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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 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 = m g .

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

T = m g = ( 5.00 kg ) ( 9.80 m/s 2 ) = 49.0 N .

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.

Figure shows mass m hanging from a rope. Two arrows of equal length, both labeled T are shown along the rope, one pointing up and the other pointing down. An arrow labeled w points down. A free body diagram shows T pointing up and w pointing down.
When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T , that force must be parallel to the length of the rope, as shown. By Newton’s third law, the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along 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] .

Figure a shows the muscle structure of a human finger. Broad muscles at the base are labeled extensor muscles. These are attached to the extensor tendons. Tendons along the length of the finger are labeled flexor tendons. Arrows labeled T are shown from the upper part of the finger towards the base. Figure b shows a bicycle. Arrows labeled T are shown from the centre of the back wheel to the seat bar, from the seat bar to the handle bar and from the handle towards the back of the bicycle.
(a) Tendons in the finger carry force T from the muscles to other parts of the finger, usually changing the force’s direction but not its magnitude (the tendons are relatively friction free). (b) The brake cable on a bicycle carries the tension T from the brake lever on the handlebars to the brake mechanism. Again, the direction but not the magnitude of T is changed.

What is the tension in a tightrope?

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

Figure shows a man at the centre of a tightrope which is supported by two poles. The rope sags under his weight and makes an angle of 5 degrees with the horizontal at each pole. Arrows labeled TL and TR point roughly to the left and right respectively and are parallel to the rope. Arrow labeled w points straight down from the man. These three arrows are also shown in a free body diagram.
The weight of a tightrope walker causes a wire to sag by 5.0 ° . The system of interest is the point in the wire at which the tightrope walker is standing.


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 w and the two tensions T L (left tension) and T 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 L and T 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 L and T 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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