From 7bc77cb1ed33745c720952c92b3b2747c5cbf2df Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Sat, 3 Feb 2018 11:01:52 +0530
Subject: Added new code

---
 3886/CH3/EX3.9/Ex3_9.sce | 13 +++++++++++++
 1 file changed, 13 insertions(+)
 create mode 100644 3886/CH3/EX3.9/Ex3_9.sce

(limited to '3886/CH3/EX3.9/Ex3_9.sce')

diff --git a/3886/CH3/EX3.9/Ex3_9.sce b/3886/CH3/EX3.9/Ex3_9.sce
new file mode 100644
index 000000000..7d2158b17
--- /dev/null
+++ b/3886/CH3/EX3.9/Ex3_9.sce
@@ -0,0 +1,13 @@
+//determine equilibriant
+//two 40 kN forces have no moment about the pulley centre hence can be considered acting at pulley centre
+//Accordingly
+Rx=20*cosd(45)-30*cosd(60)-50*cosd(30)+40*cosd(20)-40*sind(30)  //kN  (towards left)
+Ry=-20*sind(45)-20+20-30*sind(60)-50*sind(30)-40*sind(20)-40*cosd(30)  //kN  (Downwards)
+R=sqrt(Rx^2+Ry^2)  //kN
+alpha=atand(Ry/Rx)  //degree
+//Taking moment about A
+MA=20*4-20*4+30*6*sind(60)+50*2*sind(30)-50*2*cosd(30)+40*3*cosd(20)-40*3*sind(30)
+//assume that the resultant intersects AB at a distance x from A,then
+x=MA/Ry  //m
+printf("Equilibriant is equal and opposite to resultant.\nR=%.2f kN\nalpha=%.2f degree\nx=%.3f m\nAs shown in fig.3.18 (a)",R,alpha,-x)
+
-- 
cgit