Analysis Of Engineering Systems - Homework 1

David Clark MAE 488 Homework #1

1.50) Input program: disp('The disp('The disp('The disp('The

value of b is:'),b=log(0.5)/(-5500) organism died (in years):'),t=log(90/100)/(-b) +1% estimate is:'),t1=log(90/100)/(-1.01*b) -1% estimate is:'),t2=log(90/100)/(-0.99*b)

Output: The value of b is: b= 1.2603e-004 The organism died (in years): t= 836.0170 The +1% estimate is: t1 = 827.7396 The -1% estimate is: t2 = 844.4616

1.59) Input program: x=[0:0.4:1.2] y=[6.3 22 60 215] p=polyfit(x,log(y),1); m=p(1,1); b=exp(p(1,2)); f=b*exp(m*x); mu=mean(y); J = sum((f-y).^2); S= sum((y- mu).^2); r2 = 1-J/S; disp('The m value is:'),m disp('The b value is:'),b disp('The J values are:'),J disp('The S values are:'),S disp('The r^2 values are:'),r2

David Clark MAE 488 Homework #1

Output: x= 0

0.4000

0.8000

1.2000

y= 6.3000 22.0000 60.0000 215.0000 The m value is: m= 2.8984 The b value is: b= 6.4244 The J values are: J= 77.4488 The S values are: S= 2.7351e+004 The r^2 values are: r2 = 0.9972

2.3) Input Program: m=10; phi=25*((2*pi)/360); vo=2; u=0.3; f1=100; g=9.81; a1=(f1-m*g*sin(phi)-u*m*g*cos(phi))/m disp('A1 is positive, so it will not come to a stop') f2=50; a2=(f2-m*g*sin(phi)-u*m*g*cos(phi))/m disp('A2 is negative, so it will come to rest') t = roots([a2 vo]); disp('The amount of time until the block comes to rest is:'),t Output: a1 = 3.1869 A1 is positive, so it will not come to a stop a2 = -1.8131 A2 is negative, so it will come to rest

David Clark MAE 488 Homework #1

The amount of time until the block comes to rest is: t= 1.1031 a1 = 3.1869 A1 is positive, so it will not come to a stop a2 = -1.8131 A2 is negative, so it will come to rest The amount of time until the block comes to rest is: t= 1.1031

2.24 Input %fbd of m2 gives: m2a2=-m2*g*sin(phi)+T-F %fbd of m2 also gives: F=Ia2*r^2 %fbd of m1 gives: m1a1=m1*g*sin(beta)-T %combine 3 equations to yield: %[m1+m2+I/r^2]a=m1*g*sin(beta)-m2*g*sin(phi) %m1+m2+I/r^2 is the equivalent mass Output

David Clark MAE 488 Homework #1

2.33 Input %fbd for sum of the forces x: m*ax=u*Nb-m*g*sin(theta) %fbd for sum of the forces y: Na + Nb - m*g*cos(theta) = 0 %fbd for the moments around G: -Na*La+Nb*Lb-u*Nb=0 %solve 2 for Na: Na=m*g*cos(theta)-Nb %-(m*g*cos(theta))*La+Nb*Lb-u*Nb=0 %-(m*g*cos(theta))*La+Nb(Lb-u)=0 %Nb=((mc*g*cos(theta))*La)/(Lb-u) %a=u*((g*cos(theta)*La)/(Lb-u))-g*sin(theta) %a=(23.544*u*cos(theta))/(4.6-u)-9.81*sin(theta) Output

David Clark MAE 488 Homework #1

2.34) Input: v=inline('t*(3.5*t*exp(1*t)-1*9.81)-300'); disp('The function for velocity is a combination of the equation in the') disp('book and and v=v(0)+a*t') disp(v) disp('The time until 300 m/s is reached is:'),fzero(v, 2) for k = 1:5 v=inline(strcat('t*(3.5*t*exp(', int2str(k), '*t)-1*9.81)-300')); R(k)=fzero(v, 10); end plot(1:5, R(1,:),'o') Output: The function for velocity is a combination of the equation in the book and and v=v(0)+a*t Inline function: v(t) = t*(3.5*t*exp(1*t)-1*9.81)-300 The time until 300 m/s is reached is: ans = 2.6124

David Clark MAE 488 Homework #1