Tightrope problem

[549]

Hi,

If a person is standing on a tightrope which is anchored on both sides to support beams, causing the rope the be pushed down at an angle of ø, what happens to ø if the support beams are moved so now they are twice the original distance apart?

Apparently ø does not change, but the way I calculated it was that if Tension is not changing, and cosø is d/T, and d is now doubled, then won't cosø=2d/T, which would mean that ø has decreased??

Thanks!

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[Isoprop]

cos(phi) does not equal d/T. If I multiply T by both sides of the equation, I get

Tcos(phi) = d

This says that the x component of the tension is equal to the horizontal distance. Units show that this cannot be the case because the left side would be SI units of newtons and the right side would be units of meters.

Since the tightrope walker is in equilibrium, the forces add up to zero. Therefore the y components of the tension should equal the weight of the person. Let's assume the person is in the middle of his tightrope walk. The tension of the rope is equal on both sides of the man. Then,

2Tsin(phi) = mg

Distance of the rope is nowhere in this equation. The angle only depends on tension of the rope and weight of the person.

 

[549]

Okay- yeah I later realized that it was stupid to put Force and distance in the same equation. Thanks for the extra clarification!