Question

The mechanism shown is for a transfer conveyor. The driving link rotates counterclockwise at a constant rate of 25 rpm, and the box has a mass of 10 kg as shown. The mass of the driving link and the coupler are negligible. Conveyor friction is adequate to prevent box slipping in the position shown. The mass of the conveyor link is 15 kg and the center of gravity is at its midspan. The mass moment of inertia of the conveyor link relative to the center of gravity is 4 kg-m². For β = 30֠, graphically determine the following:

1. Linear acceleration of the center of gravity of the conveyor link,
2. Rotational acceleration of the conveyor link
3. Inertial force and torque of the conveyor link
4. Pin forces
5. Torque required to drive the mechanism.
   

Answer 1

MATLAB Code

clc;
clear all;
%input data=
L1=50;L2=10;L3=50;L4=20;theta_1=90;theta_2=30;w2=2*pi*25/60;alpha_2=0;
%Position Equations
syms theta_3 theta_4
A=L2*cosd(theta_2)+L3*cosd(theta_3)-L4*cosd(theta_4)-L1*cosd(theta_1);
B=L2*sind(theta_2)+L3*sind(theta_3)-L4*sind(theta_4)-L1*sind(theta_1);
[theta_3,theta_4]=vpasolve(A==0,B==0)
%Velocity Equations
syms w3 w4
C=-L2*w2*sind(theta_2)-L3*w3*sind(theta_3)+L4*w4*sind(theta_4);
D=L2*w2*cosd(theta_2)+L3*w3*cosd(theta_3)-L4*w4*cosd(theta_4);
[w3,w4]=vpasolve(C==0,D==0)
%Acceleration Equations
syms alpha_3 alpha_4
E=-L3*alpha_3*sind(theta_3)-L3*(w3^2)*cosd(theta_3)+L4*alpha_4*sind(theta_4)+L4*(w4^2)*cosd(theta_4)-L2*alpha_2*sind(theta_2)-L2*(w2^2)*cosd(theta_2);
F=L3*alpha_3*cosd(theta_3)-L3*(w3^2)*sind(theta_3)-L4*alpha_4*cosd(theta_4)+L4*(w4^2)*sind(theta_4)+L2*alpha_2*cosd(theta_2)-L2*(w2^2)*sind(theta_2);
[alpha_3,alpha_4]=vpasolve(E==0,F==0)