 23 I 24 I 34 1 2 3 4 I I Problem 4.2 Find all of the instant centers of velocity for the mechanism shown below. 2 3 4 A B C D 50˚ 150˚ BC = CD BD = 3.06"
- 3 - 2 3 4 A B C D 50˚ I 12 I 34 I 23 I 14 I 24 I 13 1 2 3 4

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- 4 - Problem 4.3 In the linkage shown below, locate all of the instant centers. 2 3 4 5 6 55˚ AB = 1.35" BD = 3.9" DE = 0.9" BC = 0.9" CF = 2.0" A B C D E F 3.3" \ Solution 2 3 4 5 6 A B C D E F I 12 I 14 I 1 6 I 2 3 I 3 4 I 3 5 I 56 I 13 I 24 I 15 I 36 I 4 5 I 25 I 26 I 46 3 5 1 2 6 4 Problem 4.4 Find all of the instant centers of velocity for the mechanism shown below.
- 5 - 2 4 5 3 B C D E 28˚ 5.0 cm AB = 8.0 cm AC = 4.5 cm BD = 13.0 cm DE = 2.9 cm A 2 4 5 3 B C D E A I 12 I 13 I 45 I 34 I 23 I 15 1 2 3 4 5 I 14 I 24 I 25 I 35 Solution

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- 6 - Problem 4.5 Locate all of the instant centers in the mechanism shown below. If link 2 is turning CW at the rate of 60 rad/s, determine the linear velocity of points C and E using instant centers. A B C D E 2 4 3 AD = 3.8" AB = 1.2" BC = 3.0" CD = 2.3" CE = 1.35" EB = 2.05" 125˚ Velocity Analysis The two points of interest are on link 3. To find the angular velocity of link 3, use I 13 and I 23 . Then 1 v I 23 = 1 ω 2 × r I 23 / I 12 = 1 3 × r I 23 / I 13 Therefore, 1 3 = 1 2 r I 23 / I 12 r I 23 / I 13 = 60 1.2 4.07 = 17.7 rad / s Because the instant center I 23 lies between I 12 and I 13 , 1 ω 3 is in the opposite direction of 1 ω 2 . Therefore, 1 ω 3 is counterclockwise. Then, 1 v C 3 = 1 3 × r C / I 13 1 v C 3 = 1 3 r C / I 13 = 17.7 2.11 = 37.3 in /s and 1 v E 3 = 1 3 × r E / I 13 1 v E 3 = 1 3 r E / I 13 = 17.7 3.25 = 57.5 in/s The directions for the velocity vectors are shown in the drawing.
- 7 - B C E 2 43 D A I 14 I 12 I 23 I 34 I 24 I 13 1 2 3 4 v C 3 v E 3 v B 3

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- 8 - Problem 4.6 Locate all of the instant centers in the mechanism shown below. If the cam (link 2) is turning CW at the rate of 900 rpm, determine the linear velocity of the follower using instant centers. A B 2 3 70˚ R AB = 1.5" R = 0.75" 103˚ Instant Centers
- 9 - Velocity of the Follower Convert the angular velocity from “rpm” to “rad/s” 1 2 900(2 ) 900 94.25 / 60sec rpm rad s CW π ω === At the point 23 I the linear velocity of follower and cam is same.

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## This note was uploaded on 02/20/2011 for the course MEC 411 taught by Professor Shudong during the Winter '11 term at Ryerson.

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ch04 - Solutions to Chapter 4 Exercise Problems Problem 4.1...

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