A steel bar weighing 200 kg is located at the bottom of the reservoir and adheres tightly to it.

A steel bar weighing 200 kg is located at the bottom of the reservoir and adheres tightly to it. The depth of the reservoir is 1 m, the height of the blank is 0.2 m. What force must be applied to tear the blank off the bottom? Atmospheric pressure is 1.01 · 10 5 Pa.

Since the height of the blank is 0.2 m, its upper plane is at a depth:
1 – 0.2 = 0.8 m.
The pressure at depth is determined by the formula:
p = ρ * g * h, where ρ is density, g is gravitational acceleration, h is depth. Then:
p = 1000 * 10 * 0.8 = 8000 Pa.
Let’s find the volume of the blank (steel density 7900 kg / m: 3):
V = m / ρ = 200/7900 = 0.025 m ^ 3.
Then the surface area will be:
S = V / h = 0.025 / 0.2 = 0.125 m ^ 2.
The force will be equal to:
F = m * g + p * S = 200 * 10 + 0.125 * 8000 = 2000 + 1000 = 3 KPa.