Equilibrium and Free-body Diagrams

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 F_x = Σ F_y = Σ F_z = 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."

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, F_Land F_R, 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 F_Land F_R .

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 τ_Land τ_R are also in opposite directions, where counterclockwise is defined as the positive direction. Using Equation 2

-τ_L + τ_R = 0 = -r_L F_L + r_R F_R. (Equation 5)

Equivalently,

F_L = F_R r_R/r_L. (Equation 6)

Using Equation 1

F_L + F_R = 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 F_L and F_R 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.

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.