Question
Asked Jul 1, 2019
At the instant represented, the velocity of point A of the 1.89-m bar is 3.6 m/s to the right. Determine the speed VB of
point B and the magnitude of the angular velocity w of the bar. The diameter of the small end wheels may be neglected.
0.46 m-
67
B
1.89 m
Answers:
m/s
VB
rad/s
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At the instant represented, the velocity of point A of the 1.89-m bar is 3.6 m/s to the right. Determine the speed VB of point B and the magnitude of the angular velocity w of the bar. The diameter of the small end wheels may be neglected. 0.46 m- 67 B 1.89 m Answers: m/s VB rad/s

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check_circleExpert Solution
Step 1

Calculate the angle of inclination of the bar with respect to horizontal.

 

0.46 m (0.46 m x sin230)
sin B=
1.89 m
=0.1483
Bsin(0.1483)
=8.53°
Here, angle of inclination of the bar with respect to horizontal is 6
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0.46 m (0.46 m x sin230) sin B= 1.89 m =0.1483 Bsin(0.1483) =8.53° Here, angle of inclination of the bar with respect to horizontal is 6

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Step 2

Write the relation for velocity of point A of the bar using the right hand rule.

 

 

V 3.6i m/s
Write the relation for velocity of point B of the bar using the right hand rule
v3 V (sin 23i - cos23°j) m/s
Write the relation for angular velocity of the bar using the right hand rule.
k rad/s
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V 3.6i m/s Write the relation for velocity of point B of the bar using the right hand rule v3 V (sin 23i - cos23°j) m/s Write the relation for angular velocity of the bar using the right hand rule. k rad/s

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Step 3

Calculate the position vector of point B with resp...

Гз4 3 Гз — Гд
[(1.89 mx cosp(-) + (1.89 mx sin )-(oi+ oj) m
=1.89(-cospisin Bj) m
Write the expression for the velocity of point B of the bar
Vз 3D Vд+V34
3= V+()XIg)
[(3.6i m/s)
3(sin 23°i - cos 23°j) m/s=\ +(k rad/s x 1.89 (-cospi + sin Bj) m)j
3(0.391 - 0.92j 3.61 + (krad/sx1.89(-cospi+ sin Aj))
3 (0.39i -0.92) 3.61+ (k rad/sx1.89(-cos8.53°j+sin 8.53°j))
v3 (0.39i-0.92 3.61-1.869j-0.28i
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Гз4 3 Гз — Гд [(1.89 mx cosp(-) + (1.89 mx sin )-(oi+ oj) m =1.89(-cospisin Bj) m Write the expression for the velocity of point B of the bar Vз 3D Vд+V34 3= V+()XIg) [(3.6i m/s) 3(sin 23°i - cos 23°j) m/s=\ +(k rad/s x 1.89 (-cospi + sin Bj) m)j 3(0.391 - 0.92j 3.61 + (krad/sx1.89(-cospi+ sin Aj)) 3 (0.39i -0.92) 3.61+ (k rad/sx1.89(-cos8.53°j+sin 8.53°j)) v3 (0.39i-0.92 3.61-1.869j-0.28i

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