A marble released sideways inside a deep bowl will move around the rim of the bowl and constantly lose speed until it comes to rest. Why does it come to rest at the bottom of the bowl and not at the top edge? Why does it come to rest at all? It's because of the same concept that allows overhanging balconies to remain in place and not come crashing to the ground, like the one in the image below. It is because of the concept of equilibrium which we will discuss in this article. There are many different types of equilibrium and countless examples, but we will discuss the basics to help you grasp this fundamental physical concept.
Fig. 1. An overhanging balcony that is seemingly defying gravity. It is actually being supported because all of the support structures in the interior of the building are in equilibrium, Wikimedia Commons CC BY-SA 3.0
Equilibrium Definition
There are two conditions that are required for an object to be in equilibrium:
- No net force is acting on the object.
- No net torque is acting on the object.
So we can provide a basic physical definition of equilibrium as follows:
Objects or systems that are in equilibrium have no net force and no net torque acting on them.
This means that the motion of objects in equilibrium will not change with time and they will also keep the same amount of energy. Force is a familiar concept but torque may be new to you. Torque is a type of force that tends to cause a rotation. Torque \(\tau\) is given by the equation
\[\tau=Fd\]
where \(F\) is the force perpendicular to the pivot (\(\mathrm{N}\)) and \(d\) is the perpendicular distance to the pivot (\(\mathrm{m}\)). Thus, torque is measured in \(\mathrm{N\,m}\) rather than in \(\mathrm{N}\) like force. The diagram below shows how you can apply a force to a spanner to cause a torque.
Fig. 2: A spanner can be used to apply a torque to another object. Source: via Wikimedia commons, CC0.
Let's study an example that includes both of these quantities, force and torque, to gain a better understanding of equilibrium. Consider a seesaw with two twins sitting at equal distances on either side, as shown below.
Fig. 3: If twins (represented by squares in this diagram though), who weigh the same, sit on either side of a seesaw at equal distances from the center of balance, the system will be in equilibrium.
The downwards force due to gravity (which is the combined weight of the twins and their seesaw) is balanced by the upwards force at the pivot of the seesaw so the net force is zero. If we presume that they both weigh the same, then the torque due to either child will be equal and in opposite directions, so the net torque will be zero. The net force and the net torque on the system are both zero so it is in equilibrium.
Equilibrium Expression
A system is said to be in equilibrium if it has the two following properties:
- The linear momentum \(p\) of its center of mass is constant.
- The angular momentum \(L\) about its center of mass, or any other point, is constant.
These two conditions can also be represented by the following expressions:
\( \begin{align} \vec{p}&=\mathrm{constant} \\ \vec{L}&=\mathrm{constant} \end{align} \)
In situations in which the constants in these equations are equal to zero, the system is said to be in static equilibrium. For example, the seesaw in the example above has no translational motion or rotational motion either (from the reference frame in which we are observing it), so it is in static equilibrium. When a system has a constant velocity or a constant angular velocity (or both), it is said to be in dynamic equilibrium. An example of a system in dynamic equilibrium is a car traveling along a road at a constant velocity. In this situation, the driving force is equal to the drag force on the car. Also, the weight of the car is balanced by the reaction force from the road. The net force is zero and the car is in equilibrium even though it is moving.
Fig. 4. There is no net force acting on a car driving at a constant velocity so it is in equilibrium.
Equilibrium Formula
Newton's second law, in its linear momentum form, is given by the following equation:
\[\vec{F}_{\mathrm{net}}=\frac{\Delta \vec{p}}{\Delta t}\]
in which \(\vec{F}_{\mathrm{net}}\) is the net force on a system and \( \Delta \) represents a change in the variable that it is next to. If an object is in equilibrium, then the expression above tells us that its linear momentum must be constant. We know that if \(\vec{p}\) is constant then \(\frac{\Delta \vec{p}}{\Delta t}\) is zero and hence the net force must be zero,
\[\vec{F}_{\mathrm{net}}=0\]
and we have arrived back at what we stated at the start - the net force on an object in equilibrium is zero. Similarly for rotational motion, we can relate the net torque on a system to its angular momentum using the following equation:
\[\tau_{\mathrm{net}}=\frac{\Delta L}{\Delta t}\]
The net torque on an object is equal to the rate of change of the object's angular momentum. This is Newton's second law applied to angular momentum. Again, we know that if \(L\) is constant then \(\frac{\Delta L}{\Delta t}\) is zero and so the net torque must be zero.
\[\tau_{\mathrm{net}}=0\]
We can thus state the two requirements for a system to be in equilibrium:
- The vector sum of all of the forces acting on the body must be zero.
- The vector sum of all of the external torques acting on the body, measured about any point, must be zero.
We have arrived again at our two conditions for equilibrium that were stated at the start of the article!
Fig. 5: The forces acting on an object in equilibrium must be balanced.
The diagram above shows a block being pushed along a table with a rough surface. For this example, let us suppose that it is moving at a constant velocity. There are four forces acting on the block:
- \( F \) is the pushing force that is moving the block along the table.
- \( F_k \) is the frictional force due to the rough table.
- \( W \) is the weight of the block.
- \( N \) is the reaction force from the table acting on the block.
We know from our requirement for an object in equilibrium that the vector sum of the forces on an object must be zero. This means that the force in every direction is zero - the forces in opposite directions balance each other out. This leads us to the equations:
\[ \begin{align} F&=F_{k} \\ W&=N \end{align} \]
The requirements for equilibrium can be very useful in finding unknown forces!
We can also use the requirement for equilibrium that the net torque must be zero to find unknown quantities for systems in equilibrium. Consider again the seesaw from above. Imagine that one of the twins was replaced by their older brother, who happens to weigh twice as much. He sits at a distance from the center of the seesaw so that it remains balanced. How could we find this distance? We know the equation for torque to be
\[\tau=Fd\]
The force has doubled due to the weight of the older brother being double which means that he must sit at half the distance for the torque to be the same as before!
You should have come across a vector sum before, it means that you must add up the forces and torques while taking into account their directions. This can be done by adding arrows, head to tail, pointing in the direction of the force or torque, with the length depending on the magnitude. This is shown below.
Fig. 6. Forces (or torques) can be added by representing them as vectors. Source: via Wikimedia commons, public domain.
Stable Equilibrium
You might have heard of a stable equilibrium before, but make sure not to get it confused with static equilibrium! Systems in stable equilibrium have the property that if they are displaced a small amount from their static equilibrium position by a force, they will return to this state of static equilibrium after the force has subsided.
Consider two tall hills next to each other with a ball placed in the divot between them as illustrated in the figure below.
Fig. 7. A ball in a divot between two hills is in stable equilibrium.
If you gave the ball a little push in either direction, it would roll up the hill, reach a certain point and roll back again (as long as you did not push it hard enough to get to the top of the hill). It would then move back and forth between either side of its equilibrium position, with the frictional force due to the ground slowing it down until it stopped at the equilibrium position (if there was no frictional force it would oscillate back and forth across the equilibrium position forever). The ball is in stable equilibrium because the force - gravity in this case - acts to bring the ball back to equilibrium when it is displaced. When it reaches the bottom it is in equilibrium because
- the net force on the ball is zero,
- and the net torque on the ball is zero.
You can probably guess what will happen to a system in unstable equilibrium. If a system in unstable equilibrium is displaced a small amount by a force, the object will no longer be in equilibrium when the force is removed.
Consider a ball placed so that it is balancing nicely on top of a single hill.
Fig. 8: A ball at the top of a hill is in stable equilibrium.
This time, if you gave the ball a push in either direction, it would just roll down the hill and would not return to the top. The ball is in unstable equilibrium because once you give the ball a small displacement, the force - again gravity - acts to move the ball away from its equilibrium position. The ball is initially in equilibrium because
- the net force on the ball is zero,
- and the net torque on the ball is zero.
Equilibrium Examples
The conditions for equilibrium above can be used to simplify many situations and solve many problems in terms of simple equations.
A \(50 \, \mathrm{kg}\) gymnast stands on the end of a uniform balancing beam, which weighs \(200 \, \mathrm{kg} \). The beam is \(5\,\mathrm{m}\) long and is kept in place by two supports which are each \(1.5\,\mathrm{m}\) from either end. This is shown in the image below. What is the reaction force at either support?
If an object is uniform, its mass is uniformly distributed so its center of mass will be at the center.
Fig. 8. A gymnast stands right on the end of a balancing beam that is held up by two supports.
The beam must be in equilibrium as it does not move - meaning that its translational and angular momentum are both constant. This means that the net force and the net torque on the beam are zero. The upwards reaction force must be equal to the downwards force equal to the weight of both the beam and the gymnast. Weight is given by:
\[W=mg\]
where \(m\) is the mass \(\mathrm{kg}\) and \(g\) is the gravitational field strength (\(9.81\,\mathrm{m}/\mathrm{s}^{2}\) for the surface of Earth). Thus, we can write the equation:
\[ \begin{align} F_{1}+F_{2}&=50g+200g \\ &=250g \\ &=2450\,\mathrm{N} \end{align} \]
in which \(F_{1}\) and \(F_{2}\) are the reaction forces at supports 1 and 2 respectively.
We also know that the net torque about any point on the beam must be zero. We can use the equation given above for torque and equate the anticlockwise and clockwise torques about the point where support 1 meets the beam. The distance from support 1 to the center of mass of the beam is \(1.0\,\mathrm{m}\), to support 2 is \(2.0\,\mathrm{m}\) and to the gymnast is \(3.5\,\mathrm{m}\). Using these values, we arrive at the following equation:
\[(200g\times1.0)+(50g\times3.5)=2.0\times F_{2}\]
which can be rearranged to find \(F_{2}\):
\[F_{2}=1\,840 \,\mathrm{N}\]
This value can be used with the equation we found by considering the forces on the beam to get \(F_{1}\):
\[F_{1}=2\,450-F_{2}=610\,\mathrm{N}\]
The diagrams below show five different situations. A uniform rod is held in place so that it can rotate about a pivot, which is represented by point P in the figure below. A force equal to the weight of the rod is applied at varying places and in different directions. State for each case, 1 to 5, whether the system will be in equilibrium or not. Note that the weight of this rod acts through its center since it is uniform.
- The system is not in equilibrium. The force acts at a distance from the pivot that is greater than the weight of the rod (downward force) and so causes a greater moment, meaning there is a net torque in the anticlockwise direction.
- The system is in equilibrium. The force acts through the center of mass and is equal to the weight of the rod so there is no net force on the rod.
- The system is not in equilibrium. This is the same as situation 1 but the force is at a slight angle. The angle to the horizontal would have to be equal to \(30^{\circ}\) for the torques to be equal but it is clearly a lot greater than this.
- The system is not in equilibrium. The applied force and the weight of the rod both cause a clockwise moment so there is a net torque in this direction.
- The system is not in equilibrium. The force acts through the pivot so results in no torque. There is no upward force to balance the weight of the rod so there is a net force in the downward direction.
Equilibrium - Key takeaways
- Systems that are in equilibrium have no net force and no net torque acting on them.
- A system in equilibrium has a constant linear momentum and angular momentum.
- When the linear and angular momentums of a system are equal to zero, the system is in static equilibrium.
- When the linear and angular momentums of a system are equal to a constant, the system is in dynamic equilibrium.
- If a system in stable equilibrium is moved a small amount from equilibrium, it will return to equilibrium.
- If a system in unstable equilibrium is moved a small amount from equilibrium, it will no longer be in equilibrium and will not return to being so.
References
- Fig. 1: Duerig-AG Theather-Fribourg copyright Duerig-AG (https://commons.wikimedia.org/wiki/File:Duerig-AG_Theater-Fribourg_copyright_Duerig-AG.jpg) by Theg2e (no author page), under CC BY-SA 3.0 License
- Fig. 2: Torque force equivalence at one meter leverage (https://commons.wikimedia.org/wiki/File:Torque_force_equivalence_at_one_meter_leverage.svg) by Zoiros, CC0
- Fig. 6: Addition af vektorer (https://commons.wikimedia.org/wiki/File:Addition_af_vektorer.png) by Bixi at Danish Wikibooks, Public domain.
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