A wedge can be used to lift heavy objects of weight w by applying a horizontal f

Question

A wedge can be used to lift heavy objects of weight w by applying a horizontal force P The mechanical advantage of the wedge (PIW) is determined by its angle 6. For this problem assume that the floor and wall are smooth, that the wedge is massless, and that the coefficient of static friction between the wedge and the block is us. W For this problem: 1. solve for the minimum PIW needed to move the wedge from rest as a function of 0, and and us. 2. create a MATLAB program named q02.m which solves for the P/W ratio 3. create a plot og P/W vs. 0. Add axis labels, a grid, and save the plot (from your code) to a file named thetavsPW.png. 4. create a plot og P/W vs. us. Add axis labels, a grid, and save the plot (from your code) to a file named musVsPW.png. 5. use your program (of from the plots) to fill in the following table. Input for Input for u X(P/W 1,2,3,4,... 50 0.2 0.05,0 1,0,15,...,0.9 15 x represents either 6 or us

Explanation / Answer

cartesian format:

%Define the vectors

A=[1 2 3];

B = [-1 -2 -1];

%Take the cross product

A_x_B = cross(A,B);

fprintf(

'AxB=[

%1.4f %1.4f %1.4f] '

, A_x_B)

%Check MATLAB's cross product operator by calculating the cross product

explicitly...

x=1; y=2; z=3;

% Define coordinate index definitions

A_x_B_exp = [A(y)*B(z)-A(z)*B(y) -(A(x)*B(z)-A(z)*B(x)) A(x)*B(y)-A(y)*B(x)];

fprintf(

'A x B calculated explicitly= [ %1.4f %1.4f %1.4f] '

, A_x_B_exp

%Use the trigonometric form of the cross product operator to find the

magnitude of the C vector.

% First, find the magnitude of th

eA&B

vectors using the norm function.

A_mag = norm(A);

B_mag = norm(B);

fprintf(

'Magnitude of vecto

r A = %1.4f '

, A_mag)

fprintf('Magnitude of vector B = %1.4f ', B_mag)

% Then, find the angle between vectors A and B.

theta = 180/pi * acos(dot(A,B)/(A_mag * B_mag));

fprintf(

'Angle between vectors A an

d B = %1.4f deg '

, theta)

% Finally, solve for the magnitude of the C vector.

C_mag = A_mag * B_mag * sin(theta * pi/180);

fprintf('Magnitude of vector C = %1.4f ', C_mag)

%Solve for the direction of the C vector

cross(A,B)/norm(cross(A,B));

% or C/C_mag wher

e C = cross(A,B)

alpha = acos(0.8944) * 180/pi;

beta = acos(-0.4472) * 180/pi;

gamma = atan(0) * 180/pi;

fprintf(

'alpha = %1.4f deg from +x '

, alpha)

fprintf(

'beta = %1.4f deg from +y '

, beta)

fprintf(

'gamma = %1.4f deg from +x '

, gamma)

% This Matlab code can be used for simply supported beam with single point
% load or uniformly distributed to find the
% * Support reaction
% * Maximum Bending Moment
% * Shear force diagram
% * Bending Moment daigram
clc; clear; close all
disp('Simply Supported Beam');

% Data input section
disp(' ');
L = input('Length of beam in meter = ');
disp(' ');disp('Type 1 for point load, Type 2 for udl')
Type = input('Load case = ');

if Type == 1
disp(' ');
W = input('Load applied in kN = ');
disp(' ');
a = input('Location of Load from left end of the beam in meter = ');
c = L-a;

R1 = W*(L-a)/L; % Left Support Reaction.
R2 = W*a/L; % Right Support Reaction.

else
    disp(' ');
    W = input('Uniformly distributed load in kN/m = ');
    disp(' ');
    b = input('Length of udl in meter = ');
    disp(' ');
    cg = input('C.G of udl from left end of the beam in meter = ');
    a = (cg-b/2);
    c = L-a-b;
    
    R1 = W*b*(b+2*c)/(2*L); % Left Support Reaction.
    R2 = W*b*(b+2*a)/(2*L); % Right Support Reaction.
end
  
% Discretization of x axis.
n = 1000; % Number of discretization of x axis.
delta_x = L/n; % Increment for discretization of x axis.
x = (0:delta_x:L)'; % Generate column array for x-axis.

V = zeros(size(x, 1), 1); % Shear force function of x.
M = zeros(size(x, 1), 1); % Bending moment function of x.

% Data processing section
if Type == 1
for ii = 1:n+1
    % First portion of the beam, 0 < x < b
    V(ii) = R1;
    M(ii) = R1*x(ii);
    
    % Second portion of the beam, b < x < L
    if x(ii) >= a
        V(ii) = R1-W;
        M(ii) = R1*x(ii)-W*(x(ii)-a);
    end
end
x1 = a;
Mmax = W*a*(L-a)/L;
else
   for ii = 1:n+1
    % First portion of the beam, 0 < x < a
    if x(ii) < a
    V(ii) = R1;
    M(ii) = R1*x(ii);
    elseif a <= x(ii) && x(ii)< a+b
    % Second portion of the beam, a < x < a+b
        V(ii) = R1-W*(x(ii)-a);
        M(ii) = R1*x(ii)-W*((x(ii)-a)^2)/2;
    elseif x(ii) >= (a+b)
    % Second portion of the beam, a+b < x < L
        V(ii) = -R2;
        M(ii) = R2*(L-x(ii));
    end
   end
   x1 = a+b*(b+2*c)/(2*L);
   Mmax = W*b*(b+2*c)*(4*a*L+2*b*c+b^2)/(8*L^2);
end

disp(' ');disp (['Left support Reaction' ' = ' num2str(R1) ' ' 'kN'])
disp(' ');disp (['Left support Reaction' ' = ' num2str(R2) ' ' 'kN'])
disp(' ');disp (['Maximum bending moment' ' = ' num2str(Mmax) ' ' 'kNm'])

figure
subplot(2,1,1);
plot(x, V, 'r','linewidth',1.5); % Grafica de las fuerzas cortantes.
grid
line([x(1) x(end)],[0 0],'Color','k');
line([0 0],[0 V(1)],'Color','r','linewidth',1.5);
line([x(end) x(end)],[0 V(end)],'Color','r','linewidth',1.5);
title('Shear Force Diagram','fontsize',16)
text(a/2,V(1),num2str(V(1)),'HorizontalAlignment','center','FontWeight','bold','fontsize',16)
text((L-c/2),V(end),num2str(V(end)),'HorizontalAlignment','center','FontWeight','bold','fontsize',16)
axis off

subplot(2,1,2);
plot(x, M, 'r','linewidth',1.5); % Grafica de momentos flectores;
grid
line([x(1) x(end)],[0 0],'Color','k');
line([x1 x1],[0 Mmax],'LineStyle','--','Color','b');
title('Bending Moment Diagram','fontsize',16)
text(x1+1/L,Mmax/2,num2str(roundn(Mmax,-2)),'HorizontalAlignment','center','FontWeight','bold','fontsize',16)
text(x1,0,[num2str(roundn(x1,-2)) ' m'],'HorizontalAlignment','center','FontWeight','bold','fontsize',16)

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