📝 Fluid force on vertical surface integral (24 MCQs)

📖 Explanation: The student has set the depth function as $h(x) = 6 - x$ which is incorrect. The depth should be measured from the water surface. With the top edge at 2m, and the positive x-axis pointing downward from the water surface, the depth at a point $x$ below the top is $2 + x$. The width is correctly given as 3. The limits should be from $x=0$ at the top to $x=4$ at the bottom. The correct integrand is $\rho g (2+x) \cdot 3$.

📖 Explanation: The fluid force depends on the depth of the submerged surface and the distribution of pressure. For a given area and geometry, the configuration that places the centroid of the plate at the greatest average depth will result in the largest total force. In this case, orienting the triangle such that its centroid is at the maximum possible depth maximizes the average pressure and hence the total force. The area being constant is not the only factor; the moment of area (depth times area) is what determines the force.

📖 Explanation: For a semicircular plate with the diameter at the surface, the submerged portion extends from depth 0 to depth R. The width $w(y)$ at depth $y$ is $2\sqrt{R^2 - y^2}$, and the depth function is $h(y) = y$. The total force is $F = \rho g \int_0^R y \cdot 2\sqrt{R^2 - y^2} \, dy$. Option A correctly identifies the depth as $y$ (measured from the surface downward) and the width from the geometry of a semicircle. Option B incorrectly measures depth from the bottom. Option C integrates from -R to R, which is incorrect because the plate is submerged only from y=0 to y=R.

📖 Explanation: The force on a rectangular plate of width $b$ and height $h$ with its top at the surface is $F = \frac{1}{2} \rho g b h^2$. When the gate is lowered by 2 ft, the new total depth for the bottom is h' = 6+2 = 8 ft, but the top is now at 2 ft, so the force becomes F' = \rho g b \int_2^8 y \, dy = \frac{1}{2} \rho g b (8^2 - 2^2) = \frac{1}{2} \rho g b \cdot 60. The ratio F'/F = [60]/[36] = 5/3. Rechecking: original force $F = \int_0^6 \rho g y (4) dy = 2 \rho g y^2|_0^6 = 72 \rho g$. New force F' = \int_2^8 \rho g y (4) dy = 2 \rho g (64 - 4) = 120 \rho g. Ratio = 120/72 = 5/3 ≈ 1.667. Option B is 9/4 = 2.25. The correct factor is 5/3.

📖 Explanation: The total hydrostatic force on a vertical plane surface is given by $F = \rho g \bar{h} A$, where $\bar{h}$ is the depth of the centroid. Substituting $\rho g = 10,000$ N/m³, $\bar{h} = 10$ m, and $A = 4$ m² gives $F = 10,000 \times 10 \times 4 = 400,000$ N. This result is independent of the shape of the plate, as long as the area and centroid depth are known, which is a direct Medium of the hydrostatic force formula. It does not require integrating for each shape.

📖 Explanation: The total hydrostatic force on a submerged plane surface is $F = \rho g \bar{h} A$. It depends only on the area $A$, the depth of the centroid $\bar{h}$, and the fluid density. It does not depend on the shape, perimeter, or orientation (as long as orientation is vertical). Since both plates have the same area and same centroid depth, the forces are equal. Distractors like perimeter or shape are common misconceptions, but the fundamental principle is that pressure acts on area, and the centroid depth captures the variation of pressure over the area.

📖 Explanation: The top edge is at depth 2 m, so the bottom edge is at depth 2 + 3 = 5 m. Using an axis with $y=0$ at the top of the plate and positive downward, the depth at a distance $y$ below the top is $2+y$. The width is constant at 2 m. The force is $F = \int_0^3 \rho g (2+y) \cdot 2 \, dy$. Option A uses limits 2 to 5 but then uses $y$ as depth and width 2, which would be correct if the limits were from 2 to 5 and the integrand was $\rho g y \cdot 2$. So A is also equivalent but the variable of integration is different. However, B is the more standard setup. Option C assumes depth decreases from top, which is wrong. Option D integrates over 0 to 5, which would place the top at the surface.

📖 Explanation: In the original orientation, the wider part is at the top where pressure is lower. In the inverted orientation, the wider part is at the bottom where pressure is higher. Since pressure increases linearly with depth, having more area at greater depths results in a larger total force. The area of the trapezoid remains the same, but the distribution of that area over depth changes. The centroid depth is also greater in the inverted case because the area is concentrated lower. Thus, the force increases. This highlights that force depends on the first moment of area, not just area.

📖 Explanation: For a square plate of side $a$, with top edge at depth $d$, the force is $F = \rho g a \int_d^{d+a} y \, dy = \rho g a (ad + \frac{a^2}{2})$. If $a=2$, $F = \rho g \cdot 2 (2d + 2) = 4\rho g (d+1)$. If $d$ doubles to $2d$, F' = 4\rho g (2d+1). The ratio F'/F = \frac{2d+1}{d+1}. This is not constant; it depends on d. For large d, the ratio approaches 2. For small d, it approaches 1. Thus, the factor is not a simple integer. The option 'Depends on the value of d' is correct. Many students incorrectly assume a factor of 4 or 2 without considering the additive constant from the plate height.

