PROBLEM 15.40
Collar B moves upward with a constant velocity of 1.5 m/s. At the instant
when θ = 50°, determine (a) the angular velocity of rod AB, (b) the velocity
of end A of the rod.
SOLUTION
Draw a diagram showing the motion of rod AB.
Plane motion
vA = v A
=
Translation
25°
vB = 1.5 m/s
+
Rotation
vB/A = vB/A
50°
vB = v A + v B/A
[1.5 m/s↑] = [v A
25°] + [vB/A
50°]
Draw the velocity vector diagram.
Interior angles of the triangle.
90° − 25° = 65°
90° − 50° = 40°
25° + 50° = 75°
vB = 1.5 m/s
Law of sines.
vB/A
vB
vA
=
=
sin 75° sin 40° sin 65°
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reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1054
PROBLEM 15.40 (Continued)
(a)
Angular velocity of AB.
sin 65°
(1.5 m/s) = 1.4074 m/s
sin 75°
v
1.4074 m/s
= B /A =
l AB
1.2 m
vB /A =
AB
(b)
ω AB = 1.173 rad/s
Velocity of end A.
vA =
sin 40°
(1.5 m/s)
sin 75°
vA = 0.998 m/s
25°
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.
1055