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+// SAMPLE PROBLEM 6/1
+clc;funcprot(0);
+// Given data
+W=3220;// lb
+v=44;// m/s (30 mi/hr)
+s=200;// ft
+mu=0.8;// The effective coefficient of friction between the tires and the road
+g=32.2;// The acceleration due to gravity in ft/sec^2
+d_G=24;// inch
+d_BG=60;// inch
+d_GA=60;// inch
+
+// Calculation
+abar=v^2/(2*s);// ft/sec^2
+theta=atand(1/10);// degree
+W_h=W*cosd(theta);// lb
+W_v=W*sind(theta);// lb
+mabar=(W/g)*abar;// lb
+// SigmaF_x = m*abar_x
+F=mabar+W_v;// lb
+function[X]=reaction(y)
+ X(1)=(y(1)+y(2)-W)-0;
+ X(2)=((d_GA*y(1))+(F*d_G)-(y(2)*d_BG))-0;
+endfunction
+y=[1000,1000];
+z=fsolve(y,reaction);
+N_1=z(1);// lb
+N_2=z(2);// lb
+FbyN_2=F/N_2;
+printf("\nThe friction force under the rear driving wheels,F=%3.0f lb \nThe normal force under each pair of wheels,N_1=%4.0f lb & N_2=%4.0f lb",F,N_1,N_2);
+// Alternative solution
+// SigmaM_A=m*abar*d
+// SigmaM_A=m*abar*d
+N_2=((mabar*d_G)+((d_GA*W_h)+(d_G*W_v)))/(d_BG+d_GA);// lb
+// SigmaM_B=m*abar*d;
+N_1=((W_h*d_BG)-(d_G*W_v)-(mabar*d_G))/(d_BG+d_GA);// lb
+printf("\nALTERNATIVE SOLUTION:The normal force under each pair of wheels,N_1=%4.0f lb & N_2=%4.0f lb",N_1,N_2);