Statics and Dynamics Quiz: Centroids

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CentroidsQuestion 1 of 3

A thin plate has the shape of a right triangle with base $b$ along the x-axis and height $h$ along the y-axis, with the right-angle vertex at the origin. A student claims that to find $\bar{x}$ of the triangle, they can place a centroid at $b/3$ from the y-axis (measured from the right-angle vertex). A second student claims the centroid is at $2b/3$ from the y-axis. A third student says the centroid location depends on which vertex is chosen as the reference. Which statement correctly resolves this dispute?

The first student is correct:

\bar{x} = b/3

measured from the right-angle vertex at the origin, because the centroid of a triangle is always located one-third of the base from the vertex where the altitude meets the base.

The second student is correct:

\bar{x} = 2b/3

measured from the right-angle vertex, because the centroid lies two-thirds of the way from any vertex to the opposite midpoint, and the base midpoint is at

b/2

, giving

\bar{x} = (2/3)(b) = 2b/3

Both the first and second students are partially correct depending on which vertex is used as the reference:

b/3

from the right-angle vertex equals

2b/3

from the opposite vertex, and both refer to the same physical point—so the third student's claim about reference-dependence is also valid.

The first student is correct only if the right angle is at the origin; otherwise the centroid shifts to

b/2

because a right triangle's centroid coincides with the midpoint of the hypotenuse when measured along the base direction.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Centroids

Practice Centroids in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Centroids, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The first student is correct: $\bar{x} = b/3$ measured from the right-angle vertex at the origin, because the centroid of a triangle is always located one-third of the base from the vertex where the altitude meets the base.
The second student is correct: $\bar{x} = 2b/3$ measured from the right-angle vertex, because the centroid lies two-thirds of the way from any vertex to the opposite midpoint, and the base midpoint is at $b/2$ , giving $\bar{x} = (2/3)(b) = 2b/3$ .
Both the first and second students are partially correct depending on which vertex is used as the reference: $b/3$ from the right-angle vertex equals $2b/3$ from the opposite vertex, and both refer to the same physical point—so the third student's claim about reference-dependence is also valid. (correct answer)
The first student is correct only if the right angle is at the origin; otherwise the centroid shifts to $b/2$ because a right triangle's centroid coincides with the midpoint of the hypotenuse when measured along the base direction.

Explanation: When locating a centroid, always anchor your measurement to a specific reference point — the numerical value alone is meaningless without context. The centroid of any triangle lies one-third of the way from each side (or equivalently, two-thirds of the way from each vertex to the opposite side's midpoint). For this right triangle with the right-angle vertex at the origin, integrating gives

\bar{x} = b/3

measured from the origin. But measured from the far vertex at

x = b

, that same point sits at

b - b/3 = 2b/3

from that vertex. Both values describe the identical physical location — they simply use different origins. This is exactly what answer C captures, and it also validates the third student's intuition that reference choice matters for the numerical value. Answer A is partially right in its result (

\bar{x} = b/3

from the origin is correct) but wrong in its reasoning. The centroid is not "where the altitude meets the base" — it's at one-third of the base from the nearest parallel side, located via the full centroid formula. The faulty reasoning could mislead you on non-right triangles. Answer B arrives at

2b/3

from the right-angle vertex, which is numerically incorrect for that reference point. The error is misapplying the two-thirds rule without specifying "from which vertex" — the centroid is

2/3

of the median's length from a vertex, not

2/3

of the base itself from the right-angle corner. Answer D invents a rule that doesn't exist: a right triangle's centroid never "shifts to

b/2

" regardless of orientation. Study tip: On centroid problems, always state your reference axis explicitly. When you see two different numbers for the same shape, ask whether they might be measured from opposite ends before assuming a contradiction.

Question 2

A uniform thin plate has a composite area consisting of a full circle of radius $R$ centered at the origin, plus a rectangle of width $2R$ and height $R$ attached directly below the circle so that the rectangle's top edge is tangent to the bottom of the circle (top edge at $y = -R$ , rectangle extends from $y = -R$ to $y = -2R$ , centered on the y-axis). The centroid of this composite shape is located at $\bar{y}$ below the center of the circle. Which expression gives the correct $\bar{y}$ (negative, measured downward from the circle's center)?

$\bar{y} = \frac{\pi R^2(0) + 2R^2(-1.5R)}{\pi R^2 + 2R^2} = \frac{-3R^3}{(\pi+2)R^2} = \frac{-3R}{\pi+2} \approx -0.585R$ , placing the rectangle centroid at $y = -1.5R$ (midpoint between $y=-R$ and $y=-2R$ ). (correct answer)
$\bar{y} = \frac{\pi R^2(0) + 2R^2(-R)}{\pi R^2 + 2R^2} = \frac{-2R^3}{(\pi+2)R^2} = \frac{-2R}{\pi+2} \approx -0.390R$ , placing the rectangle centroid at $y = -R$ (the top edge of the rectangle rather than its midpoint).
$\bar{y} = \frac{\pi R^2(-R) + 2R^2(-1.5R)}{\pi R^2 + 2R^2} = \frac{-R^3(\pi + 3)}{(\pi+2)R^2} = \frac{-R(\pi+3)}{\pi+2} \approx -1.195R$ , placing the circle centroid at $y = -R$ (the bottom of the circle) rather than at the center $y = 0$ .
$\bar{y} = \frac{\pi R^2(0) + 2R^2(-1.5R)}{\pi R^2 + 2R^2 + \pi R^2} = \frac{-3R^3}{(2\pi+2)R^2} = \frac{-3R}{2(\pi+1)} \approx -0.364R$ , adding an extra $\pi R^2$ to the denominator as if counting the circle area twice.

