Freudenstein's equation for a four-bar linkage is: (a) (k_{1} cos phi+k_{2} cos theta+k_{3}-cos (theta-phi)=0) (b) (k_{1} cos

Question:

Freudenstein's equation for a four-bar linkage is:

(a) \(k_{1} \cos \phi+k_{2} \cos \theta+k_{3}-\cos (\theta-\phi)=0\)

(b) \(k_{1} \cos \phi+k_{2} \cos \theta+k_{3}+\cos (\theta-\phi)=0\)

(c) \(k_{1} \cos \phi+k_{2} \cos \theta+k_{3}-\cos (\theta-\phi)=1\)

(d) \(k_{1} \cos \phi+k_{2} \cos \theta+k_{3}+\cos (\theta-\phi)=1\)

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: