From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 3792/CH6/EX6.10/Ex6_10.sce | 34 ++++++++++++++++++++++++++++++++++
 1 file changed, 34 insertions(+)
 create mode 100644 3792/CH6/EX6.10/Ex6_10.sce

(limited to '3792/CH6/EX6.10/Ex6_10.sce')

diff --git a/3792/CH6/EX6.10/Ex6_10.sce b/3792/CH6/EX6.10/Ex6_10.sce
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+// SAMPLE PROBLEM 6/10
+clc;funcprot(0);
+// Given data
+l=4;// ft
+W=40;// The weight of the slender bar in N
+theta=30;// degree
+k=30;// The stiffness of the spring in lb/in
+ABbar=24;// inch
+BDbar=24;// inch
+h=-2;// inch
+g=32.2;// The acceleration due to gravity in ft/sec^2
+
+
+// Calculation
+// (a)
+// T=[[(1/2)*m*v^2]+((1/2)*I_G*omega^2)];
+// T=1.449*omega^2;
+T_1=0;// ft-lb
+U_12=0;// ft-lb
+V_1=0;// ft-lb
+V_2=W*((2*cosd(theta))-2);// ft-lb
+// We now substitute into the energy equation and obtain
+omega=sqrt(((T_1+V_1+U_12)-(V_2))/1.449);// rad/sec
+// (b)
+x=ABbar-18;// ft
+V_1=0;// ft-lb
+V_3=(1/2)*k*(x^2)/12;// ft-lb
+// T=(1/2)*I_A*omega^2;
+// T_3=0.828*v_B^2;
+U_13=0;// ft-lb
+// The final gravitational potential energy is
+V_3p=W*h;// ft-lb
+v_B=sqrt(((T_1+V_1+U_13)-(V_3+V_3p))/0.828);// ft-lb
+printf("\n(a)The angular velocity of the bar,omega=%1.2f rad/sec \n(b)The velocity with which B strikes the horizontal surface,v_B=%1.2f ft/sec",omega,v_B);
-- 
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