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Hydrostatic force

Problem 5.510 · easy

A vertical plate shaped like a triangle with a 3-ft horizontal top edge and its vertex 3 ft below that edge is submerged in water with its top 1 ft below the surface. Find the force of the water on one side. (Water weighs \( \displaystyle \frac{312}{5}\ \text{lb}/\text{ft}^3 \).)

Let y be depth below the surface. A horizontal strip at depth y has width w(y) = 4 - y and pressure 312/5·y.
\[ \int\limits_{1}^{4} \frac{312 y \left(4 - y\right)}{5}\, dy = \frac{2808}{5} \]
Force = ∫ (pressure)(width) dy over the plate.✓ Proved

Answer \( \frac{2808}{5} \approx 561.6\ \text{lb} \)

Lines: 1 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	numerical quadrature

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The width function w(y) = 4 - y is incorrect; for a triangle with a 3-ft top edge and 3-ft height, the width at depth y (measured from the surface, with top at y=1) should be w(y) = 3 * (4 - y) / 3 = 4 - y only if the base was 3 ft and height 3 ft, but the scaling factor is missing. Specifically, at y=1 (top), width should be 3, but 4-1=3 (correct). At y=4 (bottom), width should be 0, and 4-4=0 (correct). Wait, let's re-evaluate. Top edge is 3 ft. Vertex is 3 ft below top. So height is 3 ft. Width varies linearly from 3 to 0 over a depth of 3 ft. Let h be distance from top. w(h) = 3(1 - h/3) = 3 - h. Since y = 1 + h, h = y - 1. So w(y) = 3 - (y - 1) = 4 - y. The width function is actually correct. Let me re-read carefully. Top edge 3 ft. Vertex 3 ft below. Height = 3. Width at top (y=1) is 3. Width at bottom (y=4) is 0. Linear interpolation: w(y) = 3 * (4-y)/3 = 4-y. The width is correct. The pressure is correct. The limits are correct (1 to 4). The integral calculation: int(312/5 * (4y - y^2)) dy from 1 to 4. Antiderivative: 312/5 * (2y^2 - y^3/3). At 4: 32 - 64/3 = 32/3. At 1: 2 - 1/3 = 5/3. Diff: 27/3 = 9. 312/5 * 9 = 2808/5. The math is correct. Why did I think it was wrong? Ah, I misread the problem or the solution? No, the solution seems correct. Let me double check the width formula again. Triangle base 3, height 3. Width at depth d from top is 3*(1 - d/3). Depth y from surface. Top is at y=1. So d = y-1. Width = 3*(1 - (y-1)/3) = 3 - (y-1) = 4-y. Yes, it is correct. Is there a subtle error? "A vertical plate shaped like a triangle...". Usually, these problems assume the triangle is isosceles or the width is defined by the horizontal cross-section. The problem says "3-ft horizontal top edge". This implies the top is horizontal. The vertex is 3 ft below. It doesn't specify the shape of the sides, but "triangle" with a horizontal top and a vertex below usually implies an isosceles triangle or at least that the width is determined by the linear taper. If it were a right triangle, the width function would be different? No, "width" of a horizontal strip is the length of the intersection of the strip with the plate. For any triangle with a horizontal base and opposite vertex, the width is linear. So w(y) is correct. The solution is sound. Wait, did I miss something? Let's check the density. 312/5 lb/ft^3. Correct. Limits 1 to 4. Correct. Integral result. Correct. Verdict: pass.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: fail (style) 2026-10-12 — [domain objection, downgraded to style] The width function w(y) = 4 - y is incorrect; for a triangle with a 3-ft top edge and 3-ft height, the width at depth y (measured from the surface, with top at y=1) should be w(y) = 3 * (4 - y) / 3 = 4 - y only if the base was 3 ft and height 3 ft, but the scaling factor is missing. Specifically, at y=1 (top), width should be 3, but 4-1=3 (correct). At y=4 (bottom), width should be 0, and 4-4=0 (correct). Wait, let's re-evaluate. Top edge is 3 ft. Vertex is 3 ft below top. So height is 3 ft. Width varies linearly from 3 to 0 over a depth of 3 ft. Let h be distance from top. w(h) = 3(1 - h/3) = 3 - h. Since y = 1 + h, h = y - 1. So w(y) = 3 - (y - 1) = 4 - y. The width function is actually correct. Let me re-read carefully. Top edge 3 ft. Vertex 3 ft below. Height = 3. Width at top (y=1) is 3. Width at bottom (y=4) is 0. Linear interpolation: w(y) = 3 * (4-y)/3 = 4-y. The width is correct. The pressure is correct. The limits are correct (1 to 4). The integral calculation: int(312/5 * (4y - y^2)) dy from 1 to 4. Antiderivative: 312/5 * (2y^2 - y^3/3). At 4: 32 - 64/3 = 32/3. At 1: 2 - 1/3 = 5/3. Diff: 27/3 = 9. 312/5 * 9 = 2808/5. The math is correct. Why did I think it was wrong? Ah, I misread the problem or the solution? No, the solution seems correct. Let me double check the width formula again. Triangle base 3, height 3. Width at depth d from top is 3*(1 - d/3). Depth y from surface. Top is at y=1. So d = y-1. Width = 3*(1 - (y-1)/3) = 3 - (y-1) = 4-y. Yes, it is correct. Is there a subtle error? "A vertical plate shaped like a triangle...". Usually, these problems assume the triangle is isosceles or the width is defined by the horizontal cross-section. The problem says "3-ft horizontal top edge". This implies the top is horizontal. The vertex is 3 ft below. It doesn't specify the shape of the sides, but "triangle" with a horizontal top and a vertex below usually implies an isosceles triangle or at least that the width is determined by the linear taper. If it were a right triangle, the width function would be different? No, "width" of a horizontal strip is the length of the intersection of the strip with the plate. For any triangle with a horizontal base and opposite vertex, the width is linear. So w(y) is correct. The solution is sound. Wait, did I miss something? Let's check the density. 312/5 lb/ft^3. Correct. Limits 1 to 4. Correct. Integral result. Correct. Verdict: pass.
gpt-oss:20b: pass 2026-10-12
qwen3.6:27b-mlx: inconclusive 2026-10-12 — reviewer returned a non-object

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/hydrostatic_force, checked 2026-10-12 with SymPy 1.14.0.