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Consider a person holding a mass on a rope as shown in [link] .

An object of mass m is attached to a rope and a person is holding the rope. A weight vector W points downward starting from the lower point of the mass. A tension vector T is shown by an arrow pointing upward initiating from the hook where the mass and rope are joined, and a third vector, also T, is shown by an arrow pointing downward initiating from the hand of the person.
When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T size 12{T} {} , that force must be parallel to the length of the rope, as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton’s third law. 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.

Tension in the rope must equal the weight of the supported mass, as we can prove using Newton’s second law. If the 5.00-kg mass in the figure is stationary, then its acceleration is zero, and thus F net = 0 size 12{F rSub { size 8{"net"} } =0} {} . The only external forces acting on the mass are its weight w size 12{w} {} and the tension T size 12{T} {} supplied by the rope. Thus,

F net = T w = 0 size 12{F rSub { size 8{"net"} } =T - w=0} {} ,

where T size 12{T} {} and w size 12{w} {} are the magnitudes of the tension and weight and their signs indicate direction, with up being positive here. Thus, just as you would expect, the tension equals the weight of the supported mass:

T = w = mg size 12{T=w= ital "mg"} {} .

For a 5.00-kg mass, then (neglecting the mass of the rope) we see that

T = mg = ( 5.00 kg ) ( 9 . 80 m/s 2 ) = 49.0 N size 12{T= ital "mg"= \( 5 "." "00"" kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) ="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.

Flexible connectors are often used to transmit forces around corners, such as in a hospital traction system, a finger joint, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. Only its direction changes, and it is always parallel to the flexible connector. This is illustrated in [link] (a) and (b).

The internal structure of a finger with tendon, extensor muscle, and flexor muscle is shown. The force in the muscles is shown by arrows pointing along the tendon. In the second figure, part of a bicycle with a brake cable is shown. Three tension vectors are shown by the arrows along the brake cable, starting from the handle to the wheels. The tensions have the same magnitude but different directions.
(a) Tendons in the finger carry force T size 12{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 size 12{T} {} from the handlebars to the brake mechanism. Again, the direction but not the magnitude of T size 12{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] .

A tightrope walker is walking on a wire. His weight W is acting downward, shown by a vector arrow. The wire sags and makes a five-degree angle with the horizontal at both ends. T sub R, shown by a vector arrow, is toward the right along the wire. T sub L is shown by an arrow toward the left along the wire. All three vectors W, T sub L, and T sub R start from the foot of the person on the wire. In a free-body diagram, W is acting downward, T sub R is acting toward the right with a small inclination, and T sub L is acting toward the left with a small inclination.
The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of interest here is the point in the wire at which the tightrope walker is standing.


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 w size 12{w} {} and the two tensions T L size 12{T rSub { size 8{L} } } {} (left tension) and T R size 12{T rSub { size 8{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 L size 12{T rSub { size 8{L} } } {} and T R size 12{T rSub { size 8{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 L size 12{T rSub { size 8{L} } } {} and T R size 12{T rSub { size 8{R} } } {} . Thus, the magnitude of those forces must be equal so that they cancel each other out.

Questions & Answers

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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