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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+//Example 4.7
+Fx=2.7*10^5;//Force exerted in the x-direction (N)
+Fy=3.6*10^5;//Force exerted in the y-direction (N)
+m=5*10^6;//Mass of the barge (kg)
+a=7.5*10^-2;//Acceleration (m/s^2)
+theta=53.1;//Angle (deg)
+F_app=sqrt(Fx^2+Fy^2);//Resultant applied force (N)
+theta=atand(Fy/Fx);//Direction of resultant applied force (deg)
+F_net=m*a;//Net external force (N)
+F_D=F_app-F_net;//Drag force (N)
+printf('Drag force exerted by water = %0.1e N',F_D)
+printf('\nDirection of the drag force is opposite to that of the applied force.')
+//Direction of drag force, theta_F_D=53 deg south of west
+//also it may be noted that if theta_F_D is to be measured from the same reference axis as theta
+//theta_F_D=theta+180;
+//if theta_F_D>360
+// theta_F_D=theta_F_D-360;
+//end
+//Openstax - College Physics
+//Download for free at http://cnx.org/content/col11406/latest