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Hydrostatic force

Problem 5.440 · easy

A vertical plate shaped like a triangle with its vertex at the top and a 6-ft horizontal base 3 ft below the vertex 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) = 2*y - 2 and pressure 312/5·y.
\[ \int\limits_{1}^{4} \frac{312 y \left(2 y - 2\right)}{5}\, dy = \frac{8424}{5} \]
Force = ∫ (pressure)(width) dy over the plate.✓ Proved

Answer \( \frac{8424}{5} \approx 1684.8\ \text{lb} \)

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 (error) — The width function w(y) = 2y - 2 is incorrect. At the top of the plate (y=1), the width should be 0, but 2(1)-2 = 0 is correct. However, at the bottom of the plate (y=4), the width should be 6, but 2(4)-2 = 6 is also correct. Wait, let's re-evaluate. The vertex is at depth y=1. The base is at depth y=4. The width at depth y is proportional to the distance from the vertex. Distance from vertex is (y-1). At y=4, distance is 3, width is 6. So width w(y) = (6/3)*(y-1) = 2(y-1) = 2y-2. This is actually correct. Let me re-read the prompt carefully. 'A vertical plate shaped like a triangle with its vertex at the top and a 6-ft horizontal base 3 ft below the vertex is submerged in water with its top 1 ft below the surface.' Vertex at top. Base 3 ft below vertex. Top of plate (vertex) is 1 ft below surface. So vertex is at y=1. Base is at y=1+3=4. Width at y=1 is 0. Width at y=4 is 6. Linear interpolation: w(y) = m(y-1). 6 = m(3) => m=2. w(y) = 2(y-1) = 2y-2. The integral bounds are 1 to 4. The pressure is rho*y. The integral is correct. Why did I think it was wrong? Let me check the calculation. Integral of (312/5) * y * (2y-2) from 1 to 4. = (312/5) * Integral(2y^2 - 2y) dy. Antiderivative: 2/3 y^3 - y^2. Evaluated at 4: 2/3(64) - 16 = 128/3 - 48/3 = 80/3. Evaluated at 1: 2/3 - 1 = -1/3. Difference: 80/3 - (-1/3) = 81/3 = 27. Force = (312/5) * 27 = 8424/5. The math is correct. The setup is correct. Is there a subtle error? 'Vertex at the top'. Usually this implies an isosceles triangle unless specified otherwise, but for hydrostatic force, only the width function matters. The width function derived is correct for any triangle with that height and base width, assuming symmetry or just that the width varies linearly from 0 to 6. The problem doesn't state it's isosceles, but 'a triangle with its vertex at the top and a 6-ft horizontal base' usually implies the standard orientation where the width is symmetric or the 'width' refers to the horizontal cross-section. If it's not isosceles, the width function might be different? No, for a triangle, the width at a given height is linear. If the vertex is at the top, the width starts at 0 and ends at 6. The function w(y) = 2y-2 is correct regardless of left/right symmetry, as long as we are integrating the total width. The solution seems sound. Wait, let me look closer at the sentence. 'A horizontal strip at depth y has width w(y) = 2*y - 2'. This is correct. 'pressure 312/5·y'. This is correct. The integral bounds 1 to 4 are correct. The result is correct. Why would this fail? Maybe the prompt implies the vertex is at the surface? No, 'top 1 ft below the surface'. Okay, I will mark it as pass.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-09
qwen3.6:27b-mlx: fail (error) 2026-10-09 — The width function w(y) = 2y - 2 is incorrect. At the top of the plate (y=1), the width should be 0, but 2(1)-2 = 0 is correct. However, at the bottom of the plate (y=4), the width should be 6, but 2(4)-2 = 6 is also correct. Wait, let's re-evaluate. The vertex is at depth y=1. The base is at depth y=4. The width at depth y is proportional to the distance from the vertex. Distance from vertex is (y-1). At y=4, distance is 3, width is 6. So width w(y) = (6/3)*(y-1) = 2(y-1) = 2y-2. This is actually correct. Let me re-read the prompt carefully. 'A vertical plate shaped like a triangle with its vertex at the top and a 6-ft horizontal base 3 ft below the vertex is submerged in water with its top 1 ft below the surface.' Vertex at top. Base 3 ft below vertex. Top of plate (vertex) is 1 ft below surface. So vertex is at y=1. Base is at y=1+3=4. Width at y=1 is 0. Width at y=4 is 6. Linear interpolation: w(y) = m(y-1). 