<h2>Tension</h2>
<p>A <strong>tension</strong> is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. The word “tension<em>” </em>comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called <em>tendons</em>.</p>
<p>Any flexible connector, such as a string, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. It is important to understand that tension is a pull in a connector. In contrast, consider the phrase: “You can’t push a rope.” The tension force pulls outward along the two ends of a rope.</p>
<p>Consider a person holding a mass on a rope as shown in the figure below.</p>
<div class="wp-caption alignnone" style="width: 323px"><img alt="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." height="450" src="https://data.ulearngo.com/assets/82eb8759-bd0d-4581-be9c-43c4f7e2495a" width="323"/><p class="wp-caption-text">When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force \(\mathbf{\text{T}}\), 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.</p></div>
<p>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 \({\mathbf{\text{F}}}_{\text{net}}=0\). The only external forces acting on the mass are its weight \(\mathbf{\text{w}}\) and the tension \(\mathbf{\text{T}}\) supplied by the rope. Thus,</p>
<div>\({F}_{\text{net}}=T-w=0,\)</div>
<p>where \(T\) and \(w\) 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:</p>
<div>\(T=w=\text{mg}.\)</div>
<p>For a 5.00-kg mass, then (neglecting the mass of the rope) we see that</p>
<div>\(T=\text{mg}=\left(\text{5.00 kg}\right)\left(9\text{.}{\text{80 m/s}}^{2}\right)=\text{49.0 N}.\)</div>
<p>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.</p>
<p>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 the (a) and (b) parts of the figure below.</p>
<div class="wp-caption alignnone" style="width: 368px"><img alt="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." height="450" src="https://data.ulearngo.com/assets/63d472b7-2384-4d71-956e-055b940854b4" width="368"/><p class="wp-caption-text">(a) Tendons in the finger carry force \(\mathbf{\text{T}}\) 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 \(\mathbf{\text{T}}\) from the handlebars to the brake mechanism. Again, the direction but not the magnitude of \(\mathbf{\text{T}}\) is changed.</p></div>
<div class="ns-school-emp">
<h3>Example: What Is the Tension in a Tightrope?</h3>
<div class="ns-school-defn">
<p>Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in the figure below.</p>
<div class="wp-caption alignnone" style="width: 450px"><img alt="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." height="234" src="https://data.ulearngo.com/assets/2bc4ae3c-fdca-4058-92fb-239ef1e10994" width="450"/><p class="wp-caption-text">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.</p></div>
<p><strong>Strategy</strong></p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p><strong>Solution</strong></p>
<p>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.</p>
<div class="wp-caption alignnone" style="width: 450px"><img alt="A vector T sub L making an angle of five degrees with the negative x axis is shown. It has two components, one in the vertical direction, T sub L y, and another horizontal, T sub L x. Another vector is shown making an angle of five degrees with the positive x axis, having two components, one along the y direction, T sub R y, and the other along the x direction, T sub R x. In the free-body diagram, vertical component T sub L y is shown by a vector arrow in the upward direction, T sub R y is shown by a vector arrow in the upward direction, and weight W is shown by a vector arrow in the downward direction. The net force F sub y is equal to zero. In the horizontal direction, T sub R x is shown by a vector arrow pointing toward the right and T sub L x is shown by a vector arrow pointing toward the left, both having the same length so that the net force in the horizontal direction, F sub x, is equal to zero." height="157" src="https://data.ulearngo.com/assets/6b92c325-3098-43a9-860f-243e0a2f71f0" width="450"/><p class="wp-caption-text">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 \(w\).</p></div>
<p>Consider the horizontal components of the forces (denoted with a subscript \(x\)):</p>
<p>\({F}_{\text{net}x}={T}_{\text{L}x}-{T}_{\text{R}x}.\)</p>
<p>The net external horizontal force \({F}_{\text{net}x}=0\), since the person is stationary. Thus,</p>
<p>\({F}_{\text{net}x} = 0 = {T}_{\text{L}x}-{T}_{\text{R}x}\)<br/>
 \({T}_{\text{L}x} = {T}_{\text{R}x}.\)</p>
<p>Now, observe the figure above. You can use trigonometry to determine the magnitude of \({T}_{\text{L}}\) and \({T}_{\text{R}}\). Notice that:</p>
<p>\(\begin{array}{lll}\text{cos}(5.0^{\circ}) &amp; = &amp; \cfrac{{T}_{\text{L}x}}{{T}_{\text{L}}} \\ {T}_{\text{L}x} &amp; = &amp; {T}_{\text{L}}\text{ cos}(5.0^{\circ}) \\ \text{cos}(5.0^{\circ}) &amp; = &amp; \cfrac{{T}_{\text{R}x}}{{T}_{\text{R}}} \\ {T}_{\text{R}x} &amp; = &amp; {T}_{\text{R}}\text{ cos}(5.0^{\circ}) \end{array}\)</p>
