Suppose you and your friend are walking in the park when you come across a see-saw. It's been a long time since either of you played on a see-saw, so you decide to give it a shot. Knowing that you are different in mass, however, you decide to try out an experiment. Letting go from a standing position, the see-saw dips in the direction of your friend who is a bit more heavy-set than you. At this point, you wonder how you could make it so that the see-saw balances. After trial and error, you discover that the see-saw balances once your friend moves closer to the pivot of the see-saw. What is the physical explanation for this? The answer depends on the relationship between force and torque. In this article, we'll define both terms before discussing the difference and relationship between them. Then, we'll derive torque from force and discuss how torque measurements are made.
Fig. 1 - Two friends sitting at opposite ends of a seesaw.
Equations of Force and Torque
Motivated by the fact that finding an explanation for how you and your friend managed to balance the see-saw, it's a good idea to start our discussion by reviewing the concept of force. In physics, we define a force as follows:
A force is an action exerted upon an object in order to change its state of motion either from rest or of uniform forward motion in a straight line.
The above definition implies that whenever an object undergoes a change in its state of motion, forces are involved. We can express this idea quantitatively via Newton's second law,
\[\vec{F} = m\vec{a},\]
where \(\vec{F}\) is the force acting on the object, \(m\) is its mass, and \(\vec{a}\) is the object's acceleration. Both force and acceleration are vector quantities with magnitude and direction. For this reason, we denote their symbols with a small arrow. To use Newton's second law to solve physics problems, we think of the following memory aid:
\[\text{Force} = \text{Mass} \times \text{Acceleration}.\]
Thinking back on the see-saw, we've noted that both you and your friend are pivoting about a fixed point: the center of the see-saw. So you're not quite moving in a straight line; rather, you are rotating around an axis. The presence of rotation is essential to the concept of torque, which we define as follows:
A torque is an action exerted on an object in order to change its state of torsion either from rest or of uniform angular motion around an axis.
The word torque comes from the Latin term torquere, which means to twist. Using the above definition as a reference, we can readily appreciate why physicists like to think of torque as the rotational analog of linear force. In fact, torque satisfies its own equation, which we call Newton's second law for rotation:
\[\vec{\tau} = I\vec{\alpha}.\]
In the above \(\vec{\tau}\) denotes torque, \(I\) is the moment of inertia, and \(\alpha\) (the Greek letter alpha) is the angular acceleration. Just as before, we can summarize the above equation in words as follows:
\[\text{Torque} = \text{Moment of Inertia} \times \text{Angular Acceleration}.\]
If you haven't encountered the concept of moment of inertia before, don't worry. All you need to know is that, like mass, the moment of inertia represents an object's tendency to resist changes in its state of motion. Unlike mass, however, the moment of inertia of an object varies according to its geometry. In engineering applications, the moment of inertia of an object is something you'd look up on a table.
The good news is that we can solve many physics problems involving torque with a simpler equation:
\[\begin{align}\tau &= Fr_\perp \\ &= Fr\sin(\theta).\end{align}\]
In using the above equation, a clockwise rotation corresponds to a positive torque while a counter-clockwise rotation corresponds to a negative torque.
The symbol we introduced in this expression, \(r_\perp\) is called the lever arm. The lever arm is the perpendicular distance from the rotation axis to the line that extends from the applied force, as shown in the image below. In this image, the blue position vector represents the rod that is rotated when a force, represented by the green vector, acts on the end of it. The angle \(\theta\) (the Greek letter theta) is the angle between the position and force vectors. We find the lever arm by extending the line from the force vector out so that we can draw the perpendicular line from the rotation axis to the extended force line. This is the lever arm. Some physics problems we can solve with this equation include finding the torque needed to open a door or tightening a bolt with a wrench and—you guessed it—finding the point of equilibrium of a see-saw.
Fig. 2 - The torque is found by multiplying the magnitude of the applied force by the lever arm.
Another name for the lever arm is the moment arm. While these two terms mean the same thing, we'll stick to using only the former in this article.
Notice that since the force vector is at an angle with respect to the position vector, the force vector has a component parallel to the position vector and a component perpendicular to the position vector. The parallel component does not contribute to the torque since it is in the same direction as the position vector; only the perpendicular component contributes to the torque. Imagine using a wrench to tighten a bolt. If you pull outward on the wrench, the bolt will not turn because the force is parallel to the wrench. You must apply a perpendicular force on the wrench to get the bolt to turn.
With the above in mind, now is a good time to work out quantitatively why you and your friend found static equilibrium after your friend moved closer to the seesaw lever. The example below is meant to guide you through the physics involved.
Fig. 3 - Two people sitting on a seesaw. The pivot point is at the center of the seesaw
Two people sit on a seesaw. On the left is a \(50\,\mathrm{kg}\) person and on the right is a \(75\,\mathrm{kg}\) person. The seesaw is \(1.2\,\mathrm{m}\) in total with a pivot at the center point. The left person sits \(0.6\,\mathrm{m}\) away from the pivot. How far will the \(75\,\mathrm{kg}\) person have to sit so that the seesaw is in static equilibrium?
For the seesaw to be in static equilibrium, the net torque on the system must be zero. Let's consider the torque from each person on the seesaw, choosing the center of the seesaw to be the origin. Notice that the lever arms for each person will simply be the distance they are away from the center of the seesaw because when the seesaw is balanced, the force of gravity pulls down on them at a \(90^{\circ}\) angle. So, the torque from the person on the left is
\[\begin{align*}\tau_1&=r_\perp F_1\\&=-d_1 m_1 g,\end{align*}\]
and the torque from the person on the right is
\[\begin{align*}\tau_2&=r_\perp F_2\\&=d_2 m_2 g.\end{align*}\]
We can then solve for the second distance by setting the net torque equal to zero:
\[\begin{align*}\sum\tau&=0\\\tau_1+\tau_2&=0\\-d_1m_1g+d_2m_2g&=0\\d_1m_1g&=d_2m_2g\\d_1m_1&=d_2m_2\\d_2&=\frac{d_1m_1}{m_2}\\&=\frac{(50\,\mathrm{kg})(0.6\,\mathrm{m})}{75\,\mathrm{kg}}\\&=0.4\,\mathrm{m}.\end{align*}\]
Difference between Force and Torque
If you compare the definitions of force and torque we gave above, you may notice that they share similar features. The same applies to their equations. In both cases, we compute the quantity we're looking for by taking the product of two things. But, by asking ourselves about what we're multiplying in each case, we arrive at the fundamental difference between force and torque. Whereas force applies to forward motion in a straight line, torque applies to rotational motion. Indeed, this is why we talk about angular acceleration when thinking about torque.
But how is angular acceleration different from linear acceleration? Well, recall that we define acceleration as the rate of change of velocity with respect to time:
\[\vec{a} = \frac{\Delta \vec{v}}{\Delta t}.\]
Contrast this with angular acceleration, which we define as the rate of change of angular velocity with respect to time:
\[\vec{\alpha} = \frac{\Delta \vec{\omega}}{\Delta t}.\]
As is the case, angular velocity \(\omega\) (Greek letter omega) is related to velocity via the following expression:
\[\omega = \frac{v}{r}.\]
If we insert this into the definition of angular acceleration,
\[\begin{align} \vec{\alpha} &= \frac{\Delta \vec{\omega}}{\Delta t} \\ &= \frac{\Delta (\vec{v}/r)}{\Delta t} \\ &= \frac{1}{r}\frac{\Delta \vec{v}}{\Delta t} \\ &= \frac{\vec{a}}{r} \end{align}\]
provided that \(r\) remains fixed. So, whenever the rotating object is rigid, we have the following simple relationship between linear acceleration and angular acceleration:
\[\vec{a} = r\vec{\alpha}.\]
This relation implies another important piece of information regarding the difference between force and torque: their difference in units.
