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Free-body diagrams in code

A free-body diagram isolates one chosen system and shows the external forces acting on it. It is not a literal picture of the scene. Its purpose is to turn an interaction description into a complete vector sum that can be used in Newton's second law.

Choose the system first

Before drawing an arrow, state what body or collection of bodies is inside the system boundary. A rope tension is external to a box-only system but internal to a combined box-and-rope system. The same interaction changes classification when the boundary changes.

The free-body diagram contains only forces exerted on the chosen system by objects outside it. It does not include forces the system exerts on other objects. Newton's third-law partner to a floor's normal force on a box acts on the floor and belongs on the floor's diagram, not the box's.

Once the forces are identified,

\[\sum\mathbf F_{\mathrm{external}}=m\mathbf a\]

applies to the selected constant-mass body in an inertial frame.

Forces are interactions, not motion labels

Velocity and acceleration arrows are useful in kinematics, but they are not force arrows. “Force of motion” is not an additional interaction. A sliding object can move right while its net force points left, causing it to slow down. It can also move at constant nonzero velocity while net force is zero.

Every force should have an identifiable source. Common interactions include:

Labels such as “centripetal force” usually name the inward net-force role, not an extra interaction to add after gravity, tension, or normal force has already been counted.

A force record

import numpy as np

mass_kg = 2.0
force_records = [
    {"target": "crate", "source": "Earth", "vector": np.array([0.0, -19.62])},
    {"target": "crate", "source": "floor", "vector": np.array([0.0, 19.62])},
    {"target": "crate", "source": "hand", "vector": np.array([4.0, 0.0])},
]
print("Forces in newtons, in laboratory x/y axes:", force_records)

The target and source make the direction of the third-law partner unambiguous. The interaction label supports domain checks without replacing the numerical vector.

def net_force(system, records):
    total = np.zeros(2)
    for force in records:
        if force["target"] in system and force["source"] not in system:
            total += force["vector"]
    return total


print("Net external force (N):", net_force({"crate"}, force_records))

For an extended body, the application point is also needed when calculating torque, even though translational acceleration depends only on net force.

Weight and normal force are not automatically equal

Near Earth's surface, weight is approximately

\[\mathbf W=m\mathbf g\]

and points downward. A normal force is perpendicular to a contact surface. Its magnitude is whatever the contact constraint requires within the contact model; it is not always \(mg\).

For a box resting on a horizontal floor with no other vertical forces and zero vertical acceleration,

\[N-mg=0\quad\Longrightarrow\quad N=mg\]

In an accelerating elevator, with upward positive,

\[N-mg=ma_y\]

so the normal force differs from weight whenever \(a_y\ne0\). During free fall with no contact, the normal force is zero even though weight remains.

Resolve components after identifying whole forces

On an incline of angle \(\theta\), choose one axis tangent to the surface and one axis normal to it. Weight remains one downward vector, whose components are

\[W_{\parallel}=-mg\sin\theta,\qquad W_{\perp}=-mg\cos\theta\]

with signs for axes chosen uphill and outward from the surface. The normal force points outward along the normal axis. If the box has no normal acceleration and no other normal forces,

\(N=mg\cos\theta\).

The components of weight are not two new forces. They are one force expressed in a useful basis. Adding the original weight vector and both components would count gravity twice.

angle = np.deg2rad(30)
uphill = np.array([np.cos(angle), np.sin(angle)])
normal = np.array([-np.sin(angle), np.cos(angle)])
weight = np.array([0.0, -mass_kg * 9.81])
parallel_weight = np.dot(weight, uphill)
normal_weight = np.dot(weight, normal)
assert np.allclose(weight, parallel_weight * uphill + normal_weight * normal)
print("Weight components (N):", parallel_weight, normal_weight)

Friction requires a contact state

Static friction adjusts to prevent relative slipping up to a limit:

\[|f_s|\le\mu_sN\]

It is not automatically equal to \(\mu_sN\). First solve for the friction needed to maintain the no-slip condition, then check whether it lies within the allowed interval. If not, the contact slips and a kinetic-friction model may use

\[|f_k|=\mu_kN\]

directed opposite relative sliding velocity. Friction opposes relative motion or its tendency at the contact, not necessarily the object's center-of-mass velocity in an arbitrary frame.

