Waldron cap. 3

Solutions to Chapter 3 Exercise Problems Problem 3.1 In the figure below, points A and C have the same horizontal coordinate, and ω3 = 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 ω3 A 2 C 3 45˚ B3 , B4

AC = 1 in BC = 3 in r = 2.8 in

Position Analysis: Draw the linkage to scale.
2 in 3 A 2 C b 2 , b4 b3 B 45.0° Ov Velocity Polygon 30 in/sec AC = 1 in BC = 3 in

Velocity Analysis:

1

v B3 = 1vB 3 / A 3 = 1ω 3 × rB 3 / A 3 ⇒ 1 vB 3 = 1 ω3 rB 3 / A 3 = 30(2.2084) = 66.252 in / sec

1

v B4 =1 v B2 = 1v B3 + 1vB4 / B3

(1)

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1

v B2 = 1v B2 / C2 = 1ω 2 × rB2 / C2

Now,
1 1 1

v B3 = 66.252in / sec in the direction of rB / A

v B2 = 1ω 2 × rB / C (⊥ to rB / C ) v B4 / B3 is on the line of AB

Solve Eq. (1) graphically with a velocity polygon. From the polygon,
1

v B4 / B3 = 15.63in / sec

Also,
1

ω2 =

1v

B2 / C 2

rB / C

=

68.829 = 22.943 rad / sec 3

From the directions given in the position and velocity polygons
1

ω 2 = 22.943 rad / sec CW

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Problem 3.2 If ω2 = 10 rad/s CCW, find the velocity of point B3.

E CA = 1.5" DE = 2.5" CD = 4.0" AB = 1.6" Α 18˚
1ω 2

110˚

3 Β

4

2 45˚ D

C

Position Analysis Draw the linkage to scale.

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Velocity Analysis

1

v A2 = 1 v C2 + 1 v A2 / C2 = 0 + 1ω2 × rA / C ⇒ 1 v A2 = 1ω2 rA / C = 10(1.5) = 15 in/s v E3 = 1 v A3 + 1 v E3 / A3 = 1 v A2 + 1 v E3 / A3

(1)

1

1

v E3 = 1 v E4 + 1 v E3 / E4 v E4 = 1 v D4 + 1 v E4 / D4 = 0 + 1ω4 × rE / D

1

Now,
1

v A2 = 15 in/s (⊥ to rA / C ) v E3 / A3 = 1ω3 × rE / A (⊥ to rE / A ) v E4 / D4 = 1ω4 × rE / D (⊥ to rE / D )
3

1

1

and

1 ω3 = ω4 , need to get ω3 to find v B .

Define the point F where AF ⊥ DF in position polygon.

1

v F3 = 1 v A3 + 1 v F3 / A3 v F3 = 1 v F4 + 1 v F3 / F4 v F4 = 1 v F3 + 1 v F4 / F3 v F4 = 1 v F4 / D4