# 5.6 Common forces  (Page 4/11)

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Consider a person holding a mass on a rope, as shown in [link] . If the 5.00-kg mass in the figure is stationary, then its acceleration is zero and the net force is zero. The only external forces acting on the mass are its weight and the tension supplied by the rope. Thus,

${F}_{\text{net}}=T-w=0,$

where T and w are the magnitudes of the tension and weight, respectively, and their signs indicate direction, with up being positive. As we proved using Newton’s second law, the tension equals the weight of the supported mass:

$T=w=mg.$

Thus, for a 5.00-kg mass (neglecting the mass of the rope), we see that

$T=mg=\left(5.00\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=49.0\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}$

If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 N, providing a direct observation and measure of the tension force in the rope.

Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a tendon, or a bicycle brake cable. If there is no friction, the tension transmission is undiminished; only its direction changes, and it is always parallel to the flexible connector, as shown in [link] .

## What is the tension in a tightrope?

Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in [link] .

## Strategy

As you can see in [link] , the wire 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 that have the same direction 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 $\stackrel{\to }{w}$ and the two tensions ${\stackrel{\to }{T}}_{\text{L}}$ (left tension) and ${\stackrel{\to }{T}}_{\text{R}}$ (right tension). It is reasonable to neglect the weight of the wire. The net external force is zero, because the system is static. We can use trigonometry to find the tensions. One conclusion is possible at the outset—we can see from [link] (b) that the magnitudes of the tensions ${T}_{\text{L}}$ and ${T}_{\text{R}}$ must be equal. We know this because there is no horizontal acceleration in the rope and the only forces acting to the left and right are ${T}_{\text{L}}$ and ${T}_{\text{R}}$ . Thus, the magnitude of those horizontal components of the forces must be equal so that they cancel each other out.

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