# Waldron cap. 3

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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 toscale.
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)

- 33 -

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

- 34 -

Problem 3.2 If ω2 = 10 rad/s CCW, find thevelocity 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.

- 35 -

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 vD4 + 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 / D41

1

1

Solve Eq. (1) graphically with a velocity polygon.

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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,
1

v B3 = 9.4 in/s

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Problem 3.3 If ω2 = 100 rad/s CCW, find vB4.

AD = 1.8" CD = 0.75" AE = 0.7" CF =0.45" FG = 1.75" CB = 1.0" DB = 1.65"

G 125˚ B4 3 C F 4 A D 95˚

2 E

Position Analysis

- 38 -

Velocity Analysis

1

v G2 = 1 v A2 + 1 v G2 / A2 = 0 + 1ω2 × rG / A v G3 = 1 v C3 + 1 v G3 / C3 = 1 v C4 + 1ω3 × rG / C v G3 = 1 v G2 + 1 v G3 / G2

(1)

1

1

1

v C4 = 1 v C3 = 1 v D4 + 1 v C4 / D4 = 0 + 1ω4 × rC / D

Now,
1

v G2 = 1ω2 rG /A = 100(3.44) = 344 in/s (⊥ to rG / A )

1

ω3 = 1ω2
1

v G3 / C3 = 1ω3 rG / C = 100(2.65) = 265 in/s (⊥ to rG / C )

1

vG3 / G2 is on the line of EG v C4 / D4 = 1ω4 × rC / D (⊥ to rC / D )

1

Solve Eq. (1) graphically with a velocity polygon.

- 39 -

To find the point “ b4 ” use velocity polygon

1

v C4 / D4 = 612.14 in / s

- 40 -

Problem 3.4 If ω2 = 50 rad/sCCW, find vD4.

B BC = CD BD = 3.06" 3 150˚ C A 50˚ D 4

2

Position Analysis

- 41 -

Velocity Analysis

1

v B2 = 1 v A2 + 1 v B2 / A2 = 0 + 1ω2 × rB / A
1

(1)

v B3 = 1 v D3 + 1 v B3 / D3 = 1 v D4 + 1ω3 × rB / D

1

v B3 = 1 v B2 + 1 v B3 / B2

1

v D4 = 1 v D3

Now,

1

v B2 = 1ω2 rB / A = 50(2.39) = 119.5 in/s (⊥ to rB / A )

1ω3 = 1ω2

1

v B3 / D3 = 1ω3 rB / D = 50(3.06) = 153 in/s (⊥ to rB / D )

1

v B3 / B2 is on the line of AB

1

v D4 = 1 v D3

(// to rD / A )

- 42 -

Solve Eq. (1) graphically with a velocity polygon.

From the velocity polygon

1

v D4 = 164.34 in / s

- 43 -

Problem 3.5 Determine the velocity and acceleration of point B on link 2.

3 in 2

Y
ω 4 = 1...