Explanation: When finding the centroid of a composite shape, your formula is

\bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i}

, where each

\bar{y}_i

is the centroid of that individual sub-shape — not its edge, not its total height, but its geometric center. Here, the circle has area

\pi R^2

and its centroid sits exactly at the origin, so

\bar{y}_{\text{circle}} = 0

. The rectangle has width

2R

, height

R

, and area

2R^2

. Since it spans from

y = -R

y = -2R

, its centroid is at the midpoint:

\bar{y}_{\text{rect}} = \frac{-R + (-2R)}{2} = -1.5R

. Plugging in:

\bar{y} = \frac{\pi R^2(0) + 2R^2(-1.5R)}{\pi R^2 + 2R^2} = \frac{-3R}{\pi + 2} \approx -0.585R

. That's answer A, which is correct. Answer B uses

-R

as the rectangle's centroid — that's the rectangle's top edge, not its center. This is the most common trap: confusing where a shape starts with where its centroid is. Answer C places the circle's centroid at

y = -R

, which would only be true if the circle were centered at the bottom of itself — the circle is centered at the origin, so its centroid is at

y = 0

. Answer D correctly identifies both centroids but inflates the denominator by adding an extra

\pi R^2

, as though the circle were counted twice. Your go-to strategy: always sketch the shape, mark each sub-shape's geometric center, and double-check your denominator equals the sum of all areas exactly once.

Question 3

A composite line (wire frame) consists of three straight segments forming a right triangle: a horizontal segment from $(0,0)$ to $(6, 0)$ , a vertical segment from $(6, 0)$ to $(6, 8)$ , and a hypotenuse from $(6, 8)$ to $(0, 0)$ . A student wants to find the centroid of this wire frame (not the enclosed area). Which of the following is the correct $\bar{y}$ -coordinate of the wire-frame centroid?

$\bar{y} = \frac{8}{3} \approx 2.67 \text{ units}$ , computed as the area centroid of the enclosed right triangle, since for a uniform wire frame the centroid coincides with the area centroid of the enclosed region.
$\bar{y} \approx 2.67 \text{ units}$ , computed by averaging the midpoint y-coordinates of the three segments without weighting by length: $\bar{y} = \frac{0 + 4 + 4}{3} = 2.67$ , treating each segment as equally significant regardless of its length.
$\bar{y} = 3.0 \text{ units}$ , computed as the length-weighted average of the midpoint y-coordinates: horizontal segment (midpoint $y=0$ , length 6), vertical segment (midpoint $y=4$ , length 8), and hypotenuse (midpoint $y=4$ , length 10), giving $\bar{y} = \frac{0(6)+4(8)+4(10)}{6+8+10} = \frac{72}{24} = 3.0$ . (correct answer)
$\bar{y} \approx 3.56 \text{ units}$ , computed by weighting only the two non-horizontal segments (vertical and hypotenuse) since the horizontal segment contributes zero to the y-moment and should be excluded from both the numerator and denominator: $\bar{y} = \frac{4(8)+4(10)}{8+10} = \frac{72}{18} = 4.0$ — but then scaled by the fraction of total length they represent.

Explanation: When finding the centroid of a wire frame (a system of line segments), you must treat it as a composite of one-dimensional elements. Each segment contributes to the centroid in proportion to its length — not its area, and not equally regardless of size. The formula is

\bar{y} = \frac{\sum \bar{y}_i L_i}{\sum L_i}

, where

\bar{y}_i

is each segment's midpoint y-coordinate and

L_i

is its length. For this triangle, the three segments are: horizontal (midpoint

y = 0

, length

= 6

), vertical (midpoint

y = 4

, length

= 8

), and hypotenuse (midpoint

y = 4

, length

= \sqrt{6^2 + 8^2} = 10

). Plugging in:

\bar{y} = \frac{0(6) + 4(8) + 4(10)}{6 + 8 + 10} = \frac{0 + 32 + 40}{24} = \frac{72}{24} = 3.0

. That confirms C is correct. A is wrong because the centroid of a wire frame is not the same as the area centroid of the enclosed region. Area centroids weight by area elements, not arc length — these are fundamentally different calculations. B makes a classic weighting error: averaging midpoint y-values equally across all three segments ignores the fact that longer segments carry more "mass" and must be weighted accordingly. D fabricates a rule by excluding the horizontal segment from the denominator because its y-contribution is zero — but every segment contributes its length to the total, regardless of where its midpoint sits. Study tip: Whenever a problem says "wire frame" or "thin rod," immediately switch to length-weighted averaging. Confusing wire-frame centroids with area centroids is one of the most common traps in composite-body problems.