Tension and ideal ropes

An ideal massless rope under tension pulls away from the attached body along the rope. In a massless rope passing over an ideal massless frictionless pulley, tension magnitude is commonly treated as uniform. A massive rope, accelerating pulley, axle friction, or elastic stretch invalidates that shortcut.

A rope cannot normally push. If a calculation produces negative tension under a tension-only model, the assumed taut configuration is inconsistent; the rope may go slack and the equations must change.

Worked incline example

A block of mass \(m\) slides down a fixed incline with kinetic coefficient \(\mu_k\). Choose downhill as positive. Figure 1(a) below shows this motion on a 30-degree incline; Figure 1(b) isolates the block and keeps only the three external forces. The velocity arrow belongs to the scene, not the force sum. With no normal acceleration,

\[N=mg\cos\theta\]

and the tangent equation is

\[mg\sin\theta-\mu_kmg\cos\theta=ma\]

Therefore

\[a=g(\sin\theta-\mu_k\cos\theta)\]

Mass cancels under these ideal assumptions. A negative result means acceleration points uphill relative to the chosen positive direction; it does not by itself show the block's velocity direction. The kinetic model also presumes the block is already sliding.

From diagram to equations

Use Figure 1(b) as the input to the force inventory below: Earth supplies weight, and the incline supplies normal force and friction. The uphill friction arrow explains the minus sign in the downhill equation. Resolve the downward weight into the chosen basis when summing components; do not add those components as extra force records.

force_n = net_force({"crate"}, force_records)
acceleration = force_n / mass_kg
print("Acceleration (m/s^2):", acceleration)
assert np.allclose(acceleration, [2.0, 0.0])

Do not calculate an unknown contact force from a guessed formula before writing the appropriate component of Newton's law. The constraint equation, such as zero normal acceleration, supplies the additional relationship.

Multiple bodies

Draw or encode one free-body diagram per independently moving body unless the bodies are combined into one system. When separate, an interaction pair appears once on each body's diagram with opposite directions. When combined, that pair is internal and drops out of the system-level force sum, though it may still matter for stress or individual motion.

Coupled systems require shared geometric constraints. Two masses linked by an ideal taut rope can have related accelerations, but the sign relation depends on coordinate choices and pulley geometry. Force equations alone do not invent the constraint.

Audit checks

A free-body diagram is successful when another reader can reconstruct the force inventory and equations without relying on the artwork. A structured force list keeps that same reasoning visible in code.

A worked Python notebook

Resolve the weight of a 2 kilogram crate along an inclined surface. The question for this calculation is precise: how does downhill weight component change when we change incline angle, 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.

This is one component of weight, not the net downhill force if friction is present. 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 1.0472 rad. 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 incline angle, expressed in rad. The output is downhill weight component, 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.

\[f(q) = 19.62 \sin{\left(q \right)}\]

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 0.523599 rad, the expression gives 9.81 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("2*9.81*sin(q)", locals={"q": q})
reference_input = 0.5235987755982989
reference_output = float(expression.subs(q, reference_input))
print("Input:", reference_input, "rad")
print("Predicted downhill weight component:", 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 0.523599 rad, gives 9.81 N. The upper endpoint gives 16.9914 N. Read these together with the assumptions: they describe different choices of incline angle 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, 0.5235987755982989, 1.0471975511965979])
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} rad -> {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 incline angle; the vertical axis shows downhill weight component. 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, 1.0471975511965979, 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("Incline angle (rad)")
ax.set_ylabel("Downhill weight component (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 downhill weight component change near the reference input? Its units are the output unit divided by the input unit. For this calculation, the derivative at 0.523599 rad is 16.9914 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.

\[f'(q) = 19.62 \cos{\left(q \right)}\]

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 = 0.5288347633542819
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 0.528835 rad. The tangent estimate is 9.89897 N, while the full expression gives 9.89883 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 = 1.0471975511965979
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 incline angle inside the stated interval, predict whether downhill weight component 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: This is one component of weight, not the net downhill force if friction is present. 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(1.0471975511965979) - evaluate(0.0))
for points in [101, 201, 401]:
    grid = np.linspace(0.0, 1.0471975511965979, 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

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