Tension

Tension isn't just the feeling you have when you're about to take a test. In regards to physics, tension is a type of force. The tension force acts similarly to other applied forces, such as if you were to pull a box across the floor. However, instead of using your hands to pull the box, you would pull the box with a rope, cord, chain, or similar object for it to count as tension. Because tension is similar to an applied force, it has no specific equation or formula. An example of tension is when a dog pulls on the leash while you take him for a walkthe leash pulls you forward with a tension force. 

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    Tension Definition

    The suspense is killing me! What is tension? Tension is a type of contact force exerted by use of a rope or cord.

    In physics, we define tension as the force that occurs when a rope, cord, or similar item pulls on an object. There are two forces at opposite sides of the rope creating the tension.

    Tension is a pulling force (because you can't push with a rope) and acts in the direction of the rope. We consider tension a contact force since the rope has to touch the object to exert a force on it.

    Tension in Physics

    One thing to note is that a rope under tension applies the same force to each attached object. For instance, when we mentioned walking a dog, we described how the dog pulling on the leash would apply a tension force on you. If we were only interested in the forces acting on you, that is all we would care about. But what if we also wanted to know the forces acting on the dog? We would notice that as the dog pulls on the leash, there is a force holdingor pullinghim back too. The tension force pulling you forward is the same (has the same magnitude) as the tension force holding him back. As seen below, we can apply two arrows across the leash to show these two forces.

    The Forces of Tension

    Tension Results from Interatomic Electric Forces. Interatomic electric forces are the cause of all contact forces. For tension, the rope is made up of many atoms and molecules that are bonded together. As the rope becomes tight under the force, one of the bonds between atoms is stretched farther apart on a microscopic level. The atoms want to stay close in their natural state, so the electric forces holding them together increase. All these tiny forces add together to create one tension force. This principle helps the arrows in Figure 1 make more senseif the dog and person are pulling outward on the leash, the forces keeping the leash together are directed towards the leash.

    Tension Equation

    There is no equation specific to tension force like there is for friction and spring forces. Instead, we need to use a free-body diagram and Newton's Second Law of Motion to solve the tension.

    Solve for Tension Using a Free-Body Diagram and Newton's Second Law

    Free-body diagrams help us visualize the forces acting on an object. For a box pulled along the floor by a rope, as shown in the figure below,

    Tension Rope Pulling Box StudySmarterFig. 2 - A rope pulling a box

    we would include arrows for all forces acting on the box.

    Tension Free-Body Diagram of Box StudySmarterFig. 3 - Here are all the forces acting on the box.

    This figure includes all forces that could be in play in this situation, including friction \(F_\text{f} \), gravity \(F_g\), normal \(F_\text{N} \), and tension \(T\).

    Remember: Always draw tension force arrows away from the object. Tension is a pulling force, so the force will always be directed outward.

    Newton's Second Law of Motion states that the acceleration of an object depends on the force acting on the object and the mass of the object

    The following equation,

    $$\sum \vec F =m\vec a\mathrm{,}$$

    is a result of Newton's Second Law.

    This equation applies to each direction, so typically, we want to include one for the \(y\)-direction and one for the \(x\)-direction. In our example in the figures above, there isn't any tension acting in the \(y\)-direction, so to solve for tension we can focus on the \(x\)-direction, where we have a friction force acting to the left and tension acting to the right. Choosing the right to be positive, our resulting equation looks like this:

    $$-F_\text{f} + T =ma\mathrm{.}$$

    Then we can rearrange to solve for tension:

    $$T=ma+F_\text{f} \mathrm{.}$$

    If the box is on a frictionless surface, the friction force is zero, so the tension would equal the box's mass times the box's acceleration.

    Examples of Tension

    In your physics problems, you may see many real-life scenarios involving tension such as:

    • Cars towing trailers
    • Tug of War
    • Pulleys and Ropes
    • Gym Equipment

    These may seem very different scenarios, but you will use the same method to solve each. Below are some problems you might see and strategies to solve them.

    Rope Between Two Objects

    Now, let's mix things up and do an example with two objects connected by a rope.

    Tension Example Rope Between Two Objects StudySmarterFig. 4 - Rope between two objects.

    The above figure shows a rope between two boxes and one pulling box 2 to the right. As we mentioned with the dog leash, the tension acting on box 1 is the same as on box 2 since it's the same rope. Therefore, in the figure, we labeled them both the same \(T_1 \).

    In any problem, we can choose which object, or group of objects, to analyze in a free-body diagram. Let's say we wanted to find \(T_1 \) and \(T_2 \). We might want to start by looking at box 1 because it's the simpler side, with only one unknown we're looking for. The following figure shows the free-body diagram for box 1:

    Tension Example Free-Body Diagram for Mass 1 StudySmarterFig. 5 - Free-body diagram of box 1.

    Since the tension acts only in the \(x\)-direction, we can disregard the forces acting in the \(y\)-direction. Picking right as positive, Newton's Second Law equation would look like this:

    $$-F_{\text{f}1} +T_1 = m_1 a\mathrm{.}$$

    We can then rearrange variables to solve for \(T_1 \)

    $$T_1 = m_1 a + F_{\text{f}1}\mathrm{;}$$

    to find \(T_2 \), we could look at the forces only on box 2, shown here:

    Tension Example Free-Body Diagram for Mass 2 StudySmarterFig. 6 - Free-body diagram of box 2.

