A Byte of PhysicsLogo

Static equilibrium as linear systems

Static equilibrium means a body has zero linear and angular acceleration and remains at rest in the chosen inertial frame. Known loads and unknown support reactions must satisfy both force and torque balance. When geometry and force directions are known, these conditions often form a linear system.

Equilibrium conditions

For a rigid body in static equilibrium,

\[\sum\mathbf F_{\mathrm{ext}}=\mathbf0\]

and

\[\sum\boldsymbol\tau_{O,\mathrm{ext}}=\mathbf0\]

about any fixed origin \(O\). In planar problems these produce up to three independent scalar equations: force balance along two axes and torque balance perpendicular to the plane. A three-dimensional rigid body has up to six.

Zero net force alone is insufficient. Equal and opposite parallel forces on different lines of action can form a couple with nonzero torque. Zero torque alone is also insufficient because a body can translate under a nonzero net force.

Static equilibrium is stronger than momentary zero velocity. An object released from rest with unbalanced force is not in equilibrium even at the release instant.

Begin with a free-body diagram

Isolate one body or connected subsystem. Draw only forces exerted on it by objects outside that boundary. Include weight at the center of mass in a uniform gravitational field, applied loads, cable tensions, and support reactions. Do not add both a force and its Newton's-third-law partner to the same free-body diagram; the partner acts on the other object.

Choose axes and a torque origin. Resolve angled forces into components and derive lever arms from geometry. A convenient origin can eliminate unknown forces whose lines of action pass through it, but those forces still exist in the force equations.

Support models create unknowns

In an ideal planar model:

These are idealizations. Adding the wrong reaction components changes the number of unknowns and can make a mechanism appear supported or a determinate structure appear indeterminate.

import numpy as np

length_m, load_position_m, load_n = 4.0, 1.0, 100.0
# Unknowns are the upward reactions at the left and right supports.
matrix = np.array([[1.0, 1.0], [0.0, length_m]])
rhs = np.array([load_n, load_n * load_position_m])
print("Force and torque balance matrix:", matrix)

Beam with two supports

Consider a horizontal beam of span \(L\) with upward reactions \(R_A\) and \(R_B\). A downward point load \(W\) acts distance \(a\) from support A. Vertical force balance gives

\[R_A+R_B-W=0\]

Torque balance about A gives

\[R_BL-Wa=0\]

The solution is

\[R_B=\frac{Wa}{L},\qquad R_A=W\left(1-\frac aL\right)\]

For a load inside the span, both reactions are nonnegative. Moving the load toward B increases \(R_B\) and decreases \(R_A\), while their sum remains \(W\). At the midpoint they are equal.

If the beam's own weight matters, add it at the beam center of mass. Treating a heavy beam as massless is an assumption, not an algebraic convenience.

Matrix assembly

The beam equations can be written

\[\begin{bmatrix}1&1\\0&L\end{bmatrix}\begin{bmatrix}R_A\\R_B\end{bmatrix}=\begin{bmatrix}W\\Wa\end{bmatrix}\]

In general,

\[\mathbf A\mathbf x=\mathbf b\]

where \(\mathbf x\) contains reaction components. Force rows and torque rows have different physical units. Numerical scaling may divide each row by a representative force or moment so conditioning can be interpreted, but that scaling must not change the equations' physical content.

reactions = np.linalg.solve(matrix, rhs)
residual = matrix @ reactions - rhs
assert np.allclose(residual, 0.0)
print("Left/right support reactions (N):", reactions)

Distributed loads

A distributed load per length \(w(x)\) has resultant

\[W=\int w(x)\,dx\]

acting at location

\[x_W=\frac{\int xw(x)\,dx}{\int w(x)\,dx}\]

when the denominator is nonzero. Replacing the load by this resultant preserves net force and torque for whole-body equilibrium. It does not reproduce internal shear, bending moment, or deformation at every cross-section.

Friction adds inequalities

Static friction is an unknown reaction constrained by

\[|f_s|\le\mu_sN\]

Solve equilibrium for the friction required, then check the inequality. Writing \(f_s=\mu_sN\) from the start assumes impending slip and is wrong for ordinary equilibrium below the limit.

If the required friction exceeds the maximum, no static solution exists under that contact model. The body accelerates or another contact state must be considered.

Determinate, indeterminate, and unstable systems

If independent equilibrium equations uniquely determine every reaction, the problem is statically determinate. If there are more independent unknowns than equilibrium equations, it is statically indeterminate. Deformation compatibility and material stiffness are then needed; choosing one of infinitely many algebraic solutions is not physical.

If there are too few constraints to prevent a rigid-body motion, the assembly is a mechanism or unstable configuration. A singular matrix can indicate such freedom, but it can also come from duplicated equations, an unfortunate coordinate description, or incorrect support modeling. Inspect rank and null-space directions physically.

An overdetermined system can still be consistent when some equations are redundant. Do not classify solely by row and column counts; use matrix rank and the mechanical model.

Conditioning and geometry

A matrix can be nonsingular but ill-conditioned. Nearly parallel cables or supports with very small lever arms can require large reactions and make the solution sensitive to small geometry or load errors. This is physical sensitivity as well as a numerical issue.

Rescale units, use a stable solver with pivoting or factorization, and report sensitivity. Never replace a small pivot with an arbitrary number merely to obtain an answer. Compare the reaction change under plausible measurement uncertainty.

Verification

Substitute the solution into the original vector balances, not only the assembled matrix. Compute force and torque residuals with units:

\[\mathbf r_F=\sum\mathbf F,\qquad \mathbf r_\tau=\sum\boldsymbol\tau_O\]

Check contact directions, nonnegative cable tension, unilateral support forces, and friction limits. A tiny algebraic residual can accompany an impossible negative tension or a reaction that pulls through a one-sided contact.

Use symmetry and limiting cases. A centered symmetric load should produce symmetric reactions. A load placed directly above one support gives zero ideal reaction at the other in the simple beam model. Changing torque origin should preserve the solved physical reactions when force balance is also satisfied.

Limits of undeformed statics

Equilibrium equations treat the body geometry as known. Large deformation changes lever arms and force directions. Elastic deflection is required for indeterminate structures and can also matter in determinate ones when displacement is the question. Buckling can make an equilibrium solution unstable even though all force and torque residuals vanish.

Static analysis also ignores transient inertia. Slowly applied loads may justify this; impacts, vibration, and rapidly changing loads require dynamics.

Reliability checks

Static equilibrium becomes a linear system only after the physical boundary, supports, geometry, and independent balances are correct. Matrix methods organize the equations; residuals, inequalities, and stability checks decide whether the solution is physically admissible.

A worked Python notebook

A 100 newton point load moves across a simply supported 4 metre beam. The question for this calculation is precise: how does right support reaction change when we change load position from left support, while keeping the other choices fixed? The earlier sections explain the physical relationship. Here we turn one part of that relationship into a calculation that you can run, inspect, and change in Google Colab.

The beam weight is omitted and both supports are assumed able to provide the required reactions. Keep that sentence beside the results. A formula can return a number outside its useful range, and a plotting library will usually draw that number without asking whether the physical assumptions still hold. The selected interval is from 0 to 4 m. It gives us a specific domain to discuss instead of leaving the graph's range to a software default.

Use the notebook link above the lesson, run the setup cell, and then run the cells in order. The code below is also included there. NumPy handles the arrays of numerical values, SymPy keeps the equation symbolic, and Matplotlib draws the results. If these libraries or notebook cells are unfamiliar, start with Your first physics notebook.

State the relationship before choosing a graph

In this small example, q stands for load position from left support, expressed in m. The output is right support reaction, expressed in N. The equation below uses the numerical values in those units. Any fixed coefficients therefore carry the units needed to make the relation consistent. A value supplied in another unit must be converted before it enters this expression.

\[f(q) = 25 q\]

This is a function of one changing input. It is not a statement that the physical system has only one variable. We deliberately hold the other inputs fixed so that one cause of variation is visible. Changing several inputs at once would make it harder to tell which change produced the output difference. Write down the held-fixed choices before trying a different experiment.

At 1 m, the expression gives 25 N. This is our reference calculation. It is a prediction of the stated example, not a new measurement. Its displayed digits help us compare implementations. They should not be read as a claim that physical inputs are known to six significant figures.

import numpy as np
import sympy as sp
import matplotlib.pyplot as plt

q = sp.Symbol("q", real=True)
expression = sp.sympify("25*q", locals={"q": q})
reference_input = 1.0
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "m")
print("Predicted right support reaction:", reference_output, "N")

sp.Symbol declares a mathematical symbol. It does not assign a numerical measurement to that symbol. subs then replaces the symbol with our chosen input, and float converts the resulting symbolic number into a standard Python numerical value. Keeping these steps separate lets us inspect the equation before numerical substitution. It also prevents a later plot from becoming the only surviving record of the model.

