Question

use matlab:

A cable of length Lc supports a beam of length Lb so that it is horizontal when the weight W is attached to the beam end. The tension force T in the cable is given by the following equation: T=(Lb*Lc*W)/(D*sqrt(Lc^2-D^2)) Where D is the distance of the cable attachment point to the beam pivot. 1. Create a function M-file that evaluates the tension T as a function of D. In this function M-file the parameters Lb, Lc, and W should be declared global. 2. Create a script M-file that creates these global variables and: a) Prompts for and accepts the following values from the Command Window : W = 400N, Lb= 5 m, and Lc = 3 m (the weight is in Newtons, and the lengths are in meters) b) Uses these values and the fminbnd function to compute and display in the Command Window the value of D that minimizes the tension, T, and the minimum tension value. c) Uses fzero to determine by how much the value of D can vary from it

Answer 1

(a). The M-file:

x = 10; W = 400;
Lb = 3;
Lc = 5;
D = [0:0.01:Lb];
T = Lb*Lc*W./(D.*sqrt(Lb^2-D.^2));
[minT, k] = min(T)
minD = D(k)
Tension_at_min=T(k)
Distance_at_min=D(k)
%plotting
Dplot = [1.5:0.001:2.2];
upper = 1.1*minT
Tplot = Lb*Lc*W./(Dplot.*sqrt(Lb^2-Dplot.^2));
plot(Dplot,Tplot,[1.5,2.2],[upper,upper])
xlabel('D (mm)');
ylabel('T (N)')
grid

minT =  1333.3
k =  213
minD =  2.1200
Tension_at_min =  1333.3
Distance_at_min =  2.1200
upper =  1466.7

(b), In Command Window:

minT =  1333.3
k =  213
minD =  2.1200
Tension_at_min =  1333.3
Distance_at_min =  2.1200
upper =  1466.7

(c) The upper tension value is 1.1(1333) = 1467 N. The intersection of the two lines on the plot gives the solution, which is approximately D = 1.6 m (1.62 is a more accurate value)

I hope its helped. Let me know whether the part (c) is correct or not.