Recall that the unit of force is the newton:
\[1 \;\mathrm{N} = 1 \;\mathrm{kg}\cdot\frac{\mathrm{m}}{\mathrm{s}^2}.\]
This makes sense since force is mass times acceleration, mass has units of kilograms, and acceleration has units of meters per second squared. On the other hand, torque is moment of inertia times angular acceleration. The units of moment of inertia are kilograms times meters squared, where the latter comes from the shape of the object. From the above, we see that angular acceleration is the product of distance from the axis of rotation times acceleration. Combining these two facts allows us to determine the units of torque:
\[\begin{align}\left[\tau \right] &= \left[I \right]\cdot \left[ \alpha \right] \\&= \left[I \right]\cdot \left[ \frac{a}{r} \right] \\&= \left(\mathrm{kg} \cdot \mathrm{m}^2 \right) \cdot \left(\frac{1}{\mathrm{m}} \cdot \frac{\mathrm{m}}{\mathrm{s}^2}\right)\\&= \mathrm{kg} \cdot \frac{\mathrm{m}^2}{\mathrm{s}^2} \\&= \mathrm{N} \cdot \mathrm{m}.\end{align}\]
Thus, force and torque are also different because they don't have the same units.
You may have noticed that torque has the same units as translational work. However, we never say that torque has units of Joules because torque by itself is not a measure of energy.
Relationship between Force and Torque
We already saw above that torque is the rotational analog of linear force. For that reason, we can say that the relationship between force and torque is one of logical dependence. By this, we mean that, as a physical quantity, we may derive torque from the concept of force. We'll do just that in the next section. Before that, we'll highlight a key feature common both to force and torque: the fact that Newton's second law has a rotational analog.
For objects experiencing translational motion, we find the equations of motion from Newton's second law which states that the net force acting on the object is equal to the mass multiplied by the acceleration,
\[\sum \vec{F}=m\vec{a}.\]
The rotational analog of Newton's second law for a rotating object is that the net torque acting on the object is equal to the moment of inertia multiplied by the angular acceleration,
\[\sum \vec{\tau}=I\vec{\alpha}.\]
When we write Newton's second law for an object in translational motion, we first consider all of the forces contributing to the net force. In a similar way, for Newton's second law in rotational motion we must consider all of the forces applying torque to the object. Static and dynamic equilibrium apply to both force and torque, meaning that when an object is in rest or in motion with constant velocity/angular velocity, the net force/torque is equal to zero.
Both force and torque are also vector quantities with magnitude and direction. We defined the magnitude of torque above by the equation:
\[\begin{align}\tau &= Fr_\perp.\end{align}\]
When considering torque as a vector, we must use the vector definition instead:
\[\vec{\tau}=\vec{r}\times\vec{F}.\]
This equation tells us that the torque vector is equal to the cross product of the force and position vectors. The direction of the torque vector is along the axis of rotation, and its sign depends on the rotation direction; counterclockwise and clockwise rotation correspond to positive and negative torque respectively, assuming that the positive direction is upward.
A \(3\,\mathrm{kg}\) thin disk of radius \(0.2\,\mathrm{m}\) is rotating about the axis through its center. The moment of inertia for the disk is \(I=\frac{1}{2}mR^2.\) A force of \(50\,\mathrm{N}\) is applied tangentially to the edge of the disk. What is the magnitude of the angular acceleration of the disk?
First, let's find the magnitude of the torque from the given force. Since the force is applied tangentially to the disk, the angle between the force vector and the radial position vector is \(90^\circ\). So, the magnitude of the torque is:
\[\begin{align*}\tau&=r_\perp F\\&=R\sin\theta F\\&=(0.2\,\mathrm{m})\sin{(90^\circ)}(50\,\mathrm{N})\\&=10\,\mathrm{N}\cdot\mathrm{m}.\end{align*}\]
Now, we can solve for the angular acceleration:
\[\begin{align*}\tau&=I\alpha\\\alpha&=\frac{\tau}{I}\\&=\frac{\tau}{\frac{1}{2}mR^2}\\&=\frac{10\,\mathrm{N}\cdot\mathrm{m}}{\frac{1}{2}(3\,\mathrm{kg})(0.2\,\mathrm{m})^2}\\&=166.7\,\mathrm{\frac{rad}{s}}.\end{align*}\]
Derivation of Torque from Force
Now, let's see how we derive the equation for torque starting from Newton's second law with forces. For this derivation, we will consider the rotation of a particle of mass \(m\) about a rotation axis. We will begin by substituting the relation between acceleration and angular acceleration, \(\vec{a}=\vec{\alpha}\times\vec{r},\) into Newton's second law of translational motion:
\[\begin{align} \vec{F} &= m\vec{a} \\ &= m\vec{\alpha}\times\vec{r}. \end{align}\]
The vector operations of the cross product and dot product are beyond the scope of AP Physics 1. However, they are necessary for the derivation of torque from Newton's second law. You can think of the math below as a primer for the more rigorous derivations one encounters in university physics courses.
