What is the velocity of the crosshead - crank:
In the slider crank mechanism shown in Figure, the crank is rotating clockwise at 120 revolution/minute. What is the velocity of the crosshead when the crank is in the 60o phase?
We have, crank AB = 200 mm = 0.2 metres.
ω AB = 120 revolutions / minute
= 120 × 2 π /60
= 4 π rad / sec
As B rotates about A, vB = AB . ω AB = 0.2 × 4 π = 0.8 π = 2.51 m / sec , and vB is perpendicular to AB.
Crosshead C moves along the horizontal line as shown.
We know vB fully, its magnitude and direction.
Consider Δ ABC,
AB /sin α = BC /sin 60o = AC / sin (120o - α)
∴ sin α = (AB / BC ). sin 60o = (200/900) × 0.866 = 0.1924
∴ α = 11.09o
(i) By Instantaneous Centre Method
Draw perpendiculars to vB at B and to vC at C. They meet at O.
Consider Δ BOC.
BC /sin 30o = CO /sin 71.09 o = BO / sin 78.9o
∴ OC = 0.9 ×( sin 71.09 /sin 30o) = 1.702 m
OB = 0.9 × (sin 78.9 /sin 30o) = 1.76 m
∴ ωBC = vB /OB =( 2.51/1.76) = 1.426 rad / sec
vC = ωBC × OC = 2.42 m / sec
(ii) By Velocity Diagram Method
From any point O draw a line perpendicular to AB to represent vB. From end of vB draw a line perpendicular to BC to represent vBC. Next, draw from O a line parallel to AC to represent vC.
∴ vB /sin 78.91o = vBC /sin 30o = vC /sin 71.11o
∴ vC = vB . (sin 71.11o /sin 78.91o)
= 2.42 m / sec
vC = vB. (sin 30 o / sin 78.91o)
= (2.51 × 0.5 )/0.98
= 1.2806 m / sec
∴ ωBC = vBC / BC
= 1.2806 /0.9 = 1.423 rad/sec