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+
+clc
+//given
+BG=4//in
+//solution a
+w=15//lb
+W=120//lb
+k=.720
+BD=10.08//in
+CE=BD
+DG=BD+BG
+//by equating quations 13.2 and 13.10 and reducing, we get
+w1=(W/2*(1+k))/(((W/2*(1+k)+w)*DG/(BD*w))-1)
+printf("\nWeight of ball = %.3f lb\n",w1)
+//solution b
+CD=6.5//in
+BC=12//in
+BF=10//in
+AB=12//in
+CG=(DG^2+CD^2)^(1/2)
+gama=atan(CD/DG)
+bita=asin(CD/BC)
+alpha1=asin(BF/AB)
+bita1=asin(8.5/BC)
+gama1=gama+bita1-bita
+F=((w1+W/2)*8.471*(tan(alpha1)+tan(bita1)))/(CG*cos(gama1))-(w1*tan(gama1))
+printf("F1= %.1f lb",F)
+r1=CG*sin(gama1)+1.5//radius of rotation
+N1=(30/%pi)*(F*32.2*12/(w1*r1))^(1/2)
+printf("\nr1= %.2f in\nN1= %.1f rpm",r1,N1)