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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+// SAMPLE PROBLEM 5/9
+clc;funcprot(0);
+// Given data
+n=1500;// rev/min
+theta=60;// degree
+r=5;// inch
+d_AG=10;// The distance from A to G in inch
+d_GB=4;// The distance from G to B in inch
+d_AB=14;// The distance from A to B in inch
+
+// Calculation
+v_B=(r/12)*((2*%pi*n)/60);// ft/sec
+// From the law of sines,
+beta=asind(r/(d_AB/sind(theta)));// degree
+theta_3=30;// degree
+theta_1=90-beta;// degree
+theta_2=180-theta_3-theta_1;// degree
+v_A=(v_B*sind(theta_2))/sind(theta_1);// ft/sec
+v_AB=(v_B*sind(theta_3))/sind(theta_1);// ft/sec
+ABbar=d_AB/12;// ft
+omega_AB=v_AB/ABbar;// rad/sec
+GBbar=d_GB/12;// ft
+v_GB=(GBbar/ABbar)*v_AB;// ft/sec
+// From velocity diagram
+v_G=64.1;// ft/sec
+printf("\nThe velocity of the piston A,v_A=%2.1f ft/sec \nThe velocity of point G on the connecting rod,v_G=%2.1f ft/sec \nThe angular velocity of the connecting rod,omega_AB=%2.1f rad/sec",v_A,v_G,omega_AB);
+