# Question Video: Using Integration to Calculate the Force Due to Pressure from a Fluid on the inside of a Container Mathematics

A cubical container with side length 20 cm is filled with liquid mercury. Find the fluid force exerted on one side of the cubical container, given that the density of mercury is 13,534 kg/mΒ³ and π = 9.8 m/sΒ². Give your answer to 1 decimal place.

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### Video Transcript

A cubical container with side length 20 centimeters is filled with liquid mercury. Find the fluid force exerted on one side of the cubical container, given that the density of mercury is 13,534 kilograms per meter cubed and π equals 9.8 meters per second squared. Give your answer to one decimal place.

With any problem of this kind, the key is to set up the integration to be as simple as possible before making any calculations. Letβs start by making a nice, clear diagram of the scenario. So, we have a cubical container of side length 20 centimeters. And letβs convert this into meters so that weβre working entirely in SI units. So, the side length is 0.2 meters. This container is filled to the brim with liquid mercury, which weβve represented here in magenta. And weβre asked to find the fluid force on one side of the container. Letβs call this πΉ.

The total force due to pressure on the surface is given by the pressure multiplied by the surfaceβs area; πΉ equals ππ΄. However, in this case, the pressure will increase with depth from the free surface of the mercury. So, instead, we will need to integrate the pressure function over one side of the cube. There are multiple approaches to an integration like this. One approach is to reduce the 2D integral to a 1D integral. This can be set up easily because although the surface is two-dimensional, the pressure depends only on one dimension, the depth.

Consider a thin horizontal strip of the surface on the side of the cube of height πΏπ¦. This strip also has a width of 0.2 meters, but letβs just call it π€ for now for simplicity. Since pressure varies only with depth and the strip is narrow and horizontal, we can assume that the pressure along it is approximately constant. So, the total force on this strip β letβs call this πΏπΉ β is given approximately by the approximate pressure on this strip π multiplied by the area πΏπ΄, where πΏπ΄ equals π€πΏπ¦.

The pressure π is given by the hydrostatic equation π equals πππ¦, where π is the density of the mercury, π is the acceleration due to gravity, and π¦ is the depth from the free surface of the mercury. Substituting these into our approximation for πΏπΉ, we get πΏπΉ is approximately equal to πππ¦π€πΏπ¦. In the limit as πΏπ¦ tends to zero, this approximation will become exact, giving us dπΉ equals πππ¦π€dπ¦.

We have now essentially turned the problem into a one-dimensional problem since π, π, and π€ are all constant, and weβre integrating only with respect to the depth, π¦. So, the total force πΉ will be given by the integral between some lower and upper limits π¦ one and π¦ two of πππ¦π€dπ¦. Remember that we have arranged the problems such that π¦ increases as we go down. So, the lower limit π¦ one will be at the top of the cube and the upper limit π¦ two will be at the bottom of the cube. So, what are π¦ one and π¦ two?

For the lower limit π¦ one, consider a typical point at the top of the cube. π¦ is related to pressure by the hydrostatic equation π equals πππ¦. We are assuming that above the cube is a vacuum. So, the pressure at this point is zero. Since π and π are nonzero constants, π¦ at this point must also be zero. So, π¦ one is just equal to zero, and π¦ two is just 0.2 meters below that. So, π¦ two equals 0.2.

Letβs leave the limits out of the integration for now and plug them in at the end. We can also take π, π, and π€ outside of the integration, since these are all constants. So, this gives us πππ€ times the integral between π¦ one and π¦ two of π¦dπ¦. Integrating gives us πππ€ times half π¦ squared evaluated between π¦ one and π¦ two. Taking the common factor of a half outside and evaluating between π¦ one and π¦ two gives us half πππ€ times π¦ two squared minus π¦ one squared.

Plugging in all our values gives us one-half times 13,534 times 9.8 times 0.2 times 0.2 squared minus zero squared. Plugging all these into our calculator, we get 530.5. And since we were working entirely in SI units, the unit here is simply newtons.