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authorprashantsinalkar2018-02-03 11:01:52 +0530
committerprashantsinalkar2018-02-03 11:01:52 +0530
commit7bc77cb1ed33745c720952c92b3b2747c5cbf2df (patch)
tree449d555969bfd7befe906877abab098c6e63a0e8 /3886/CH22/EX22.9/22_9.sce
parentd1e070fe2d77c8e7f6ba4b0c57b1b42e26349059 (diff)
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+//Uniform bar
+//refer fig. 22.15(a),(b),(c),(d),(e) and (f)
+beta=atand(0.6928/1.7856) //degree
+//aG=alpha*sqrt((1.7856^2)+(0.6928^2))
+I=(300*1.6^2)/(12*9.81)
+//Equating
+alpha=((300*0.6928)/(82.3160)) //rad/sec^2
+//Taking horizontal components of the forces
+NB=(300*1.7856*2.5249)/(9.81*cosd(45)) //N
+//Taking vertical components of the forces
+NA=(300)-(194.98*sind(45))+((300*0.6928*2.5249)/(9.81)) //N (Printing mistake in text book)
+printf("\nNA=%.2f N\nNB=%.2f N\nalpha=%.2f rad/sec^2",NA,NB,alpha)