An acrobat is walking on a tightrope of length L = 80.3 ft attached to supports A and B at a
distance of 80.0 ft from each other. The combined weight of the acrobat and his balancing pole is
200 lb, and the friction between his shoes and the rope is large enough to prevent him from
slipping. Neglecting the weight of the rope and any elastic deformation, use computational
software to determine the deflection y and the tension in portion AC and BC of the rope for
values of x from 0.5 ft to 40 ft using 0.5-ft increments.
From the results obtained, determine: (a) the maximum deflection of the rope and (b)
the maximum tension in the rope, and (c) the minimum values of the tension in portions
AC and BC of the rope.
Solution
% deflection is y
%AC=sqrt(y^2+x^2);
%BC=sqrt(y^2+(80-x)^2);
% length of rope = 80.3 ft. So AC+BC = 80.3 ft
% sqrt(y^2+x^2)+sqrt(y^2+(80-x)^2)=80.3
% we need to solve for y in the above equation for given values of x
y=zeros(1,80);
i=1;
for x=0.5:0.5:40
y(i)=fsolve(@(y) sqrt(y^2+x^2)+sqrt(y^2+(80-x)^2)-80.3,0.1);
i=i+1;
end
x=0.5:0.5:40;
plot(x,y)
deflection y is maximum when x=40ft, ymax = 3.4673 ft.
An acrobat is walking on a tightrope of length L = 80.3 ft attached .pdf
1. An acrobat is walking on a tightrope of length L = 80.3 ft attached to supports A and B at a
distance of 80.0 ft from each other. The combined weight of the acrobat and his balancing pole is
200 lb, and the friction between his shoes and the rope is large enough to prevent him from
slipping. Neglecting the weight of the rope and any elastic deformation, use computational
software to determine the deflection y and the tension in portion AC and BC of the rope for
values of x from 0.5 ft to 40 ft using 0.5-ft increments.
From the results obtained, determine: (a) the maximum deflection of the rope and (b)
the maximum tension in the rope, and (c) the minimum values of the tension in portions
AC and BC of the rope.
Solution
% deflection is y
%AC=sqrt(y^2+x^2);
%BC=sqrt(y^2+(80-x)^2);
% length of rope = 80.3 ft. So AC+BC = 80.3 ft
% sqrt(y^2+x^2)+sqrt(y^2+(80-x)^2)=80.3
% we need to solve for y in the above equation for given values of x
y=zeros(1,80);
i=1;
for x=0.5:0.5:40
y(i)=fsolve(@(y) sqrt(y^2+x^2)+sqrt(y^2+(80-x)^2)-80.3,0.1);
i=i+1;
end
x=0.5:0.5:40;
plot(x,y)
deflection y is maximum when x=40ft, ymax = 3.4673 ft
