rendi Bilgi Sistemi ster Problem I: An acrobat is walking on a tightrope of leng
ID: 1842012 • Letter: R
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rendi Bilgi Sistemi ster Problem I: An acrobat is walking on a tightrope of length L-20 attached to supports and B at a of 20.0 m from each other. The combined weight of the acrobat and his balancing pole is 800 N, 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, write a computer program to calculate the deflection y and the tension in portions AC and BC of the rope for values of x from 0.5 m to 10.0 m using 0.5-m increments. From the data obtained, determine (a) the maximum deflection of the rope, (by the maximum tension in the rope, (c) the smallest values of the tension in portions ACand BCof the rope. 20,0 mExplanation / Answer
The problem can be solved analytically quite simply:
The maximum deflection point is at the centre of the rope;
if the length of the cable is 20.1m
the sag is calculated as 10.05 sin(alpha), alpha = inverse cos ( 10/.10.05)=5.7176 degrees
sag = 10.05 sin(5.7176) =1 metre.
by the formulas shown: Tac = W sin (alpha)/sin(2 alpha) = W/2 (cos alpha) =W/2( 10.05)/10 = 402 N
c) smallest value of tension when man is at A and B=0
from the given formula for y:
put a=20.1
simplifies to y =1/40*sqrt( 16+8*40.2x+16.04*X^2)
alpha and beta canbe found from tan apha = x/(10.05-x) approx, or accurately as taninv y/x where y from above.
Program on calculator gives y as a function of x,
tensions also easily done from TAc T Bc formulae
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