Question

Using Instantaneous Centre of Rotation (ICR) method, find the velocity of point A for the instant shown in Figure 2. Collar B moves along the vertical rod, whereas link AB moves along the plane which is inclined at 25⁰. Ɵ = 45⁰

Answer 1

By using sine rule,

`(AB)/sinI = (BI)/sinA =( AI)/sin B`

`BI = (1.2 xx sin (60))/(sin (75)) = 1.076m`

`AI = (1.2 xx sin (45))/(sin (75)) = 0.878m`

ωAB = BI x VB = 1.076 x 1.5 = 1.614 rad/s
ωAB = AI x VA

`V_A =(omegaAB)/(AI) = 1.614/0.878 = 1.838`m/s

The velocity of point A for the given instance is 1.838 m/s.