Answer to Question #96125 in Mechanics | Relativity for d

Question

A 500-N tightrope walker stands at the center of the rope. If the rope can withstand a tension of 1 800 N without breaking, what is the minimum angle the rope can make with the horizontal?

Answer 1

Let $T$ be the tension in the rope.

$\theta$ be the angle the rope make wrt. horizontal.

Vertical component of Tension will balance the weight of tight rope Walker,

So,

$$2T\sin\theta=weight=500$$

Now,

Maximum tension allowed = 1800 N

Therefore,

$$2×1800×\sin\theta=500\\or\\\sin\theta=\frac{5}{36}\\or,\\\theta=7.983\degree$$

So minimum angle that the rope can make is $7.983\degree$