1. Collar A moves in a constant velocity of 900 mm/s to the right. At the instant when e = 30°, determine (a) the angular velocity of rod AB, (b) the velocity of collar B. (BJ 15.34 on page 721) 30° 70° B y Solution

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hello, i got the solution, however i am unable to understand, able to explain what each line means and whats the purpose of it ? for example, va vector is ? rba vector is ? thank you in advance

1. Collar A moves in a constant velocity of 900 mm/s to the
right. At the instant when 0 = 30°, determine (a) the
angular velocity of rod AB, (b) the velocity of collar B.
(BJ 15.34 on page 721)
30°
70°
B
Solution
A
With O-xy, we can express
30°
70°
v, = 90020° = 900 î (mm/s)
A
B
TRA = 300Z- 30° = 300 (0.866 î – 0.5 ĵ) (mm)
Vg = VgZ-70° = V½( 0.342î – 0.94 Ĵ) (mm/s)
Using v,
V, + öx ĩ gives
BA
A
V ( 0.342 î – 0.94 ĵ) = 900 î + Ôk × 300 (0.866 î – 0.5 ĵ)
| "АВ"
Component of î:
0.342 vB
Ov
A
150 0 = 900
(1)
ליק
Component of ĵ:
VB A
0.94 v B
+ 259.80 = 0
(2)
Solving the simultaneous equations (1) and (2) gives
B
VB
= 1017 (mm/s)
3.68 (rad/s)
Hence v = 1017Z-70° (mm/s) (ans)
ö = ék= - 3.68 k (ans)
Transcribed Image Text:1. Collar A moves in a constant velocity of 900 mm/s to the right. At the instant when 0 = 30°, determine (a) the angular velocity of rod AB, (b) the velocity of collar B. (BJ 15.34 on page 721) 30° 70° B Solution A With O-xy, we can express 30° 70° v, = 90020° = 900 î (mm/s) A B TRA = 300Z- 30° = 300 (0.866 î – 0.5 ĵ) (mm) Vg = VgZ-70° = V½( 0.342î – 0.94 Ĵ) (mm/s) Using v, V, + öx ĩ gives BA A V ( 0.342 î – 0.94 ĵ) = 900 î + Ôk × 300 (0.866 î – 0.5 ĵ) | "АВ" Component of î: 0.342 vB Ov A 150 0 = 900 (1) ליק Component of ĵ: VB A 0.94 v B + 259.80 = 0 (2) Solving the simultaneous equations (1) and (2) gives B VB = 1017 (mm/s) 3.68 (rad/s) Hence v = 1017Z-70° (mm/s) (ans) ö = ék= - 3.68 k (ans)
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