# Numerical solution of an EQM for dry friction

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Alaa Hameed 2017년 6월 4일
댓글: Walter Roberson 2017년 6월 4일
Hello,
I have the equation of motion for dry friction case, and I need a solution for it. I impliment Runge_Kutta method to do the solution and write the following .m files but it seems not working!any ideas?
%%Runge-Kutta method for solving equation of motion
% set the initial conditions and the time step
dt=.002;
u0=60/1000;d(1)=u0;
v0=0;v(1)=v0;
% a0=0;
m=2358.9;
k=107000;
zeta= .09;
wn=sqrt(k/m);
c=zeta*(2*m*wn);
wD=wn*sqrt(1-zeta^2);
t=0:dt:5;
a(1)=accel(v(1),d(1));
for i=1:length(t)-1
vel=v(i); dis=d(i);
dv1=dt*accel(vel,dis);
dx1=dt* vel;
vel=v(i)+dv1/2; dis=d(i)+dx1/2;
dv2=dt*accel(vel,dis);
dx2=dt* vel;
vel=v(i)+dv2/2; dis=d(i)+dx2/2;
dv3=dt*accel(vel,dis);
dx3=dt* vel;
vel=v(i)+dv3; dis=d(i)+dx3;
dv4=dt*accel(vel,dis);
dx4=dt* vel;
v(i+1)=v(i)+(1/6)*(dv1+2*dv2+2*dv3+dv4);
d(i+1)=d(i)+(1/6)*(dx1+2*dx2+2*dx3+dx4);
a(i+1)=accel(v(i+1),d(i+1));
end
%%For the function accel
function A = accel(V,D)
m=2358.9;k=107000;zeta= .09;
wn=sqrt(k/m);c=zeta*(2*m*wn);
mue=5;g=386.4;
A =(-1/m)*(mue*m*g*sign(V)+k*D);
end
Thanks
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Walter Roberson 2017년 6월 4일
What difficulty do you observe? What do you observe that you think should be different?

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