Let's solve the problem step by step: ### **Given:** - **Density of oil,** \(\rho = 940 \, \text{kg/m}^3\) - **Gravity,** \(g = 9.8 \, \text{m/s}^2\) - **Trapezoidal end dimensions:** - Height, \(h = 3\, \text{m}\) - Top width, \(w_\text{top} = 2\, \text{m}\) - Bottom width, \(w_\text{bottom} = 1\, \text{m}\) - Length of trough, \(L = 5\, \text{m}\) (not needed for end force calculation) --- ## **(a) Hydrostatic force on one end if full (depth = 3 m)** **Hydrostatic force (\(F\))** on a vertical surface: \[ F = \rho g \int_{0}^{h} w(y) y \, dy \] Where: - \(w(y)\) = width of the trough at depth \(y\) (from the surface down) - \(y\) = depth below the surface ### **Step 1: Find width as a function of depth** The trapezoidal end: - At \(y = 0\) (top): width = 2 m - At \(y = 3\) m (bottom): width = 1 m The sides taper linearly, so: \[ \text{Width decreases by } (2-1) = 1 \text{ m over } 3 \text{ m} \] So, at depth \(y\): \[ w(y) = 2 - \left(\frac{1}{3}\right)y \] ### **Step 2: Set up the integral** \[ F = \rho g \int_{0}^{3} \left[2 - \frac{1}{3}y\right] y \, dy \] Expand: \[ F = \rho g \int_{0}^{3} (2y - \frac{1}{3}y^2) \, dy \] ### **Step 3: Integrate** \[ \int_{0}^{3} 2y \, dy = [y^2]_{0}^{3} = 9 \] \[ \int_{0}^{3} \frac{1}{3}y^2 \, dy = \frac{1}{3} \left[\frac{y^3}{3}\right]_{0}^{3} = \frac{1}{3} \times \frac{27}{3} = 3 \] \[ F = \rho g (9 - 3) = \rho g (6) \] ### **Step 4: Plug in values** \[ F = 940 \times 9.8 \times 6 \] \[ F = 9212 \times 6 = 55,272 \text{ N} \] ### **Final Answer (a):** \[ \boxed{55,272 \text{ N}} \] --- ## **(b) Hydrostatic force if filled to depth 1.5 m** Repeat the above, but upper limit is \(y = 1.5\) m: \[ F = \rho g \int_{0}^{1.5} \left[2 - \frac{1}{3}y\right] y \, dy \] \[ F = \rho g \int_{0}^{1.5} (2y - \frac{1}{3}y^2) \, dy \] Integrate: \[ \int_{0}^{1.5} 2y \, dy = [y^2]_{0}^{1.5} = (1.5)^2 = 2.25 \] \[ \int_{0}^{1.5} \frac{1}{3}y^2 \, dy = \frac{1}{3} \left[\frac{(1.5)^3}{3}\right] = \frac{1}{3} \times \frac{3.375}{3} = \frac{1}{3} \times 1.125 = 0.375 \] \[ F = \rho g (2.25 - 0.375) = \rho g (1.875) \] \[ F = 940 \times 9.8 \times 1.875 \] \[ = 9212 \times 1.875 = 17,272.50 \text{ N} \] ### **Final Answer (b):** \[ \boxed{17,272.50 \text{ N}} \] --- ## **Summary** - (a) **55,272 N** - (b) **17,272.50 N** (rounded to two decimal places) Let me know if you need further clarification!