Question

A woman is balancing on a high wire, which is tightly strung, as shown in Fig. 9–45. The tension in the wire is

(a) about half the woman’s weight.

(b) about twice the woman’s weight.

(c) about equal to the woman’s weight.

(d) much less than the woman’s weight.

(e) much more than the woman’s weight.

Short answer

The correct option is (e).

Step-by-step solution

Concepts

At equilibrium, the net force is zero. For this problem, find the condition for vertical forces.

Explanation

Let m be the mass of the woman and T be the tension in the wire.

At equilibrium for the vertical forces,

\(\begin{aligned}{c}T\sin \theta + T\sin \theta = mg\\2T\sin \theta = mg\\\sin \theta = \frac{{mg}}{{2T}}\end{aligned}\).

You can observe that the angle of the rope with the horizontal is very small and less than \({30^ \circ }\). Then,

\(\begin{aligned}{c}\sin \theta < \sin {30^ \circ }\\\frac{{mg}}{{2T}} < \frac{1}{2}\\T > mg\end{aligned}\).

Therefore, the tension in the wire is more than the woman’s weight.