An acrobat is walking on a tightrope of length L = 20 l m attached to supports A

Question

An acrobat is walking on a tightrope of length L = 20 l m attached to supports A and B at a distance 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 tape is large enough to prevent him slipping. Neglecting the weight of the rope in large elastic deformation, write a computer Program calculate the deflection y and the tension in positions AC and BC of the rope for values of from 05 to 10.0 m using 0.5 m increments. From the data obtained, determine (a) the maximum deflection of the rope, (b) the maximum tension in the rope, (c) the smallest values of the tension in portions AC and BC of the rope.

Explanation / Answer

start=0.5 stop=3.20/4 n=15

Y(start,stop,n,j,LAB,LT )

alpha=acos(LAB-xj/LT-L)

Tita= =acos(xj/l)

Y=Lsin

delta=stop-start/n-1 where i=1,2..n

L=LT2-LAB2+

TAC=Wsin(90deg - )/sin( + )

TBC =Wsin(90deg - )/sin( + )

sorting="maximum"if xj=max(x)

sorting="minimum" if xj=min(x)

Y(start, Stop,n ,1 5,8 10,2 0,2 0.1) = (0.869 3503.104 3549.618 "maximum")
Y(start, stop,n ,1 ,8 10,2 0,2 0.1) = (0.327 1443.859 1208.898 "minimum")

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