# ch03 - Solutions to Chapter 3 Exercise Problems Problem 3.1...

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- 33 - Solutions to Chapter 3 Exercise Problems Problem 3.1 In the figure below, points A and C have the same horizontal coordinate, and ω 3 = 30 rad/s. Draw and dimension the velocity polygon. Identify the sliding velocity between the block and the slide, and find the angular velocity of link 2. 4 2 B 3 , B 4 A ω 3 3 AC = 1 in BC = 3 in r = 2.8 in C 45˚ Position Analysis: Draw the linkage to scale. AC = 1 in BC = 3 in 30 in/sec Velocity Polygon O v b 3 b 4 b 2 , A B C 2 in 3 2 45.0° Velocity Analysis: 1 v B 3 = 1 v B 3 /A 3 = 1 ω 3 × r B 3 3 1 v B 3 = 1 ω 3 r B 3 3 = 30(2.2084) = 66.252 in/sec 1 v B 4 = 1 v B 2 = 1 v B 3 + 1 v B 4 /B 3 (1)

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- 34 - 1 v B 2 = 1 v B 2 /C 2 = 1 ω 2 × r B 2 2 Now, 1 v B 3 = 66.252in/sec in the direction of r B/A 1 v B 2 = 1 ω 2 × r B/C ( to r ) 1 v B 4 /B 3 is on the line of AB Solve Eq. (1) graphically with a velocity polygon. From the polygon, 1 v B 4 3 = 15.63in/sec Also, 1 ω 2 = 1 v B 2 2 r = 68.829 3 = 22.943 rad /sec From the directions given in the position and velocity polygons 1 ω 2 = 22.943 rad/sec CW
- 35 - Problem 3.2 If ω 2 = 10 rad/s CCW, find the velocity of point B 3 . 1 ω 2 Α Β 2 3 4 E C D 45˚ 18˚ 110˚ CA = 1.5" DE = 2.5" CD = 4.0" AB = 1.6" Position Analysis Draw the linkage to scale.

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- 36 - Velocity Analysis 222 2 2 111 1 1 1 /2 / 2 / 0 10(1.5) 15 in/s ACA C A C A A C ωω =+ = == = vvv r v r (1) 333 3 2 3 3 1 1 // EAE AA E A = + v v 343 4 / EEE E 444 4 1 /4 / 0 EDE D E D ω = + × r Now, 2 1 15 /s A in = v / (t o ) AC r 33 11 /3 / / EA vr r 44 / / () E D ED to r and 34 = , need to get 3 to find 3 1 B v . Define the point F where AF DF in position polygon. 3 / FAF A 4 / FFF F 434 3 / F 4 / FF D = vv Solve Eq. (1) graphically with a velocity polygon.
- 37 - After finding point “f3”, construct the velocity image to find the point “b3” a line to AB through the point “a” a line to BF through the point “f3” fine the point “b3” From the polygon, 3 1 9.4 in/s B = v

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- 38 - Problem 3.3 If ω 2 = 100 rad/s CCW, find v B 4 . A B 2 3 4 C D 4 95˚ G E F AD = 1.8" CD = 0.75" AE = 0.7" CF = 0.45" FG = 1.75" CB = 1.0" DB = 1.65" 125˚ Position Analysis Draw the linkage to scale.
- 39 - Velocity Analysis 222 2 111 1 /2 / 0 GAG A G A ω =+ = vvv r (1) 333 3 4 11 /3 / GCG C C G C = + × v r 323 2 / GGG G 43 4 4 4 1 1 1 /4 / 0 CC D C D C D == + = + × vv v v r Now, 2 2/ / 100(3.44) 344 in/s ( ) GG A G A to === vr r 32 ωω = 33 / / 100(2.65) 265 in/s ( ) G C GC to r 1 / v is on the line of EG 44 / / () CD to r Solve Eq. (1) graphically with a velocity polygon.

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- 40 - To find the point “ 4 b use velocity polygon 44 1 / 612.14 / CD in s = v
- 41 - Problem 3.4 If ω 2 = 50 rad/s CCW, find v D 4 . 2 3 4 A B C D 50˚ 150˚ BC = CD BD = 3.06" Position Analysis Draw the linkage to scale.

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- 42 - Velocity Analysis 222 2 111 1 /2 / 0 BAB A B A ω =+ = vvv r (1) 333 3 4 11 /3 / BDB D D B D = + × v r 323 2 / BBB B 43 D D = vv Now, 2 2/ / 50(2.39) 119.5 in/s ( ) BB A B A to == = vr r 32 ωω = 33 / / 50(3.06) 153 in/s ( ) BD to =⊥ r 1 / v is on the line of AB / (// ) D DD A to = r
- 43 - Solve Eq. (1) graphically with a velocity polygon. From the velocity polygon 4 1 164.34 / D in s = v

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