Question

B. (4 pts) Use the : operator to create the vectors x [2.5,1.50.5,-0.5,-1.51 and y [0.4,0.8,1.2,1.6,2], then use them in the following expression to compute vector z using element-by-element calculations C. (8 pts) A 70 lb bag of rice is being pulled by applying a force F at an angle θ as shown. The force required to drag the bag is given by: θ μ sin θ + cos θ Where u-0.35 is the friction coefficient. Determine: (a) (3 pts) Fa for θα-0°, 5°, 10°, 15°, 20°, 25°, 30°, 35°, 40°, 45°. Enter the variable theta as a vector using the : operator. (b) (5 pts) the angle 0 where F is minimum. Do it by creating a vector 0, with elements ranging from 0o to 45° and spacing of 0.01. Calculate Fb for each value of eb and then find the minimum Fand the associated 0 with MATLAB's built-in function min0. Write the command plot (theta_b, F_b) at the end of your script to plot Fb versus , Hint from the help of min function: [Y,1] = min (X) returns the index! ofthe minimum value in a vector Xlfthe values along xcontain more than one minimal element, the index of the first one is returned.

Please show matlab editor and command window

Answer 1

`Hey,

Note: Brother in case of any queries, just comment in box I would be very happy to assist all your queries

B)

clc;
clear all;
close all;
x=[2.5:-1:-1.5]
y=[0.4:0.4:2]
z=(x.^2).*(y.^3).*exp(y./x)-((x.^3).*(y.^2)+72.67).^(1/4)

MATLAB R2014a EDITOR Find Files Compare ▼ Comment Print New Open Save Breakpoints Run Run and Advance Advance Run and Indent Ele , i Find - Time C: Users PRADEEP KALRA 、Documents MATLAB Current Folder Editor-CAUsersPRADEEP KALRA!Documents\MATLABAsaveee.m saveee.m >| newtonrapsonsbypradeep.m secantbypradeep.m ×| bisectionmethodbypradeep.m This file can be published to a formatted document. For more information, see the publishing yideo or help. 1-clc: 2clear all; 3close all: 4-2F[2.51-11-1.5] 5Y [0.4:0.4:21 Name ▼ xl Z.m yytt.m yy.m d.dat wq.m WCF.m Velocity.m Command Window Details New to MATLAB? Watch this Video, see Examples or read Getting Started. Name Value Min 2.5000,1.5000,0.5000,.1 0.4000,0.8000,1.2000 04000 -2.4751,-0.9775,1.840. -2.874 0.400000000000000 0.800000000000000 1.200000000000000 1.600000000000000 2.000000000000000 -2.475092796602314-0.977457127981578 1.840502116812555 -2.874743742571364 1.971266226278113 script Ln 6 Col 58 O Type here to search ENG 1200 PM US 02/27/19

C)

clc;
clear all;
close all;

u=0.35;
theta=[0:5:35];

l=numel(theta);

for i=1:l
F(i)=(70*u)./((u.*sind(theta(i)))+cosd(theta(i)));

end
disp('THE F(THETA) VALUES FOR THETA ARE=');
disp(F);

%FOR THETA NOW FROM 5 TO 35 AND SPACING OF 0.01.

theta1=5:0.01:35;

l1=numel(theta1);
disp('IN NEW O.O1 SPACED VECTOR THE MINIMUM AND MAXIMUM VALUE OF F AND THETA ARE');
for i=1:l1
F(i)=(70*u)./((u.*sind(theta1(i)))+cosd(theta1(i)));

end
disp('THE MINIMUM VALUE OF F=');

[ m,i]=min(F);
disp(m);

disp('THE VALUE OF THETA AT MINIMUM F=');
disp(theta1(i));

disp('THE MAXIMUM VALUE OF F=');
[ma,j]= max(F);
disp(ma);

disp('THE VALUE OF THETA AT MAXIMUM F=');
disp(theta1(j));
plot(theta1,F);

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