📖 Explanation: For a circular plate of radius $R$, if the top is at depth $R$ and bottom at $2R$, the force is $F = \int_R^{2R} \rho g y \, 2\sqrt{R^2 - (R-y)^2} \, dy$. If flipped with top at depth 0 and bottom at $2R$, the force is F' = \int_0^{2R} \rho g y \, 2\sqrt{R^2 - (R-y)^2} \, dy. The difference is the integral from 0 to $R$, which is exactly the force on a semicircle with diameter at the surface. That force is known to be $\rho g \pi R^3/2$? Let's compute: For a full circle, centroid depth is at center. When top is at depth R, centroid is at $1.5R$, so force $F = \rho g (1.5R)(\pi R^2) = 1.5 \rho g \pi R^3$. When top is at surface, centroid is at $R$, so force F' = \rho g R (\pi R^2) = \rho g \pi R^3. So F' / F = 2/3. Thus new force is $2/3 F$, not listed. The correct option is not given, but if forced, option 'C' for same is incorrect. The correct factor is 2/3. Let's re-evaluate: The question might have top at depth 0 and bottom at $2R$ means the centroid is at $R$. The original has centroid at $1.5R$. So ratio = $R/(1.5R) = 2/3$. So new force = $2/3$ F.

📖 Explanation: The student's statement is only correct for a horizontal plate where the force is exactly the weight of the fluid column above it. For a vertical plate, the pressure is due to the depth of the fluid, but the force is the integral of pressure over the area. The 'weight of fluid above' concept applies to horizontal surfaces because the area is perpendicular to the pressure. For vertical surfaces, the pressure distribution varies with depth, and the force is not simply the weight of the fluid above the projected area. The force on a vertical plate is actually the integral of pressure, and is often less than the weight of a hypothetical column of fluid above it because the pressure acts horizontally on the vertical surface.

📖 Explanation: For a rectangular side of width $b$ and height $h$, with the water level at the top, the force is $F = \frac{1}{2} \rho g b h^2$. If the density doubles (\rho' = 2\rho) and the height halves (h' = h/2), the new force is F' = \frac{1}{2} (2\rho) g b (h/2)^2 = \frac{1}{2} \cdot 2 \rho g b \cdot \frac{h^2}{4} = \frac{1}{4} \rho g b h^2 = \frac{1}{2} F. So the new force is half the original force, i.e., 500 N. Distractors like 1000 N ignore the change in depth, 2000 N ignore the square dependence on height, and 250 N overcompensates.

📖 Explanation: The center of pressure is the point where the resultant force acts. For a rectangular plate with its top at the free surface, the center of pressure is located at a distance $\frac{2}{3}h$ from the surface. For $h = 8$ m, $\frac{2}{3} \times 8 = 5.33$ m. This is deeper than the centroid (which is at 4 m) because the pressure increases with depth, shifting the resultant downward. This concept is crucial in engineering design to locate the point where the resultant force acts, which is not at the centroid. Option A (4 m) is the centroid depth, a common mistake. Option C (6 m) is arbitrary. Option D (2.67 m) is $\frac{1}{3}h$, which would be for a different loading distribution.

📖 Explanation: The fluid force on a submerged surface depends only on the area and the depth of the centroid, not on the orientation of the surface in the plane perpendicular to the pressure direction. Rotating the plate about a vertical axis does not change the projected area or the depth of any point, so the pressure distribution remains identical. The force remains the same. This is because pressure is a scalar and acts normal to the surface; rotating the surface does not change the pressure at each point. This is a common conceptual test to see if students understand that force depends only on the submerged area and depth distribution, not on the planar orientation.

📖 Explanation: When the bottom side is longer, more area is at greater depths where pressure is higher. When inverted, the longer side is at the top where pressure is lower, and the shorter side is at the bottom. The area is the same, but the distribution of that area is shifted. The centroid depth of the trapezoid depends on the lengths of the bases. For a trapezoid with longer base at the bottom, the centroid is deeper than when the longer base is at the top. Since force is proportional to centroid depth, the force increases when the longer base is at the bottom. This shows that the distribution of area matters, not just the area itself.

📖 Explanation: The pressure varies linearly from 5 kPa to 15 kPa over the height of the plate. The average pressure is (5 + 15)/2 = 10 kPa. The force is average pressure times area. The height $h$ is found from the pressure difference: $\Delta P = \rho g h = 10$ kPa = 10,000 Pa. With $\rho g = 9810$ N/m³, $h = 10,000/9810 \approx 1.02$ m. Area = $2 \times 1.02 = 2.04$ m². Force = $10,000 \times 2.04 \approx 20,400$ N ≈ 20 kN. Option A is correct. Option B (40 kN) would result if using maximum pressure times area. Option C (10 kN) uses top pressure times area. Option D (30 kN) uses an incorrect average. This problem requires converting pressure to depth and using the linear distribution.

