Answer to Question #195368 in Mechanical Engineering for muhammed mudhasir

Question

a crank and slotted lever quick return motion mechanism, the distance between the fixed centers is 370 mm and the length of the driving crank is 150 mm.Find the inclination of the slotted bar with the vertical in the extreme position and the time ratio of cutting stroke to the return stroke. If the length of the slotted bar is 450 mm,crank angle of 45°from inner dead centre position.

Answer 1

a. $cos (\frac{\beta}{2})=\frac{150}{370}$

$\frac{\beta}{2}=cos^{-1}\frac{150}{370}$

$\frac{\beta}{2}=66.08$

$\beta=66.08*2=132.16^0$

The inclination of the slotted bar with the vertical in the extreme position

$\frac{\alpha}{2}=90-\frac{\beta}{2}=90-66.08=23.92^0$

b. The time ratio of cutting stroke to the return stroke

$\frac{Time \space of\space cutting \space stroke }{Time \space of\space return \space stroke}=\frac{360-\beta}{\beta}=\frac{360-132.16}{132.16}=1.72397$