A quick return mechanism shown as Figure 1 Link 4 and Link 6

A quick return mechanism shown as Figure 1. Link 4 and Link 6 are frames. Crank 5 is a driver with a constant angular velocity omega_s of 20 rpm CCW. The angle of Link 5 theta_5 equals 45 degree and angle Link 3 theta_3 equals 15.75 degree. Determine the angular velocity omega_3 of Link 3. Determine the position of the slider on the Link 4. Determine the velocity of the slider on the Link 4.

Solution

The drawing and the dimension is not clear anyway to my best understanding of the figure, I am anawering.

The angle of lever 3 when lever 5 is at 90 degree to it is one extreme point and the same on the other sire is the other extreme point. considering the length of link 5 and link 6 , when the 90 degree is formed, the sin inverse of ( 250/450 ) gives the angle of link 3 and that is 33.7 deg.

similarly the othe rside the lever 3 moves by 33.7 deg to the right side.

The total angle moved is = 67.4 deg per rotation

For 20rpm the anglular velocity is 20 x 67.4 /60 = 22.46 deg/sec

At this point of consideration, the slider in link 3 is at 374mm from the reference ( obtained by taking the cosine of the angle)

Now consider the slider in link 4 , and analyse the triangle of the slider, the centre of rotation and the perpenticular line from the bottom reference point, we get the angle subtended at the rotating point by the slider is 37.07deg ( taking cosine angle as x/376, where x is the distance the slide has moved ).

we get the slider has moved 226 mm on one side, hence the total mmovement is 2x 226mm

The velocity of the slider is 2x 226x 20 rpm/60 = 151mm/sec.

The slider positions are 226mm on either side of the normal through the bottom reference point.