Topic: Fluid Mechanics
An inclined plate 2 m long and 1 m wide lies with its length inclined at 45° to the surface of water and its nearest edge is 1m below it. If specific weight of water is $1000kg/m^{3}$, then total pressure force on the plate (in kg) is approximately
- A. 2150
- B. 2510
- C. 3220
- D. 3420
Answer: D. 3420
Explanation: The area of the plate is $A = 2 \times 1 = 2$ m². The depth of the center of gravity from the free surface is $\bar{h} = 1 + (L/2)\sin(45^{\circ}) = 1 + 1\cos(45^{\circ})$ (noting $\sin 45^{\circ} = \cos 45^{\circ}$). The total pressure force is $F = wA\bar{h} = 1000 \times 2 \times (1 + 1\cos 45^{\circ}) = 2000 \times (1 + 0.707) = 3414$ kg, which is approximately 3420 kg.