Tension

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 holding—or pulling—him 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 sense—if 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,

Fig. 2 - A rope pulling a box

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

Fig. 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 \(T\).

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.

Fig. 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:

Fig. 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:

Fig. 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:

Fig. 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.

Fig. 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.

Fig. 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.

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,

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.

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.

First, start with a free-body diagram.

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.

Fig. 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.

Fig. 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.

Fig. 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.

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.

Fig. 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.