A tightly stretched horizontal “high wire” is 36 m long. It sags vertically 2.1 m when a 60.0-kg tightrope walker stands at its center. What is the tension in the wire? Is it possible to increase the tension in the wire so that there is no sag?

The length of the wire when the tightrope walker stands on it is\(L = 36\;{\rm{m}}\).

The wire sags vertically by a distance of\(h = 2.1\;{\rm{m}}\).

The mass of the tightrope walker is\(m = 60\;{\rm{kg}}\).

A free body diagram (FBD) is a simplified representation of a rigid body to visualize all the force vectors acting on it.The free body shown below depicts the forces on the wire.

The forces acting on the wire are the following:

• Weight of the tightrope walker,\(W = mg\)vertically downwards
• Tension force,\({F_{\rm{T}}}\)in the wire

By symmetry, the tension forces on both sides of the wire have the same magnitude. The rope sags at an angle of\(\theta \)below the horizontal.

From the FBD,

\(\begin{array}{c}\sin \theta = \frac{{2.1\;{\rm{m}}}}{{18\;{\rm{m}}}}\\\theta = {\sin ^{ - 1}}\left( {\frac{{2.1\;{\rm{m}}}}{{18\;{\rm{m}}}}} \right)\\ = {6.70^{\rm{o}}}\end{array}\)

From the FBD, the forces along the vertical direction are given by the equilibrium condition. Applying Newton’s second law gives,

\(\begin{array}{c}\sum {{F_{\rm{y}}}} = 0\\2{F_{\rm{T}}}\sin \theta - mg = 0\\{F_{\rm{T}}} = \frac{{mg}}{{2\sin \theta }}\end{array}\)

Substitute the known numerical values in the above expression.

\(\begin{array}{c}{F_{\rm{T}}} = \frac{{\left( {60\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{2\sin \left( {{{6.70}^{\rm{o}}}} \right)}}\\ = 2519.91\;{\rm{N}}\\ \approx {\rm{2500}}\;{\rm{N}}\end{array}\)

If the tension in the wire increases, the sag angle decreases. It is impossible to increase the tension by eliminating the sag.

There will always be a vertical component of the tension force to balance the weight of the person and the cable.