    Again ignoring the \(y\)-direction, the equation for the \(x\)-direction is the following:

    $$-T_1 - F_{\text{f}2} + T_2 = m_2 a\mathrm{.}$$

    Because we know that \(T_1 \) is the same for each box, we can take the \(T_1 \) we learned from box 1 and apply it to box 2 by substitution

    $$-(m_1 a + F_{\text{f}1}) - F_{\text{f}2} +T_2 = m_2 a$$

    and then we can solve for \(T_2 \),

    $$T_2 = (m_2 + m_1 )a + F_{\text{f}1} + F_{\text{f}2}\mathrm{.}$$

    However, if we don't need to know \(T_1 \), we can always look at both boxes together as if they were one. Below, we can see what the free-body diagram looks like when you group the two boxes:

    Tension Example Free-Body Diagram for Two Masses StudySmarterFig. 7 - Free-body diagram of both boxes together.

    If we write Newton's Second Law equation for the \(x\)-direction, we get

    $$-(F_{\text{f}1} + F_{\text{f}2})+T_2 = (m_1 +m_2 )a$$

    and can rearrange it to solve for \(T_2 \),

    $$T_2 = (m_1 + m_2 )a + F_{\text{f}1} + F_{\text{f}2}\mathrm{.}$$

    We can see that this yields the same result as when we looked at the boxes separately and then pieced the equations together. Either method works to find \(T_2 \) (you can decide which is easier and use either), but sometimes the variable you need to solve for can only be found by focusing on one specific object.

    Pulling at an angle

    Now, let's do an example with everyone's favorite: angles.

    Tension Tension Force on Box at an Angle StudySmarterFig. 8 - Rope pulling at an angle.

    In the figure above, the rope pulls on the box at an angle instead of along the horizontal surface. As a result, the box slides across the surface horizontally. To solve for tension, we would use the superposition of forces to split the angled force into the part of the force that acts in the \(x\)-direction and the part of the force that acts in the \(y\)-direction.

    Tension Example Free-Body Diagram of Rope at an Angle StudySmarterFig. 9 - Free-body diagram with tension split into \(x\) and \(y\) components.

    This is shown in red in the figure of the free-body diagram above. Then we can write a separate equation for the \(x\)-direction and the \(y\)-direction according to the free-body diagram.

    \(T_x = T\cos{\theta}\) and \(T_y = T\sin{\theta}\).

    In this example, we now have some tension acting in the \(y\)-direction, so we don't want to ignore the gravitational and normal force as we did in the examples above. Since the box isn't accelerating in the \(y\)-direction, the sum of the forces in the \(y\)-direction equals zero

    $$F_\text{N} + T\sin{\theta} -F_g =0\mathrm{,}$$

    and rearranging to find \(T\) yields

    $$T=\frac{F_g - F_\text{N} }{\sin{\theta}}\\\mathrm{.}$$

    The \(x\)-direction looks similar to what we've done above, but with just the \(x\) component of the angled tension force:

    $$-F_\text{f} + T\cos{\theta} = ma\mathrm{.}$$

    Then, we rearrange to find \(T\):

    $$T=\frac{ma+F_\text{f}}{\cos{\theta}}\\\mathrm{.}$$

    Both of these results will give you the same value for \(T\), so depending on what information you're given, you can choose either to focus on just the \(x\)-direction, just the \(y\)-direction, or both.

    Free-Hanging Object

    When an object hangs from a rope, as shown below,

    Tension Free-Hanging Object StudySmarter

    Fig. 10 - Object hanging from a rope

    the only forces on it are the gravitational force pulling it down and the tension holding it up.

    This is shown in the free-body diagram below.

    Tension Free-Body Diagram of Free-Hanging Object

    Fig. 11 - Free-body diagram of an object hanging from a rope

    The resulting equation would look like the following:

    $$T-F_g =ma\mathrm{.}$$

    If we rearrange to find \(T\) and substitute \(mg\) for the gravitational force, we get

    $$T=ma +mg\mathrm{.}$$

    If the object is not accelerating, the tension and gravitational force would be equal and opposite, so \(T=mg\).

    Pulling on an Angled Surface

    When tension is applied to a box on an angled surface, we use a similar strategy as when the rope was pulling at an angle.

    Tension on an Object on an Incline StudySmarter

    Fig. 12 - Tension on an object on an incline

    First, start with a free-body diagram.

    Tension Example Free-Body Diagram on Angled Surface StudySmarter

    Fig. 13 - Free-body diagram of tension on an angled surface

    When dealing with an angled surface, remember that the normal force always acts perpendicular to the surface, and the gravitational force (weight) always acts straight down.