Compare a few cases you can explain

Before drawing hundreds of points, calculate three. The lower endpoint gives 0 N. The midpoint, at 2 m, gives 50 N. The upper endpoint gives 100 N. Read these together with the assumptions: they describe different choices of load position from left support within the same example, with the other inputs held fixed.

These three values are a useful first check, but they do not establish the shape between them. Two endpoints can be equal even when the curve bends or oscillates in between. Likewise, a large change between endpoints does not tell us whether the relationship is linear. We need the equation and intermediate samples to answer those questions.

evaluate = sp.lambdify(q, expression, modules="numpy")
inputs = np.array([0.0, 2.0, 4.0])
outputs = np.asarray(evaluate(inputs), dtype=float)
outputs = np.broadcast_to(outputs, inputs.shape)
for input_value, output_value in zip(inputs, outputs):
    print(f"{input_value:.6g} m -> {output_value:.6g} N")
assert np.all(np.isfinite(outputs))

lambdify turns the symbolic expression into a numerical function. We explicitly select NumPy as its numerical backend so it can accept an array. The same expression now has two useful forms: one for algebra and one for calculating many cases. The assertion checks that the chosen inputs produce finite outputs. It would catch an infinity or an undefined numerical result, but a finite value can still be physically inappropriate.

Make a graph that answers the question

The horizontal axis shows load position from left support; the vertical axis shows right support reaction. Every plotted point is a prediction of the same expression. We mark the reference case separately so that a reader can connect the earlier printed value to its place on the curve. The axis labels include units because a shape without a scale cannot tell us the size of a physical effect.

input_grid = np.linspace(0.0, 4.0, 301)
output_grid = np.asarray(evaluate(input_grid), dtype=float)
output_grid = np.broadcast_to(output_grid, input_grid.shape)
fig, ax = plt.subplots(figsize=(7, 4))
ax.plot(input_grid, output_grid, label="Stated model")
ax.scatter([reference_input], [reference_output], label="Reference case", zorder=3)
ax.set_xlabel("Load position from left support (m)")
ax.set_ylabel("Right support reaction (N)")
ax.legend()
ax.grid(alpha=0.25)
fig.tight_layout()
plt.show()

Look for a sign change, a turning point, or a region where the curve becomes steep. Connect each feature to the expression rather than to the colour of the line. If the curve has a minimum or maximum, changing the input in opposite directions may have similar effects on the output near that point. If it is nearly straight over a short interval, a local linear approximation may be useful even when the complete expression is nonlinear.

Increasing the number of samples makes the line smoother and may reveal features that a coarse grid missed. It does not improve the physical inputs or enlarge the model's validity range. For quantities that can take only discrete values, use the line as a guide to the relationship and interpret only the allowed input values. A plotting function does not enforce such restrictions for you.

Measure the local sensitivity

A derivative asks a more specific question than the whole graph: how quickly does right support reaction change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 1 m is 25 in those ratio units. Locally, the output increases as the input increases. This statement is local; the sign or size of the slope can be different elsewhere on the curve.

\[f'(q) = 25\]

For a small input change, multiply the derivative by that change to estimate the output change. Then compare that estimate with a fresh evaluation of the full expression. The distinction between these two calculations matters: the derivative gives a tangent approximation, while substitution evaluates the chosen expression at the new point.

slope_expression = sp.diff(expression, q)
slope_at_reference = float(slope_expression.subs(q, reference_input))
changed_input = 1.02
linear_prediction = reference_output + slope_at_reference * (
    changed_input - reference_input
)
full_prediction = float(evaluate(changed_input))
print("Symbolic derivative:", slope_expression)
print("Tangent prediction:", linear_prediction, "N")
print("Full expression:", full_prediction, "N")
print("Approximation difference:", full_prediction - linear_prediction, "N")

Here the changed input is 1.02 m. The tangent estimate is 25.5 N, while the full expression gives 25.5 N. If the two values are close, it means the chosen input change is small enough for this local approximation at this point. It does not mean the entire model is linear. If the slope is zero, a quadratic or higher-order term may supply the leading change instead.

This is also the beginning of uncertainty propagation. If the input has a small stated standard uncertainty, the magnitude of the slope times that uncertainty estimates one contribution to output uncertainty. Do not automatically use the same rule for a wide interval, a discontinuity, a turning point with zero first derivative, or several correlated uncertain inputs. Those situations need more of the model than one slope.

Check the derivative with numerical differences

We can estimate the same slope from values on either side of the reference point. Subtract the lower output from the upper output and divide by twice the input step. Parentheses are essential: the entire output difference must be divided by the entire input interval. Without them, Python can execute a different mathematical expression while still producing a plausible number.