<p>Equating \({T}_{\text{L}x}\) and \({T}_{\text{R}x}\):</p>
<p>\({T}_{\text{L}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)={T}_{\text{R}}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right).\)</p>
<p>Thus,</p>
<p>\({T}_{\text{L}}={T}_{\text{R}}=T,\)</p>
<p>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 the figure above,</p>
<p>\({F}_{\text{net}y}={T}_{\text{L}y}+{T}_{\text{R}y}-w=0.\)</p>
<p>Observing the figure above, 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, \({T}_{\text{L}}={T}_{\text{R}}=T\):</p>
<p>\(\begin{array}{lll}\text{sin}(5.0^{\circ}) &amp; = &amp; \cfrac{{T}_{\text{L}y}}{{T}_{\text{L}}} \\ {T}_{\text{L}y} = {T}_{\text{L}} \text{sin}(5.0^{\circ}) &amp; = &amp; T\text{ sin}(5.0^{\circ}) \\ \text{sin}(5.0^{\circ}) &amp; = &amp; \cfrac{{T}_{\text{R}y}}{{T}_{\text{R}}} \\ {T}_{\text{R}y} = {T}_{\text{R}} \text{sin}(5.0^{\circ}) &amp; = &amp; T\text{ sin}(5.0^{\circ})\end{array}\)</p>
<p>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:</p>
<p>\(\begin{array}{lll}F_{\text{nety}} &amp; = &amp; T_{\text{Ly}} + T_{\text{Ry}} - w = 0 \\ F_{\text{nety}} &amp; = &amp; T\text{ sin }(5.00^{\circ}) + T\text{ sin }(5.00^{\circ}) - w = 0 \\ 2T\text{ sin }(5.00^{\circ}) - w &amp; = &amp; 0 \\ 2T\text{ sin }(5.00^{\circ}) &amp; = &amp; w \end{array}\)</p>
<p>and</p>
<p>\(T=\cfrac{w}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)}=\cfrac{\text{mg}}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(5.0º\right)},\)</p>
<p>so that</p>
<p>\(T=\cfrac{\left(\text{70}\text{.}\text{0 kg}\right)\left(9\text{.}{\text{80 m/s}}^{2}\right)}{2\left(0\text{.}\text{0872}\right)},\)</p>
<p>and the tension is</p>
<p>\(T=\text{3900 N}.\)</p>
<p><strong>Discussion</strong></p>
<p>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.</p>
<p>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.</p>
</div>
</div>
<p>If we wish to <em>create</em> a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in the figure below. As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way:</p>
<div>\(T=\cfrac{w}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}.\)</div>
<p>We can extend this expression to describe the tension \(T\) created when a perpendicular force (\({\mathbf{\text{F}}}_{\perp }\)) is exerted at the middle of a flexible connector:</p>
<div>\(T=\cfrac{{F}_{\perp }}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}.\)</div>
<p>Note that \(\theta \) is the angle between the horizontal and the bent connector. In this case, \(T\) becomes very large as \(\theta \) approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., \(\theta =0\) and \(\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =0\)). (See <a href="#import-auto-id2847184">the figure below</a>.)</p>
<div class="wp-caption alignnone" style="width: 800px"><img alt="A car stuck in mud is being pulled out by a chain tied to a tree trunk. A force perpendicular to the length of the chain is applied, represented by an arrow. The tension T along the chain makes an angle with the horizontal line." height="131" src="https://data.ulearngo.com/assets/f37dc4a6-1399-416e-af10-3a9434661a47" width="800"/><p class="wp-caption-text">We can create a very large tension in the chain by pushing on it perpendicular to its length, as shown. Suppose we wish to pull a car out of the mud when no tow truck is available. Each time the car moves forward, the chain is tightened to keep it as nearly straight as possible. The tension in the chain is given by \(T=\frac{{F}_{\perp }}{2\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\left(\theta \right)}\) ; since \(\theta \) is small, \(T\) is very large. This situation is analogous to the tightrope walker shown in the example below, except that the tensions shown here are those transmitted to the car and the tree rather than those acting at the point where \({\mathbf{\text{F}}}_{\perp }\) is applied.</p></div> <div class="wp-caption alignnone" style="width: 450px"><img alt="A picture of the Golden Gate Bridge." height="337" src="https://data.ulearngo.com/assets/e5891204-2167-447e-aefb-aceb834628c5" width="450"/><p class="wp-caption-text">Unless an infinite tension is exerted, any flexible connector—such as the chain at the bottom of the picture—will sag under its own weight, giving a characteristic curve when the weight is evenly distributed along the length. Suspension bridges—such as the Golden Gate Bridge shown in this image—are essentially very heavy flexible connectors. The weight of the bridge is evenly distributed along the length of flexible connectors, usually cables, which take on the characteristic shape. (credit: Leaflet, Wikimedia Commons)</p></div>

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Tension

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Tension

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