📖 Explanation: If the vertex is at the bottom, the parabola opens upward, and its vertex is at the bottom. The depth from the surface to the vertex is h. At a depth y from the surface, the distance from the vertex is $h-y$. The equation relating width and distance from vertex is $x = k (h-y)^2$, so width is $2x = 2k (h-y)^2$. Thus, the correct width function is $w(y) = 2k (h-y)^2$. Option B is for vertex at the surface. Option C and D are incorrect forms. This question tests the ability to translate geometric descriptions into the correct mathematical functions for integration, a key skill in setting up hydrostatic force integrals.

📖 Explanation: The force on a submerged plane surface is $F = \rho g \bar{h} A$. Here $\bar{h} = 3$ m, $A = \pi (1)^2 = \pi$ m². So $F = 9810 \times 3 \times \pi \approx 9810 \times 9.4248 \approx 92,400$ N. Wait, recalc: $9810 \times 3 = 29,430$. $29,430 \times \pi = 92,457$ N. So option A is 92,400 N. Option B is 61,600 N, which is $9810 \times 2 \times \pi$ (centroid at 2m). Option C is 30,800 N, which is $9810 \times 1 \times \pi$ (centroid at 1m). Option D is 123,200 N, which is $9810 \times 4 \times \pi$ (centroid at 4m). So the correct answer is A.

📖 Explanation: The total hydrostatic force on the gate is $F = \frac{1}{2} \rho g b h^2 = \frac{1}{2} \rho g \cdot 2 \cdot 9 = 9 \rho g$ N. The center of pressure is at $\frac{2}{3}h = 2$ m from the top. Taking moments about the hinge (top), the force on the latch at the bottom (distance 3 m from hinge) must balance the moment due to the hydrostatic force. So $F_{latch} \times 3 = F \times 2$. Thus $F_{latch} = \frac{2}{3}F = \frac{2}{3} \times 9 \rho g = 6 \rho g$ N. So option B is correct. Option A is the total force, not the latch force. Option C and D are incorrect. This problem tests the understanding of center of pressure and moment equilibrium, which is a classic engineering Medium.

📖 Explanation: The triangular plate has its base at the surface, so the vertex is at depth 4 m. The width at depth y is a linear function decreasing from 6 m at y=0 to 0 m at y=4. The slope is $-\frac{6}{4} = -1.5$, so $w(y) = 6 - 1.5y = \frac{3}{2}(4-y)$. The depth is y. The limits are from 0 to 4. Thus the integral is correct. Option A is correct. Distractor B, C, D are incorrect because the student correctly identified the geometry and depth. This question tests whether students can validate a setup against the geometry, a critical skill in modeling.

📖 Explanation: Pressure increases linearly. Average pressure = (2 + 5)/2 = 3.5 atm = 3.5 × 101,325 = 354,637.5 Pa. Area = 1 × 3 = 3 m². Force = average pressure × area = 354,637.5 × 3 = 1,063,912.5 N ≈ 1,064,000 N. Option A is correct. Option B (709,000 N) uses 2.33 atm average (incorrect). Option C (1,519,000 N) uses max pressure times area. Option D (3,038,000 N) uses sum of pressures times area. This problem requires converting atm to Pa and using linear average, which is a common engineering calculation.

📖 Explanation: The force is directly proportional to the weight density of the fluid. Since the plate and depth are the same, the force in oil is $F_{oil} = F_{water} \times \frac{\rho_{oil}}{\rho_{water}} = 9810 \times \frac{8500}{9810} = 8500$ N. This is a direct Medium of the formula $F = \rho g \bar{h} A$. Option B assumes the force is independent of fluid, which is false. Option C is a reversed ratio. Option D is incorrect because 8500 is less than 9810 but still positive.

📖 Explanation: For a rectangular plate with top at depth $d$, width $b$, height $h$, force is $F = \rho g b \int_d^{d+h} y \, dy = \rho g b (dh + \frac{h^2}{2})$. If $h$ is halved, h' = h/2, the new force is F' = \rho g b (d(h/2) + \frac{(h/2)^2}{2}) = \rho g b (\frac{dh}{2} + \frac{h^2}{8}). The ratio F'/F = \frac{\frac{dh}{2} + \frac{h^2}{8}}{dh + \frac{h^2}{2}}. For large $d$ (deep), this ratio approaches 1/2. For small $d$, it approaches 1/4. Since the problem likely assumes $d=0$ (water level at top), then F' = \frac{1}{4} F = 12,500 N. Option B is correct. Option A (25,000) would be if only half the height but same average pressure, which is incorrect because average pressure also changes.

📖 Explanation: The width function $w(y) = a y$ is linear and passes through the origin. At depth $y=0$ (the surface), the width is zero. This describes a plate where the width increases with depth, i.e., a triangle with its apex at the surface and base at the bottom. Option A is correct. Option B would have width decreasing with depth, so $w(y) = a(h-y)$. Option C would have a quadratic dependence. Option D would have constant width. This question tests the ability to interpret a given integral and match it to the geometric shape of the submerged plate.