I need help with part 2 of this problem. How do I solve part 2 using MATLAB. Tha

Question

I need help with part 2 of this problem. How do I solve part 2 using MATLAB.

Thank You

Problem statement A common mechanism is a four-bar linkage, pictured below. Link 1 is fixed, Link 2 and Link 4 are cranks, and Link 3 is the coupler. They are used to generate a desired output motion (Link 4) from an input motion (Link 2). It is a one-degree of freedom mechanism; if the position of any one of the links is known, then the position of the rest can be found. For certain geometries, the motion of Link 4 is restricted. It cannot rotate completely around its pivot; it moves back and forth. These mechanisms are called crank-and-rocker. They are useful for turning rotation into translation, for example, as in a motor powering windshield wipers. Given a crank and rocker mechanism, pictured here in a specific position (theta 4 = 90 degree): Link 1 (ground) = 6'' Link 2 (red) = 2.5'' Link 3 (black) = 8'' Link 4 (blue) = 5'' Assume there is a motor driving Link 2. Determine the angle theta 2 that corresponds to the position pictured here. Determine the path length of the rocker. Download and run FinalPart2animate.p to see this mechanism in action.

Explanation / Answer

% What i did here is evaluated theta4 for all values of theta2 ie 0 to 360

% degrees and then found out the maximum and the minimum theta4 (theta4

% will oscillate between some angles at it is a crank rocker) and then the

% arc length is given by radius * angle in radians

clear all

clc

syms x

L1=6;

L2=2.5;

L3=8;

L4=5;

theta2=0:0.1:2*3.14;

for i=1:63

theta4(i,:)=eval(solve(L4*cos(x)-L2*cos(theta2(i))+sqrt(L3^2 -( L4*cos(x)-L2*cos(theta2(i)))^2)-L1==0,x));

end

max_theta4=max(abs(theta4));

min_theta4=min(abs(theta4));

arc_rad=max_theta4(1)-min_theta4(1);

arc_length=L4*arc_rad

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