CEN 100 Lecture Notes - Lecture 2: Matlab

• School: Ryerson University
• Department: Common Engineering
• Course: CEN 100
• Professor: Lamya Amleh

Document Summary

% first we write matrice a and b. % first we calculate matrix f fora=b(:,1) forb=b(:,2) forc=c(2:2,1:3) ford=reshape(forc,[],1) % since mass is same for all lights so we cosider their downward force, or their weights (done by gravity) as forceg forceg=-20*u. kg*9. 81*u. m/u. s^2. % horizontal component of ab is forceabh,horizontal component of. Bc is forcebch,vertical component of ab is forceabv,vertical component of bc is forcebcv. % forceabv will equal to the weight of the light at b, but in the opposite direction forceabv=20*u. kg*9. 81*u. m/u. s^2. % since bc is not inclined so forcebcv has to be zero forcebcv=0. % we use trigonometric ratios to calculate forceabh forceabh=forceabv/(tan(pi-pi/36)) % since the system is in equilibrium forcebch=-forceabh >> complete. 4 6 12 c) ans = 1 4 2. 6 forc = 5 4 7 ford = 5. 0. 2222 -0. 2222 0. 1111 b) row2 = -0. 2593 0. 2593 0. 0370.