Physics / Force table & vector addition
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EXPERIMENT 17 · PHYSICS LABORATORY

Force table & vector addition lab report

A worked force table & vector addition lab report with simulated observation tables, graph, calculations, uncertainty and results.

WORKED EXAMPLE · SIMULATED OBSERVATIONS

A complete example report.

Simulated demonstration data.

This report shows the method and arithmetic. It is not evidence of a physical experiment. Replace its observations with your own measurements when preparing a real lab report.

Student: __________________Student ID: __________________Course: __________________Date: __________________Partner: __________________Instructor: __________________

1. Title and aim

Vector addition and equilibrium of concurrent coplanar forces on a force table. Investigate the static mechanical equilibrium of concurrent coplanar forces using a precision circular force table, verify the parallelogram and polygon laws of vector addition by resolving forces into orthogonal Cartesian components, and determine the unknown magnitude of equal symmetric load forces using ordinary least-squares regression.

2. Abstract

Five observations were generated from the ideal teaching model over the settings listed below. A straight-line fit gave an estimate of component load mass of 0.1500 kg, differing by 0.002% from the model value. The reported scatter comes from scale rounding and any stated model approximation, not measured laboratory noise.

3. Apparatus and fixed settings

4. Theory and graph choice

In classical mechanics, forces are vector quantities possessing both magnitude and direction, obeying the axioms of vector addition rather than simple scalar arithmetic. When multiple coplanar forces act concurrently through a single common point of application, static translational equilibrium requires that their vector sum vanishes identically: Σ F = F_1 + F_2 + ... + F_n = 0. In this laboratory investigation, concurrent force equilibrium is studied using a precision circular force table equipped with a 360° graduated rim, radial low-friction pulleys, and a central ring floating around a vertical centering pin. Two known equal forces F_1 and F_2, generated by suspended standard masses M_0 = 150.0 g (0.1500 kg), are aligned symmetrically at equal half-angles +θ and -θ relative to the 0° Cartesian x-axis. Resolving each force into orthogonal components yields F_{1x} = M_0 · g · cos(θ), F_{1y} = M_0 · g · sin(θ), F_{2x} = M_0 · g · cos(θ), and F_{2y} = -M_0 · g · sin(θ). Due to the angular symmetry about the x-axis, the lateral vertical components cancel identically: Σ F_y = F_{1y} + F_{2y} = 0. The resultant force vector R acts entirely along the positive x-axis (0°) with magnitude R = R_x = F_{1x} + F_{2x} = 2 M_0 · g · cos(θ). To establish complete equilibrium, a third pulley carrying an equilibrant mass m_E is clamped diametrically opposite at 180°. By definition, the equilibrant force E must be equal in magnitude and opposite in direction to the resultant: E = -R, so E = m_E · g = 2 M_0 · g · cos(θ). Dividing through by the gravitational acceleration g yields the direct linear relationship m_E = (2 M_0) · cos(θ). By measuring the equilibrant mass m_E across five distinct half-angles θ, and plotting m_E against cos(θ), the resulting straight-line slope S = 2 M_0 allows the component mass M_0 = S / 2 to be deduced with high statistical precision, confirming the vector addition law and the validity of orthogonal component resolution.

m_E = (2 M_0) · cos(θ) + b

The calculation fits y = mx + b in SI units. The graph may display scaled axes for readability; those display factors are removed before calculating the physical result. The gradient has units kg.

5. Method

  1. Place a circular bullseye spirit level directly onto the center of the force table disk. Adjust the three leveling screws on the tripod base until the spirit level bubble is centered within the target index circle. Rotate the level through 90° to verify that the table is horizontal in all directions.
  2. Clamp the first pulley at angle +θ (e.g. 20.0°) and the second pulley at symmetric angle 360° - θ (e.g. 340.0°). Thread two radial cords from the central ring over these pulleys and attach identical total masses M_0 = 150.0 g (including the weight hangers) to both cords.
  3. Clamp the third pulley at exactly 180.0° on the table rim, diametrically opposite the 0° line of action of the resultant force. Pass the third cord over this pulley and suspend an empty weight hanger (50.0 g tare).
  4. Carefully add slotted brass weights to the equilibrant hanger until the central ring floats freely and concentrically around the centering pin without contacting it. Tap the rim of the table gently with a pencil to eliminate static bearing friction in the pulley axles.
  5. Record the total equilibrant mass m_E (weights plus hanger) to the nearest 0.1 g. Systematically alter the symmetric half-angle setting to θ = 30.0°, 40.0°, 50.0°, and 60.0°. At each setting, adjust the equilibrant mass m_E until the ring is perfectly centered, and record the new balance mass.
  6. Calculate the independent variable x = cos(θ) for each setting. Convert m_E to kilograms (y = m_E / 1000). Plot y against x, fit an ordinary least-squares regression line, determine the slope S, and compute the estimated component mass M_0 = S / 2.

6. Observations and calculation table

All observations use fixed model settings. Values are rounded only at the instrument reading stage; further calculations use full precision. Repeated physical trials are not claimed.

Table 1. Five simulated measurements; x and y are in SI units.
No.Symmetric half-angle θ (°)Equilibrant mass m_E (g)x (SI)y (SI)Fitted y (SI)Residual (SI)
120.0281.90.939690.281900.281903.8574e-06
230.0259.80.866030.259800.259804.5320e-06
340.0229.80.766040.229800.22980-4.8830e-07
450.0192.80.642790.192800.19282-2.2586e-05
560.0150.00.500000.150000.149991.4685e-05

Download the sample observations (CSV) ↓

7. Worked calculation for one observation

θ = 20.0°; x = cos(20.0°) = 0.93969; observed equilibrant mass m_E = 281.9 g = 0.28190 kg. Single-reading component mass M₀ = m_E / (2 cos θ) = 0.28190 / (2 × 0.93969) = 0.1500 kg. Linear regression of m_E against cos θ yields the reported applied mass M₀.

8. Graph, gradient and final calculation

0.140.460.180.590.220.720.260.850.300.98cos(θ)Equilibrant mass m_E (kg)

Dots show the sample observations. The line is the ordinary least-squares fit.

x̄ = 0.74291; ȳ = 0.22286
Σ(x − x̄)² = 0.12345
Σ(x − x̄)(y − ȳ) = 0.03703
m = Σ(x − x̄)(y − ȳ) / Σ(x − x̄)² = 0.30000692
b = ȳ − mx̄ = -1.8143e-05
Slope S = 0.300007 kg; M₀ = S / 2 = 0.300007 / 2 = 0.15000 kg (150.00 g)

9. Uncertainty and comparison

Percentage difference = |estimate − model value| / |model value| × 100 = 0.0023%. Percentage difference is not experimental uncertainty.

For n = 5 observations, Σresidual² = 7.6141e-10; s² = Σresidual²/(n − 2). Gradient standard error = √[s²/Σ(x − x̄)²] = 4.5343e-05. Intercept standard error = √[s²(1/n + x̄²/Σ(x − x̄)²)] = 3.4431e-05.

Propagating only the relevant fit coefficient gives a fit-only standard error of 2.2672e-05 kg. This is a one-standard-error quantity under the regression assumptions. Deterministic rounded model data do not supply a complete physical uncertainty budget.

Angular scale resolution of the circular table (±0.25°), pulley axle bearing friction and string-pulley groove contact resistance, knot stretching and non-coaxial central ring alignment, and slotted mass manufacturing tolerances contribute uncertainty. Tapping the table breaks static friction dead-bands. The regression slope standard error quantifies random ring-centering scatter, but excludes shared systematic zero-degree alignment errors.

10. Result and discussion

M_0 = 0.1500 kg
Model comparison value: 0.1500 kg

The observations follow the stated model relationship. Check whether the intercept and residual pattern suggest a systematic offset or a missing approximation. Agreement with the model does not independently verify its input constants.

11. Precautions and limitations

The simulation assumes idealized frictionless pulleys, massless cords, an infinitely thin centering pin, and perfectly concurrent lines of action through the mathematical center. Simulated readings reflect symmetric applied loads of reference mass M_0 = 0.1500 kg (150.0 g) with slotted weight set resolution of 0.1 g.

12. Conclusion

The worked example illustrates how to obtain component load mass from measurements, graph variables and a fitted coefficient. The estimate is 0.1500 kg. A real experiment must additionally evaluate the apparatus-specific uncertainties and corrections described above.