1. Determine the velocity of the ram E, for the given position, if crank OA rotates uniformly at 150 rpm.
Solution:
1
=2 = 2 (150 ) 150 = 2356.5
(a) Solve for the linear velocity of crank OA;
60
(b) Draw the link using CAD, KS = 10:
(c) The vector equation connecting known velocity vB to unknown velocity vC on the same link BC is:
= + /
(d) Sketch the velocity polygon, KV=50:
oc = vC
ac = vC/A

2. Select suitable position of pole O, draw the first velocity vector VA, considering its magnitude and
direction.
Sketch velocity vC from O, knowing only its direction, which is perpendicular to crank DC.
= (40.6280)(50) = 2031.4
Finally, sketch velocity VC/A from the tip of VA, the direction of VC/A is perpendicular to link AC
= (10.5854)(50) = 529.27
The intersection of OC and ac is the magnitude of each velocity: and
/ .
(e) “ac” in the velocity polygon is the velocity image of AC. Since B lies on the extension of link AC, hence “b” is
represented in velocity polygon by velocity image principle.
=
550
=
10.5854 450
= 12.9377
=
(12.9377)(50) = 646.885
With ab = 12.9377 mm, mark a position of “b” along the direction “ac” in the velocity polygon.
(f) Considering link BE, the relative velocity equation for it is given by:
= + /
Sketch velocity VE/B from the tip of VB, the direction of VE/B is perpendicular to link EB.
The direction of slider E is going to the right, the intersection of de and be defines the magnitude of VE.

3. Line Oe on velocity polygon represents velocity ve.
The magnitude of velocity E is: = (29.9801)(50) = 1049
Summary of Answers:
= .
= .
=
Reference: Theory of Mechanisms 14th Edition by HG Phakatkar ©2009, HG Phakatkar