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diff --git a/3862/CH2/EX2.23/Ex2_23.sce b/3862/CH2/EX2.23/Ex2_23.sce
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+++ b/3862/CH2/EX2.23/Ex2_23.sce
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+clear
+//
+
+//When the roller is about to turn over the curb, the contact with the floor is lost and hence there is no reaction from the floor. The reaction R from the curb must pass through the intersection of P and the line of action of self weight, since the body is in equilibrium under the action of only three forces (all the three forces must be concurrent). 
+
+//variable declaration
+W=2000.0               //weight of roller,N
+r=300.0                //radius of roller,mm
+h=150.0                // height of curb,mm
+OC=r-h
+AO=r
+
+alpha=acos(OC/AO)
+
+//angleOAB=angleOBA,Since OA=OB,
+angleOBA=(alpha)/2
+
+//the reaction makes 30° with the vertical
+//sum of vertical Fy & sum of horizontal forces Fx is zero
+//Assume direction of Fx is right
+//Assume direction of Fy is up
+
+R=W/cos(angleOBA)
+P=R*sin(angleOBA)
+
+printf("\n P= %0.2f  N",P)
+
+//Least force through the centre of wheel: Now the reaction from the curb must pass through the centre of the wheel since the other two forces pass through that point. Its inclination to vertical is theta = 60°. If the triangle of forces ABC  representing selfweight by AB, reaction R by BC and pull P by AC, it may be observed that AC to be least, it should be perpendicular to BC. In other words, P makes 90° with the line of action of R.
+//From triangle of forces ABC, we get 
+P=W*sin(alpha)
+printf("\n P= %0.2f  N",P)