Machine Dynamics, Solved Problems, Kinematics, Position analysis Inverted slider-crank, Problem 1

This video is part of a series of videos presenting solutions of problems related to the machine dynamics topic.

This video is related to the topic of kinematics of mechanisms.

In this problem, we are going to find the position of an inverted slider-crank mechanism.

The question here asks to write the position equations of the inverted slider-crank mechanism shown here.
In this mechanism link 2, the bar A B, is the input.
It is asked to start from the the loop-closure equation A B plus B C is equal to A D plus D C,
Or r 2 plus r 3 is equal to r 1 plus r 0.
It is also asked to find the angle theta 3 and, the modulus r 3, if r 0 is equal to 10 cm,
R 2 is equal to 15 cm, r 1 is equal to 25 cm, and theta 2, the input is equal to 60 degrees.


This mechanism is an inverted slider-crank.
It is built of three bars, the bar A B, link 2, the bar B E, link 3, and the bar A C, link 1, which is here the ground.
The fourth link here is the slider, which rotates about C.
The bar BE, is here translating relatively to the slider C.

The mobility of the mechanism is one.
Thus, one input is required.
Here, link 2 is considered as the input.
The angle theta 2 is considered given.

The vector A B, is the vector position of the bar A B.
The modulus of the vector A B is equal to r 2 the length of the bar A B.
The angle of the vector A B is the theta 2, the input.

The vector A B is the vector r 2, the vector position of link 2.

Vector loop method