An engineer needs to find the centroid of an L-shaped bracket. Which approach correctly applies the composite method?
AIntegrate dA over the entire L-shaped area using a custom boundary function
BTreat the L-shape as a rectangle minus a corner rectangle; compute x̄ = (A₁x̄₁ − A₂x̄₂)/(A₁ − A₂)
CAverage the centroids of the two rectangular arms of the L without weighting by area
DFind the midpoint of the outer boundary of the L-shape
The composite method uses subtraction: model the L-bracket as a full rectangle (area A₁, centroid x̄₁) minus the corner cutout (area A₂, centroid x̄₂). The formula is x̄ = (A₁x̄₁ − A₂x̄₂)/(A₁ − A₂), with the negative sign treating the removed area as a negative contribution. Option C is a common error — simply averaging the two arm centroids without area-weighting ignores the fact that a larger arm has more influence on the centroid location. Option D confuses a geometric boundary midpoint with the area-weighted centroid.
Question 2 Multiple Choice
A rectangular plate has a small circular hole drilled near its left edge. Compared to the unmodified rectangle, where does the centroid of the plate with the hole lie?
AIn the same location — the hole is small so its effect is negligible
BSlightly to the left — toward the removed material
CSlightly to the right — away from the removed material
DUpward, because the hole breaks the plate's vertical symmetry
Removing material from the left shifts the centroid rightward — away from where material was removed. The centroid is the balance point: taking weight off the left side shifts the balance point right. In the formula, the subtracted term (negative area × centroid of hole) removes a left-side contribution, leaving the right side relatively heavier. Option B reverses this direction. Quickly checking which direction the centroid should shift is a useful sanity check: the centroid always moves away from removed material.
Question 3 True / False
The centroid of a composite shape is found by computing the area-weighted average of the centroids of its component parts.
TTrue
FFalse
Answer: True
This is the definition of the composite method: x̄ = (Σ Aᵢ x̄ᵢ) / (Σ Aᵢ). Each component contributes its centroid location weighted by its area. A large component near one edge contributes more to pulling the overall centroid toward that edge than a small component at the same location would. This weighted-average formula is equivalent to evaluating ∫x dA / ∫dA but uses tabulated centroids for standard shapes rather than integrating from scratch.
Question 4 True / False
When using the subtraction method for a shape with a cutout, you should use the centroid of the original uncut shape as the location associated with the negative area term.
TTrue
FFalse
Answer: False
This is the most common setup error in the subtraction method. The negative area represents the removed piece, so you must use the centroid of the removed piece — not the centroid of the original complete shape. In the formula x̄ = (A₁x̄₁ − A₂x̄₂)/(A₁ − A₂), x̄₂ is the centroid location of the cutout shape itself. Think of it as the removed piece 'pulling' the centroid in its own direction with negative weight — so you need to know where that piece was located.
Question 5 Short Answer
Why does the composite centroid formula use area-weighted averaging rather than a simple average of component centroids? What does area weighting capture that a simple average would miss?
Think about your answer, then reveal below.
Model answer: Area weighting captures the physical reality that a larger piece has more influence on the balance point than a smaller piece at the same location. A simple average treats all components equally regardless of size, giving wrong results whenever components have different areas. The centroid is defined as the first moment of area divided by total area — ∫x dA / ∫dA — which is a continuous area-weighted average. For composite shapes, this reduces to Σ(Aᵢ x̄ᵢ) / Σ(Aᵢ): the area of each component is its weight in the average.
A practical example: an L-bracket with a long thick horizontal arm and a short thin vertical arm. A simple average of the two arm centroids places the composite centroid midway between them. But area weighting correctly pulls the centroid close to the heavy horizontal arm's midpoint, where most of the mass lies. This is exactly why the same composite logic — area as weight, centroid location as value — also applies to moments of inertia and to finding where distributed loads act as concentrated resultants.