The velocity of point A in Fig 652 is 40 ftsec ccw Find the
The velocity of point A in Fig. 6-52 is 40 ft/sec ccw. Find the velocities of points B, C, and D. Scale: 1 in. = 40 ft/sec.
Solution
Velocity Diaram can be drawn as above
scale 1 in = 40ft/sec
Velocity of point A is 40ft/sec = 1inch
We know that velocity of a point on a link is always perpendicular to that link.
Assuming ground wrt point A is point P and wrt B is point Q.
On measuring qb =1.75 inch and ab=1.375inch we find
Velocity of point B wrt to Q is = 1.75x40 = 70 ft/sec
Velocity of B wrt A is Vb= 1.375 x 40 = 55 ft/sec
Now Vab is known so w of link ABC is = 55/AB
AB=sqrt(2.4^2+1^2) = 2.6 ft.
w=55/2.6 = 21.15 rad/sec
Given BC = 1.4 ft
Now Velocity of C wrt to B = w x BC = 21.15 x 1.4 = 29.61 ft/sec
Velocity of C wrt to round Q is = Vb+ Vbc = 55+29.61 =84.61 ft/sec
DC = sqrt(2^2+1.4^2)
Now w of link DC is = Vc/DC = 84.61/2.44 = 34.67 ft/sec
