Problem 2 - Problem 2.32 (Ehsan Mohammadi) clc;clear %F(dx)...

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Problem 2.32 (Ehsan Mohammadi) clc;clear %F(dx) = c*dx (2.46) %F(dx) = u*m*g*sgn(dx) (2.52) %F(dx) = cd*dx^2*sgn(dx) (2.54) %F(dx) = k*pi*Bn*sgn(dx)*abs(dx) (2.57) %For all of the cases we devied both sides of the functions by the %constants so that the function is normilized. t=0:pi/100:2*pi; x=(0.4)*sin(2*pi*t); dx=0.4*2*pi*cos(2*pi*t); %Viscous-damper force figure(1) Fv=dx; plot(t,Fv); axis([0 6 -3 3]) xlabel('Time') ylabel('Force') Title('Viscous-damper force') %Dry friction force figure(2) Fd = sign(dx); plot(t,Fd); axis([0 6 -1.1 1.1]) xlabel('Time') ylabel('Force') title('Dry friction force') %Fluid-damping force figure(3) Ff=(dx/abs(dx))*dx; plot(t,Ff) axis([0 6 -.15 .15]) xlabel('Time') ylabel('Force') title('Fluid-damping force') %Hysteretic force figure(4) Fh=abs(x)*(dx/abs(dx)); plot(t,Fh) axis([0 6 0 .018]) xlabel('Time') ylabel('Force') title('Hysteretic force')
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The Viscous-damper force is the force that works against the movement. As it is depicted above when there is no movement then there is no force exerted on system. However, when there is a
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This note was uploaded on 02/13/2011 for the course ENME 361 taught by Professor Yoo during the Spring '11 term at Maryland.

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Problem 2 - Problem 2.32 (Ehsan Mohammadi) clc;clear %F(dx)...

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