    Instead of breaking the tension force into \(x\) and \(y\) components, we want to break the gravitational force into components. If we tilt our coordinate system to match the angle of the surface, as seen below, we can see that the tension acts in the new \(x\)-direction, and the normal force acts in the new \(y\)-direction. The gravitational force is the only force at an angle, so that we would split it into components following the new \(x\) and \(y\) directions, shown in red below.

    Tension Example Free-Body Diagram With Weight in Components StudySmarterFig. 14 -Free-body diagram with new coordinate system and gravitational force split into \(x\) and \(y\) components

    Then we would apply Newton's Second Law in each direction, just like any other problem.

    Hanging from Two Ropes

    When an object hangs from multiple ropes, the tension isn't equally distributed across the ropes unless the ropes are at the same angles.

    Tension Example Object Hanging From Two Ropes StudySmarterFig. 15 - Object hanging from two ropes

    We'll plug in real numbers in this example to find \(T_1 \) and \(T_2 \).

    First, we start with a free-body diagram.

    Tension Example Free-Body Diagram of Object Hanging From Two Ropes StudySmarterFig. 16 - Free-body diagram of an object hanging from two ropes

    This box isn't moving, so the acceleration is zero; thus, the sum of the forces in each direction equals zero. We chose our up and right as positive, so in the \(x\)-direction, using only the \(x\) components of the tensions, the equation would be

    $$-T_1 \cos{45^{\circ}} + T_2 \cos{60^{\circ}} = 0\mathrm{.}$$

    In the \(y\)-direction, we have the \(y\) components of the tensions and the gravitational force:

    $$T_1 \sin{45^{\circ}} + T_2 \sin{60^{\circ}} - 15\,\mathrm{kg} \times 9.81\,\mathrm{kg/m^2}=0\mathrm{.}$$

    We can solve these two equations and two unknowns algebraically any way we are comfortable. For this example, we will solve the first equation for \(T_1 \) and substitute it for the second. Solving for \(T_1 \) gives

    $$\begin{align*} \frac{1}{\sqrt{2}} T_1 &= \frac{1}{2} T_2 \\ T_1 &= \frac{\sqrt{2}}{2} T_2 \mathrm{,} \\ \end{align*}$$

    and substituting this into the second equation to find \(T_2 \) yields

    $$\begin{align*} \frac{\sqrt{2}}{2} T_2 \times \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} T_2 - 147.15\,\mathrm{N} &= 0 \\ \frac{1+\sqrt{3}}{2} T_2 &= 147.15\,\mathrm{N} \\ T_2 &= 107.72\,\mathrm{N.} \\ \end{align*}$$

    Then plugging \(T_2 \) back into the first equation to solve for \(T_1 \) gives us a final answer of

    $$\begin{align*} T_1 &= 107.72\,\mathrm{N} \times \frac{\sqrt{2}}{2} \\ T_1 &= 76.17\,\mathrm{N.} \\ \end{align*}$$

    Pulley, Incline, and Hanging Object

    The example pictured below combines much of what we discussed in each of the above examples.

    Tension Example Pulley StudySmarter

    Fig. 17 - Incline, pulley, and hanging object

    The following figure shows what the forces on each object would look like, keeping in mind that the friction force could act in the opposite direction depending on how the system moves.

    Tension Example Free-Body Diagram for Pulley StudySmarterFig. 18 - Forces shown for the scenario above

    The following are tips we learned in each of the above problems that also apply to this one:

    • We can look at one object by itself and do an individual free-body diagram and Newton's Second Law equations.
    • The rope applies the same amount of tension on each object.
    • We can choose to tilt our coordinate system. We can even have a different coordinate system for each object if we analyze the forces on each individually. In this case, we would isolate box 2 and tilt the coordinate system to match the angle of the surface, but when we look at box 1 by itself, we would keep the coordinate system standard.
    • We can split forces into an \(x\) component and a \(y\) component. In this case, once we tilted the coordinate system on box 2, we would split the box's gravitational force into components.

    Tension - Key takeaways

    • Tension is the force that occurs when a rope (or similar item) pulls on an object.
    • Tension is caused by interatomic electric forces trying to keep the atoms of the rope together.
    • There is no equation for the tension force.
    • Use free-body diagrams and Newton's Second Law to solve for tension.
    Frequently Asked Questions about Tension

    What is tension in physics?

    In physics, tension is the force that occurs when a rope, cord, or similar item pulls on an object. 

    What is an example of tension?

    An example of tension is when someone walks a dog on a leash. If the dog pulls on the leash, the leash pulls the person forward with a tension force.

    How do you measure tension?

    Tension is measured in Newtons.

    How is tension calculated?

    Tension is calculated using free-body diagrams and Newton's Second Law (which says that the sum of the forces acting on an object equals its mass times its acceleration). This lets one solve for tension using the other forces acting on an object and the object's acceleration.

    What is the force of tension?

    The force of tension is the force that occurs when a rope, cord, or similar item pulls on an object.

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    Test your knowledge with multiple choice flashcards

    If a rope pulls an object up a frictionless slope, the tension equals the mass of the object times the acceleration of the object (\(T=ma\)).

    Which of the following ways to move a box involves tension?

    When a rope is between two objects, the tension acting on the first object has the same magnitude as the tension acting on the second object.

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