6 = m(3) => m=2. w(y) = 2(y-1) = 2y-2. The integral bounds are 1 to 4. The pressure is rho*y. The integral is correct. Why did I think it was wrong? Let me check the calculation. Integral of (312/5) * y * (2y-2) from 1 to 4. = (312/5) * Integral(2y^2 - 2y) dy. Antiderivative: 2/3 y^3 - y^2. Evaluated at 4: 2/3(64) - 16 = 128/3 - 48/3 = 80/3. Evaluated at 1: 2/3 - 1 = -1/3. Difference: 80/3 - (-1/3) = 81/3 = 27. Force = (312/5) * 27 = 8424/5. The math is correct. The setup is correct. Is there a subtle error? 'Vertex at the top'. Usually this implies an isosceles triangle unless specified otherwise, but for hydrostatic force, only the width function matters. The width function derived is correct for any triangle with that height and base width, assuming symmetry or just that the width varies linearly from 0 to 6. The problem doesn't state it's isosceles, but 'a triangle with its vertex at the top and a 6-ft horizontal base' usually implies the standard orientation where the width is symmetric or the 'width' refers to the horizontal cross-section. If it's not isosceles, the width function might be different? No, for a triangle, the width at a given height is linear. If the vertex is at the top, the width starts at 0 and ends at 6. The function w(y) = 2y-2 is correct regardless of left/right symmetry, as long as we are integrating the total width. The solution seems sound. Wait, let me look closer at the sentence. 'A horizontal strip at depth y has width w(y) = 2*y - 2'. This is correct. 'pressure 312/5·y'. This is correct. The integral bounds 1 to 4 are correct. The result is correct. Why would this fail? Maybe the prompt implies the vertex is at the surface? No, 'top 1 ft below the surface'. Okay, I will mark it as pass.
qwen3.6:27b-mlx: fail (style) 2026-10-09 — [domain objection, downgraded to style] The width function w(y) = 2y - 2 is incorrect. At the top of the plate (y=1), the width should be 0, but 2(1)-2=0 is correct. However, at the bottom (y=4), the width should be 6, but 2(4)-2=6 is also correct. Wait, let's re-evaluate. The vertex is at y=1 (depth 1). The base is at y=4 (depth 4). The width at depth y is proportional to the distance from the vertex. Distance from vertex is y-1. Total height is 3. Width at base is 6. So w(y) = (6/3)*(y-1) = 2(y-1) = 2y-2. This is actually correct. Let me re-read the prompt carefully. "vertex at the top... 6-ft horizontal base 3 ft below the vertex... top 1 ft below the surface". Vertex depth = 1. Base depth = 1+3=4. Width at depth y: w(y) = (Base Width / Height) * (y - y_vertex) = (6/3)*(y-1) = 2(y-1). The solution says w(y) = 2y-2. This is correct. Why did I think it was wrong? Let me check the integral. Integral from 1 to 4 of (312/5)*y*(2y-2) dy. = (312/5) * Integral(2y^2 - 2y) dy from 1 to 4. Antiderivative: 2/3 y^3 - y^2. Evaluated at 4: 2/3(64) - 16 = 128/3 - 48/3 = 80/3. Evaluated at 1: 2/3 - 1 = -1/3. Difference: 80/3 - (-1/3) = 81/3 = 27. Force = (312/5) * 27 = 8424/5. The calculation is correct. The setup is correct. Is there a subtle error? "A horizontal strip... has width w(y)... and pressure...". The sentence says "width w(y) = 2*y - 2". This is correct. The sentence says "pressure 312/5 * y". This is correct. The integral bounds are 1 to 4. This is correct. The result is correct. Why would this fail? Maybe the prompt implies I should find an error? Let me look closer. "vertex at the top". Usually, triangles are defined with base at bottom. Yes. "submerged... top 1 ft below surface". Yes. Is the width formula derived correctly in the text? It just states it. Is it misleading? No. Is it an error? No. Wait, look at the equation line. "Integral(312*y*(2*y - 2)/5, (y, 1, 4)) = 8424/5". The CAS checked this. The sentence sets up the model. Is there a missing factor? Force = integral of pressure * area. Area of strip = w(y) dy. Pressure = rho y. So integrand is rho y w(y). The solution has 312/5 * y * (2y-2). This is correct. I see no error. I will mark it pass.
gpt-oss:20b: pass 2026-10-09

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 2026-10-09 with SymPy 1.14.0.