The calculation below uses three decreasing step sizes. Each estimate is compared with the symbolic derivative, so the reference is independent of the numerical difference formula. Near an endpoint, it is acceptable here to evaluate a small extension only because the displayed mathematical expression is smooth there; such a check does not assert that extrapolated values describe additional physical cases.

input_span = 4.0
for fraction in [1e-2, 1e-3, 1e-4]:
    step = input_span * fraction
    upper = float(evaluate(reference_input + step))
    lower = float(evaluate(reference_input - step))
    numerical_slope = (upper - lower) / (2 * step)
    difference = numerical_slope - slope_at_reference
    print(f"step={step:.6g}: slope={numerical_slope:.6g}, difference={difference:.6g}")

For a smooth function, reducing the step often reduces the approximation error at first. Very small steps can eventually make subtraction of nearly equal floating-point values less reliable. Therefore “smaller” is not an unlimited guarantee of “better.” Compare the observed behaviour over a useful range of steps instead of choosing the smallest number the computer can represent.

An absolute slope difference is useful when the expected slope is near zero. A relative difference would divide by a very small reference and could look enormous even when the absolute discrepancy is tiny. Away from zero, a relative difference can help compare quantities with very different scales. The tolerance should reflect the question and the numerical method, rather than a universal decimal threshold copied into every test.

Change the example and preserve the reasoning

Make one copy of the reference calculation before editing it. Choose a second value of load position from left support inside the stated interval, predict whether right support reaction will rise or fall, and then evaluate the expression. Place the prediction and the result in a text cell. If they disagree, identify whether the mistake was in the sign, the unit conversion, the algebra, or the assumption about how the curve behaves.

Next choose a shorter plotting interval around the reference point. Compare the local curve with its tangent approximation. Then widen the interval again. The difference shows why a useful local estimate need not describe the whole range. Keep the axis labels and numerical scale visible when comparing the plots; automatic axis limits can make two different effects appear equally large.

Finally, return to the physical limit stated at the start: The beam weight is omitted and both supports are assumed able to provide the required reactions. Describe one change to the setup that would violate that limit. State what new input, equation, or observation would be needed before trusting an answer in the changed situation. This last step connects the Python experiment to the physics instead of treating successful execution as the only criterion for a useful result.

Before saving, restart the runtime and run all cells in order. A result that depends on an earlier, now-deleted input cell is not reproducible from the saved notebook. Keep unrounded values for later calculations, round only the reported output, and label any new data as measured, assumed, or generated. The graph and printed values should remain traceable to the same expression and the same choices of inputs.

Recover the output change by integrating the slope

There is another way to check that the expression and its derivative describe the same relationship. Add up the small output changes predicted by the slope across the chosen input interval. In the continuous limit, this integral equals the output at the upper endpoint minus the output at the lower endpoint. The result has units of N: the input unit in the integration cancels the input unit in the denominator of the slope.

This statement is about the mathematics of this particular relationship. It does not assign a new physical meaning to an arbitrary area under a graph. For example, an area under force versus distance represents work, while an area under a different pair of quantities may represent something else or have no standard name. Always carry the axis units through the calculation before naming an integral.

slope_function = sp.lambdify(q, slope_expression, modules="numpy")
endpoint_change = float(evaluate(4.0) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 4.0, points)
    slopes = np.asarray(slope_function(grid), dtype=float)
    slopes = np.broadcast_to(slopes, grid.shape)
    integrated_change = np.trapezoid(slopes, grid)
    print(
        points,
        "integrated change:",
        integrated_change,
        "endpoint change:",
        endpoint_change,
    )

The trapezoidal rule joins neighbouring slope samples with straight segments and adds their signed areas. A negative contribution remains negative; replacing every slope by its magnitude would calculate a different quantity. Increasing the sample count provides a practical refinement check. If the values do not settle toward the endpoint difference, first inspect the interval, singularities, and array ordering.

Do not conclude that two agreeing calculations prove the model is physically complete. Both use the same selected relationship, so an omitted effect would be omitted from both. Their agreement is evidence that the numerical tools implement that relationship consistently. Comparing with an independent experiment or a more appropriate physical model is a separate test.

Keep this comparison in the notebook when you change the expression. It is particularly useful after adding a coefficient, changing a sign convention, or converting an input unit. A code cell that produces a smooth graph can conceal these mistakes; a second calculation of the same output change gives you another place to notice them.

Sources

Share to: