Home›Calculus 2›Hydrostatic force›Problem 5.370

Hydrostatic force

Problem 5.370 · easy

A vertical plate shaped like a triangle with a 2-m horizontal top edge and its vertex 2 m below that edge is submerged in water with its top 3 m below the surface. Find the force of the water on one side. (Water weighs \( \displaystyle 9800\ \text{N}/\text{m}^3 \).)

Let y be depth below the surface. A horizontal strip at depth y has width w(y) = 5 - y and pressure 9800·y.
\[ \int\limits_{3}^{5} 9800 y \left(5 - y\right)\, dy = \frac{215600}{3} \]
Force = ∫ (pressure)(width) dy over the plate.✓ Proved

Answer \( \frac{215600}{3} \approx 71866.7\ \text{N} \)

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) = 5 - y is incorrect. At the top of the plate (y=3), the width is 2 m, but the formula gives 2 m. At the bottom (y=5), the width is 0, and the formula gives 0. However, the slope is wrong. The width decreases from 2 to 0 over a depth of 2 m, so the slope should be -1. The formula w(y) = 5 - y has a slope of -1, which is correct. Wait, let's re-evaluate. Top edge y=3, width=2. Bottom vertex y=5, width=0. Linear interpolation: w(y) = m(y-3) + 2. w(5) = 2m + 2 = 0 => m = -1. So w(y) = -(y-3) + 2 = -y + 3 + 2 = 5 - y. The width function is actually correct. Let me re-read the problem carefully. '2-m horizontal top edge... vertex 2 m below that edge'. Top is at depth 3. Bottom is at depth 3+2=5. Width at depth y: At y=3, w=2. At y=5, w=0. w(y) = 2 * (5-y)/2 = 5-y. The width function is correct. The pressure is 9800*y. The integral is from 3 to 5. The calculation is correct. Why did I think it was wrong? Let me check the integral value. Integral of 9800(5y - y^2) from 3 to 5. Antiderivative: 9800(5y^2/2 - y^3/3). Evaluated at 5: 9800(125/2 - 125/3) = 9800(125/6) = 1225000/6 = 612500/3. Evaluated at 3: 9800(45/2 - 27/3) = 9800(45/2 - 9) = 9800(27/2) = 132300 = 396900/3. Difference: (612500 - 396900)/3 = 215600/3. The math is correct. The setup is correct. The answer is correct. So it should pass.

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-07
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The width function w(y) = 5 - y is incorrect. At the top of the plate (y=3), the width is 2 m, but the formula gives 2 m. At the bottom (y=5), the width is 0, and the formula gives 0. However, the slope is wrong. The width decreases from 2 to 0 over a depth of 2 m, so the slope should be -1. The formula w(y) = 5 - y has a slope of -1, which is correct. Wait, let's re-evaluate. Top edge y=3, width=2. Bottom vertex y=5, width=0. Linear interpolation: w(y) = m(y-3) + 2. w(5) = 2m + 2 = 0 => m = -1. So w(y) = -(y-3) + 2 = -y + 3 + 2 = 5 - y. The width function is actually correct. Let me re-read the problem carefully. '2-m horizontal top edge... vertex 2 m below that edge'. Top is at depth 3. Bottom is at depth 3+2=5. Width at depth y: At y=3, w=2. At y=5, w=0. w(y) = 2 * (5-y)/2 = 5-y. The width function is correct. The pressure is 9800*y. The integral is from 3 to 5. The calculation is correct. Why did I think it was wrong? Let me check the integral value. Integral of 9800(5y - y^2) from 3 to 5. Antiderivative: 9800(5y^2/2 - y^3/3). Evaluated at 5: 9800(125/2 - 125/3) = 9800(125/6) = 1225000/6 = 612500/3. Evaluated at 3: 9800(45/2 - 27/3) = 9800(45/2 - 9) = 9800(27/2) = 132300 = 396900/3. Difference: (612500 - 396900)/3 = 215600/3. The math is correct. The setup is correct. The answer is correct. So it should pass.
qwen3.6:27b-mlx: fail (error) 2026-10-07 — The width function w(y) = 5 - y is incorrect; for a triangle with top width 2 at depth 3 and vertex at depth 5, the width should be w(y) = 2(5 - y)/2 = 5 - y? No, similar triangles: width is proportional to distance from vertex. At depth y, distance from vertex is 5-y. Width = (2/2)*(5-y) = 5-y. Wait. Top edge is 2m wide. Vertex is 2m below top. So height of triangle is 2m. At depth y, distance from vertex is 5-y. The width at the top (y=3) is 2. The width at the vertex (y=5) is 0. Linear interpolation: w(y) = 2 * (5-y)/2 = 5-y. This part is actually correct. Let me re-read carefully. Top edge 2m horizontal. Vertex 2m below that edge. Top is 3m below surface. So vertex is at 3+2=5m depth. Width at depth y: w(y) = (2 / 2) * (5 - y) = 5 - y. This is correct. Why did I think it was wrong? Let me check the integral. Integral of 9800*y*(5-y) from 3 to 5. 9800 * [5y^2/2 - y^3/3] from 3 to 5. At 5: 5(25)/2 - 125/3 = 125/2 - 125/3 = 125/6. At 3: 5(9)/2 - 27/3 = 45/2 - 9 = 27/2 = 81/6. Difference: 125/6 - 81/6 = 44/6 = 22/3. 9800 * 22/3 = 215600/3. The calculation is correct. The setup is correct. Why would this fail? Ah, look at the sentence: "A horizontal strip at depth y has width w(y) = 5 - y". Is the width really 5-y? At y=3, w=2. At y=5, w=0. Yes. Is the pressure 9800*y? Yes. Is the integral bounds 3 to 5? Yes. Is the result correct? Yes. Wait, is there a subtle error? "Water weighs 9800 N/m^3". Pressure = rho * g * h = weight_density * depth. Correct. Force = integral P dA. dA = w(y) dy. Correct. Everything seems correct. Let me double check the similar triangles. Triangle height H=2. Top width W=2. Distance from vertex z = 5-y. Width w = W * (z/H) = 2 * (5-y)/2 = 5-y. Correct. Maybe the issue is the sentence says "w(y) = 5 - y" without deriving it, but that's not an error. Is it possible the problem implies the vertex is at the top? "vertex 2 m below that edge". No, vertex is at bottom. "top 3 m below the surface". So top is at y=3. Vertex is at y=5. Correct. I see no error. Wait, let me look closer at the prompt's instruction. "Find what is wrong". If nothing is wrong, I should pass. But usually these have a trick. Let's re-read the width. Top edge 2m. Vertex 2m below. So it's an isosceles triangle? Doesn't matter for width. Width is linear. w(y) = 5-y. Correct. Integral is correct. Result is correct. I will pass.
gpt-oss:20b: pass 2026-10-07

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 2026-10-07 with SymPy 1.14.0.