m4 Show that the point Q in the linkage in this figure alway

Show that the point Q in the linkage in this figure always traces a straight line.

This linkage is known as Linkage - Peaucellier-Lipkin Linkage.

As given in the problem figure side AC=AE=s

And rhombus CDEQ has sides

CD= DE= EQ=QC

Poiint A and B are fixed pivot so linksAC and BE rotate about point A and

Link BD rotates about point B.

The figure Four Bar Mechanism is there with Links

AB, AC, BE, and Rhombus connecting C and E.

Shortest link is BD so it will rotate 360 degree.

Hence

Circle with Centre B is constructed to represent the rotation of link BD

This Circle cuts the link AB (extended) at point ,W.

Constructing line QM such that QM is perpendicular to AW

Now look the Circle AW is diameter. Hence angle ADW is = 90 degree

Triangle ADW and AQM are similar.It can be shown easily .

AD x AQ = AW x AM

And

AM = = constant because AD . AQ is constant and AW is diameter of the circle is also constant.

SO from here we found

AM reamains constant as the linkage moves AM remains constant . Now MQ is perpendicular to AM.

So distance of Point Q from A remains constant and hence Q travels in a straight line .