SITUATION 9: A vat filled with paint is 9 m long. Its ends are in the form of trapezoid 3 m at the bottom and 5 m on the top. The vat is 4 m high and the specific gravity of paint is 0.8 . 50 . Compute the total weight of paint. \( \begin{array}{ll}\text { a. } 1960.2 \mathrm{kN} & \text { b. } 1130.1 \mathrm{kN} \\ \text { c. } 1340.1 \mathrm{kN} & \end{array} \)

To compute the total weight of the paint in the vat, we first need to find the volume of the trapezoidal prism. The area of the trapezoidal end is calculated using the formula: \[ \text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h \] where \( b_1 = 3 \, m \) (bottom) and \( b_2 = 5 \, m \) (top), and \( h = 4 \, m \) (height of the vat). Plugging in the values, we have: \[ \text{Area} = \frac{1}{2} \times (3 + 5) \times 4 = \frac{1}{2} \times 8 \times 4 = 16 \, m^2 \] The volume of the vat is then: \[ \text{Volume} = \text{Area} \times \text{Length} = 16 \, m^2 \times 9 \, m = 144 \, m^3 \] To find the weight, we first need the mass. The mass can be calculated using the specific gravity, which is the ratio of the density of the substance to the density of water (approximately \( 1000 \, kg/m^3 \)). \[ \text{Density of paint} = \text{Specific gravity} \times \text{Density of water} = 0.8 \times 1000 \, kg/m^3 = 800 \, kg/m^3 \] Now, we find the mass of the paint: \[ \text{Mass} = \text{Density} \times \text{Volume} = 800 \, kg/m^3 \times 144 \, m^3 = 115200 \, kg \] The weight (W) of the paint is obtained by multiplying the mass by gravity (assumed to be \( 9.81 \, m/s^2 \)): \[ W = m \times g = 115200 \, kg \times 9.81 \, m/s^2 = 1131.27 \, kN \] Thus, the total weight of the paint is closest to option **b. 1130.1 kN**. So, the final answer is **b. 1130.1 kN**!