What Is the Tension in a Tightrope?

As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight $\mathbf{\text{w}}$ and the two tensions ${\mathbf{\text{T}}}_{\text{L}}$ (left tension) and ${\mathbf{\text{T}}}_{\text{R}}$ (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset—we can see from part (b) of the figure that the magnitudes of the tensions $T_{\text{L}}$ and $T_{\text{R}}$ must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are $T_{\text{L}}$ and $T_R$. Thus, the magnitude of those forces must be equal so that they cancel each other out.

Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the $x$-axis and the vertical the $y$-axis.

First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.

Figure 4.18 When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is stationary. The small angle results in $T$ being much greater than $w$.

Consider the horizontal components of the forces (denoted with a subscript $x$):

The net external horizontal force $F_{\text{net}x}=0$, since the person is stationary. Thus,

Now, observe Figure 4.18. You can use trigonometry to determine the magnitude of $T_{\text{L}}$ and $T_{\text{R}}$. Notice that:

Equating $T_{\text{L}x}$ and $T_{\text{R}x}$:

Thus,

as predicted. Now, considering the vertical components (denoted by a subscript $y$), we can solve for $T$. Again, since the person is stationary, Newton’s second law implies that net $F_y=0$. Thus, as illustrated in the free-body diagram in Figure 4.18,

Observing Figure 4.18, we can use trigonometry to determine the relationship between $T_{\text{L}y}$, $T_{\text{R}y}$, and $T$. As we determined from the analysis in the horizontal direction, $T_{\text{L}}=T_{\text{R}}=T$:

Now, we can substitute the values for $T_{\text{L}y}$ and $T_{\text{R}y}$, into the net force equation in the vertical direction:

and

so that

and the tension is

Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker.