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# Equilibrium and Free-body Diagrams

### Overview

Source: Ketron Mitchell-Wynne, PhD, Asantha Cooray, PhD, Department of Physics & Astronomy, School of Physical Sciences, University of California, Irvine, CA

Equilibrium is a special case in mechanics that is very important in everyday life. It occurs when the net force and the net torque on an object or system are both zero. This means that both the linear and angular accelerations are zero. Thus, the object is at rest, or its center of mass is moving at a constant velocity. However, this does not mean that no forces are acting on the objects within the system. In fact, there are very few scenarios on Earth in which no forces are acting upon any given object. If a person walks across a bridge, they exert a downward force on the bridge proportional to their mass, and the bridge exerts an equal and opposite upward force on the person. In some cases, the bridge may flex in response to the downward force of the person, and in extreme cases, when the forces are great enough, the bridge may become seriously deformed or may even fracture. The study of this flexing of objects in equilibrium is called elasticity and becomes extremely important when engineers are designing buildings and structures that we use every day.

### Principles

The requirements for a system to obtain equilibrium are simple to write down. In equilibrium, the sum of the forces and the sum of the torques are zero:

Σ F = 0 (Equation 1)

and

Σ τ = 0. (Equation 2)

The torque τ is an angular force, defined as the cross product of the length of the lever arm from where the force is applied to the axis of rotation. That distance is denoted as r:

τ = r x F, (Equation 3)

= r F sin(θ)

where θ is the angle at which the force is applied to the lever arm. For forces perpendicular with respect to the lever arm, Equation 3 simply becomes τ = r · F.

These equations are simple enough to write down, but as the system in question becomes more complex, more forces and torques are involved, and finding the optimum configuration that satisfies equilibrium can become quite difficult. The general approach to solving Equation 1 is to decompose the forces into the x-, y- and z-directions and then to solve Equation 1 for each of the three directions (e.g.,Σ Fx = Σ Fy = Σ Fz = 0). In situations where there is only movement in the xy-plane, the torque is calculated about an axis perpendicular to that plane. This axis is arbitrarily chosen to simplify the calculations; if all the objects in the system are at rest, then Equation 2 will hold true about any axis. In three dimensions, the axis of rotation is again generally chosen such that the calculations are the simplest, which depends upon the configuration of the system. For example, choosing the rotation axis so that one of the unknown forces acts through that axis will result in zero lever arm and produce no torque (see Equation 3), making one less term appear in the torque equation. There is no single technique for solving equilibrium problems, but choosing convenient coordinate systems can greatly simplify the process of solving Equations 1 and 2.

When the objects in the system undergo equilibrium forces, some of them will compress or expand, depending upon their material and the configuration within the system. For example, when a force is exerted on a rod or spring, its length will expand proportionally to the force, given by Hooke's law:

F = k ΔL, (Equation 4)

where ΔL is the length of expansion and k is a constant of proportionality called the "spring constant."

### Procedure

1. Observe equilibrium in a static system and verify that the sum of the forces and torques is zero. Confirm the spring constants k used in the system.

1. Obtain a meter stick, two spring scales with known spring constants, two stands to suspend the springs from, two weights of different masses, and a mechanism to suspend the weights from the meter stick.
2. Secure the two stands to the table, 1 m apart.
3. Attach the springs to the stands.
4. Attach the spring to each end of the meter stick.
5. Attach the first weight to the middle of the meter stick.
6. Compute both the force and torque exerted by the weight on the meter stick and record them in Table 1.
7. Record the force exerted on each of the springs in Table 1.
8. Shift the weight to the left by 0.2 m and repeat steps 1.6-1.7.
9. Shift the weight to the left an additional 0.2 m, so the total displacement from the center of the meter stick is 0.4 m. In other words, the length of the moment arm for the spring on the left is 0.1 m, and the length of the moment arm for the spring on the right is 0.9 m.
10. Repeat steps 1.5-1.9 for the other weight.
11. Compute the percent difference of the calculated forces on the left and right springs, FL and FR, against the corresponding forces read off the spring scales.

Equilibrium is a special case in classical mechanics but is ubiquitous in everyday life, while free-body diagrams help decipher the underlying forces present.

A system is in translational equilibrium if the forces acting on it are balanced, that is, the net force is zero. Equilibrium can also be established in a rotational system if the net torque, t, is zero.

In addition to these static equilibrium cases where the systems are at rest, dynamic equilibrium implies that a system is moving but experiencing no linear acceleration a or angular acceleration, a.

Now, even if a system is in equilibrium, a multitude of individual forces or torques can be acting on it, and free-body diagrams -- composed of simple shapes and arrows -- are often implemented in order to conceptualize these forces and/or torques acting on a system.

The goal of this experiment is to understand the equilibrium of a system composed of multiple components under the influence of various forces.

Before analyzing this complex system, let's revisit the concepts of equilibrium and free-body diagrams. As mentioned earlier, equilibrium occurs in a translational system, such as a loaded spring, when the restoring force balances the gravitational weight. In a rotational system, example when weights are attached to a freely rotating beam, equilibrium is established when the torques balance one another. Note that, with respect to the axis of rotation, torque is positive for counter-clockwise rotation and negative for clockwise rotation.

In these cases, the net forces or torques are equal zero and therefore no linear or angular acceleration exists. Per Newton's First Law, since these systems are in static equilibrium they must remain at rest.

Despite the absence of a net force or torque, multiple forces are acting on the objects within these systems. Free-body diagrams, or force diagrams, are often drawn in order to understand the forces and torques acting on systems in equilibrium.

Each contributing force or torque is represented by an arrow whose size and direction fully describes the vector in question. Through vector addition, the translational system is shown to be in equilibrium. Similarly, by accounting for the torque direction with respect to the axis, the rotational system is also in equilibrium.

Now, imagine combining these systems such that a weight is attached to the center of the beam while the beam itself is suspended at its ends by two springs. The system is complex but can be understood by using two separate free-body diagrams. The translational system includes the weight and the left and right spring restoring forces, denoted as FL and FR, respectively.

Since the system is in equilibrium, the sum of magnitudes of FL and FR should be equal to the magnitude of the weight. This equation describes transitional equilibrium.

In the rotational system, instead of forces we have torques. Recall that torque is defined as the perpendicular force times the distance r that the force is applied from the axis of rotation. Since the weight is positioned at the axis of rotation, it exerts no torque on the beam. Whereas for the springs in this case, the perpendicular forces are the restoring forces and r is the respective distance from the weight.

Now again, since the system is in equilibrium, the magnitudes of these torques should be equal, and this equation illustrates rotational equilibrium.

Moving the weight away from the center causes the beam to tilt. For the translational system, the sum of the restoring forces is still equal and opposite to that of the weight. Therefore, the equation for translational equilibrium -- dealing with the magnitude of these forces -- stays the same.

For the rotational system, the tilt by an angle θ changes the forces in the spring torques to the cosine component of the respective restoring forces. The lengths of the rotational arms also change. However, the weight is still at the axis of rotation and therefore exerts no torque on the beam.

Since this system is also in equilibrium, the magnitudes of the torques applied by the springs should be the same. Cancelling out the cosine θ, results in the same rotational equilibrium formula.

Now that you understand the principles of equilibrium, let's apply these concepts to a system that experiences both forces and torques. This experiment consists of a meter stick, two spring scales, two stands, and two weights of different mass capable of being suspended from the meter stick.

To begin, place the two stands one meter apart on the table making sure they are secure. Suspend a spring scale off of each stand, and attach each end of a meter stick to the bottom of a spring scale.

Next, attach the least massive weight to the meter stick midway between the spring scales. With the system under translational and rotational equilibrium, calculate the individual forces acting on the meter stick and record them.

Read the values on each of the spring scales and record these restoring forces exerted by the springs.

Now shift the weight 0.2 m to the left making the left rotation arm 0.3 m and the right rotation arm 0.7 m. Repeat the calculation of the individual forces and the spring scale measurements.

Lastly, shift the weight to the left an additional 0.2 m and perform the force calculations and spring scale measurements. Repeat this equilibrium experiment for the more massive weight.

The individual forces acting on the meter stick consist of the gravitational force on the attached weight and the restoring forces of the springs. When looking at the free-body diagrams of the system under translational and rotational equilibrium, two equations can be used to determine the two unknown restoring forces.

The rotation arms are identical when the weight is midway between the springs. Therefore, each of the restoring forces should equal half of the weight. For the experiments when the weight is moved from the center, the restoring forces are dictated by the ratio of their respective rotation arms.

These calculated values can be compared with the restoring forces determined from the spring scale measurements. The differences between the values are within the measurement errors of the experiment. Therefore, by invoking equilibrium conditions, the restoring forces can be determined with knowledge of the mass of the weight and the length of the rotation arms.

The basic principles of equilibrium can be invaluable when engineers are designing structures that we use every day.

A bridge is always in static equilibrium while constantly experiencing large forces and torques from both its own weight and the loads moving across it. Therefore, construction of a suspension bridge, like the Golden Gate in San Francisco, requires significant structural engineering efforts to ensure that equilibrium is maintained even during the times of heavy traffic

Similarly, skyscrapers have a complex system of steel beams under tremendous forces, which altogether compose a rigid system in static equilibrium. Therefore, an understanding of the concepts behind equilibrium helps an architect decide the construction parameters, so that these structures can withstand a certain amount of torque, especially in the earthquake prone zones.

You've just watched JoVE's introduction to Equilibrium. You should now understand the principles of equilibrium and how free-body diagrams can be used to determine the forces and torques contributing to a system in equilibrium. Thanks for watching!

### Results

The representative results for the experiment can be found in Table 1. The force exerted on the two springs by the hanging mass are denoted by their locations: left and right, denoted by subscripts L and R. Since there are two unknowns in this experiment, FLand FR, two equations are required to solve for them. Thus, Equations 1 and 2 are used to solve for the two forces. The torques are used to obtain a relationship between FLand FR .

Since the force exerted by the weight is downward, the angle θ in Equation 3 is 90°,and the torque is just r · F. The torques τLand τR are also in opposite directions, where counterclockwise is defined as the positive direction. Using Equation 2

-τL + τR = 0 = -rL FL + rR FR. (Equation 5)

Equivalently,

FL = FR rR/rL. (Equation 6)

Using Equation 1

FL + FR = m g, (Equation 7)

where m is the mass of the weight and g is the gravitational constant of 9.8 m/s2. In other words, the downward force of the weight equals the sum of the forces holding up the weight and meter stick system, which is just the two springs on the left and right, which are suspending the system. With these two equations (6 and 7), the unknowns FL and FR can be calculated. These are shown in Table 1. These values are compared with the forces exerted on the springs in the last two columns of the table. Slight discrepancies are expected from measurement errors. In addition, it has been assumed that the mass of the meter stick is zero, which is incorrect, strictly speaking, but nevertheless a good approximation. This lab uses spring scales, which show how many Newtons are being applied to the spring when stretched, so it is not necessary to know the spring constant, k.

Table 1. Theoretical and experimental results.

 Mass (g) rL (cm) rR (cm) FL (N) FR (N) FL,spring (N) FR,spring (N) % diff (left) % diff (right) 100 50 50 0.5 0.5 0.45 0.45 9.9 9.9 100 30 70 0.68 0.29 0.65 0.3 4.4 3.4 100 10 90 0.9 0.1 0.85 0.1 5.5 0 200 50 50 0.98 0.98 1 1 0 0 200 30 70 1.38 0.59 1.35 0.55 2.1 7.2 200 10 90 1.8 0.2 1.85 0.2 2.7 0

### Applications and Summary

All bridges are under some amount of stress, from both their own weight and the weight of the loads moving across. Suspension bridges, like the Golden Gate, are a complex system of objects under very heavy forces and in equilibrium. The cables that hold the bridge up are elastic, and their elasticity was considered when the structural engineers designed the bridge. Similarly, skyscrapers have a complex system of steel beams under tremendous forces, which altogether compose a rigid system in static equilibrium. Elasticity plays a role in the materials used to construct buildings, as they need to be able to withstand a certain amount of flexing, especially in areas where earthquakes are prevalent. The cranes used to construct these structures are also in equilibrium, with a complex system of cables and pulleys to lift and lower the construction materials.

In this study, the equilibrium of a system composed of multiple components under various forces was observed. The effects of the elastic components were also observed using spring scales of known spring constants. The forces exerted upon the springs were computed using the two conditions necessary for equilibrium: the sum of the forces and the sum of the torques are zero. Play Video

### Cite this

JoVE Science Education Database. Physics I. Equilibrium and Free-body Diagrams. JoVE, Cambridge, MA, (2021).More

JoVE Science Education Database. Physics I. Equilibrium and Free-body Diagrams. JoVE, Cambridge, MA, (2021).

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