Now, we take the cross product of both sides with the position vector:
\[\vec{r}\times\vec{F} = m\vec{r}\times(\vec{\alpha}\times\vec{r}).\]
We can simplify this equation using the following cross-product identity:
\[\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}.\]
Using this rule gives us:
\[\vec{r}\times\vec{F} = m(\vec{r}\cdot\vec{r})\vec{\alpha}-m(\vec{r}\cdot\vec{\alpha})\vec{r} .\]
The first dot product, \(\vec{r}\cdot\vec{r},\) becomes the magnitude of the position vector squared, \(r^2.\) Since the acceleration vector, \(\vec{\alpha},\) points along the axis of rotation, and the position vector, \(\vec{r}\) points perpendicular to it along the rotating plane, the dot product between them goes to zero: \(\vec{r}\cdot\vec{\alpha}=0.\) Thus, our equation becomes:
\[\vec{r}\times\vec{F} = mr^2\vec{\alpha}.\]
We recognize \(mr^2\) as the moment of inertia of the particle about its rotation axis, so we can write:
\[\begin{align*}\vec{r}\times\vec{F} &= I\vec{\alpha}\\&=\vec{\tau}.\end{align*}\]
Thus, the torque on the particle about the rotation axis is:
\[\vec{\tau}=\vec{r}\times\vec{F}.\]
We can use another cross-product rule to arrive at the equation we used above in terms of the lever arm. The cross-product identity we use is
\[\vec{a} \times \vec{b} = \lvert \vec{a} \rvert \lvert \vec{b} \rvert \sin(\theta),\] so the equation for torque can be written:
\[\begin{align*}\vec{\tau}&=\vec{r}\times\vec{F}\\&=\lvert \vec{r} \rvert \lvert \vec{F} \rvert \sin(\theta) \\&=(r\sin(\theta))F\\&=r_\perp F.\end{align*}\]
Measurements of Force and Torque
We can make measurements of force and torque using certain instruments in a lab. To measure force, one common force gauge is the spring meter. It contains a spring and a hook to attach to the object being measured, and it outputs the force required (in newtons) to stretch the spring. Torque measurements can be made using a torque sensor. Torque sensors use a sensor or tranducer to measure the torque on an object and outputs the torque in newton meters.
Fig. 4 - A force sensor used to measure the applied force.
Force and Torque - Key takeaways
- A force is an action exerted upon an object in order to change its state of motion either from rest or of uniform forward motion in a straight line.
- A torque is an action exerted on an object in order to change its state of torsion either from rest or of uniform angular motion around an axis.
- While force has units of newtons \(\mathrm{N}\), torque has units of newton-meters, \(\mathrm{N}\cdot\mathrm{m}.\)
- Torque is related to force by the equation: \(\vec{\tau}=\vec{r}\times\vec{F}.\)
- The rotational analog of Newton's second law for a rotating object is \(\vec{\tau}=I\vec{\alpha}.\)
References
- Fig. 1 - Friends on Seesaw (https://pixabay.com/photos/clone-seesaw-outdoors-park-fun-4014107/) by MichaelMags (https://pixabay.com/users/michaelmags-11692883/) licensed by Pixabay license (https://pixabay.com/service/license/).
- Fig. 2 - Torque Lever Arm, StudySmarter Originals
- Fig. 3 - Pivot Point Seesaw, StudySmarter Originals
- Fig. 4 - Force Sensor (https://search-production.openverse.engineering/image/fa078701-c074-4811-8deb-e415ee77f6d6) by SparkFun Electronics (https://www.flickr.com/photos/41898857@N04) licensed by CC BY 2.0 (https://creativecommons.org/licenses/by/2.0/)
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