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Question Video: Calculating Total Force Exerted Due to Pressure Physics • 9th Grade

The hull of a sunken boat, laying on the seabed, is 12 m below the surface of the sea, where the sea water has an average density of 1,025 kg/m³. The hull’s surface area is 15 m². What is the total force exerted by seawater on the hull?

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Video Transcript

The hull of a sunken boat, laying on the seabed, is 12 meters below the surface of the sea, where the seawater has an average density of 1,025 kilograms per cubic meter. The hull’s surface area is 15 meters squared. What is the total force exerted by seawater on the hull?

Let’s say that this is the hull of our sunken boat lying on the seabed. The hull is a depth of 12 meters below sea level. We’ll call that 𝑑. And it’s got a surface area we’ll call 𝐴 of 15 meters squared. We want to know the total force exerted on the hull by seawater. If we imagine the column of seawater that’s above the hull, all of that is weight pushing down on the hull. The force of the seawater on the hull is spread out over its surface area. Therefore, it creates a pressure.

According to Pascal’s principle, in general, pressure is equal to force over area. When it comes to fluids, the pressure of a given fluid equals that fluid’s density 𝜌 times the acceleration due to gravity 𝑔 multiplied by the height or depth of the fluid. In our case, it’s the force 𝐹 that we want to solve for. Rearranging this equation, that force equals the pressure of the seawater on the hull multiplied by the hull’s area. We know that area, but we don’t yet know the pressure 𝑃. However, we can solve for it using this relationship. Replacing 𝑃 in our equation with 𝜌 times 𝑔 times 𝑑, we can now note that we’re given the density 𝜌 of seawater. That’s 1,025 kilograms per cubic meter. We know the depth below the surface at which the hull is located. And we also know that hull’s area 𝐴. Lastly, 𝑔, the acceleration due to gravity, is 9.8 meters per second squared.

With all these values plugged into this equation, let’s consider for a second the units involved. Notice that in the denominator of our density, we have meters cubed. That density though is multiplied by a length in units of meters and an area in units of meters squared. We can divide out then this unit of meters cubed in numerator and denominator. We’re left with units of kilograms meters per second squared. These are the units of newtons.

Indeed then, we are calculating a force. Numerically, the number we get is 1,808,100. And as we’ve seen, the units of this result are newtons. The total force exerted by seawater on the hull is 1,808,100 newtons.

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