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# A tank is 8 m long, 4 m wide, 2 m high, and contains kerosene with density $820 kg/m^3$ to a depth of 1.5 m. Find (a) the hydrostatic pressure on the bottom of the tank, (b) the hydrostatic force on the bottom, and (c) the hydrostatic force on one end of the tank.

## (a) $$\approx 12 \mathrm{kPa}$$(b) $$\approx 3.86 \times 10^{5} \mathrm{N}$$(c) force on the 4 $\mathrm{m}$ side: 36162 $\mathrm{N} \quad$ force on the 8 $\mathrm{m}$ side: 72324 $\mathrm{N}$

#### Topics

Applications of Integration

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

in this problem, we're given that there is a thing which is eight meters long, four meters wide and two meters deep and that has failed. What in fluid with 820 clever kilograms per meter, you density and it is filled with that flute tow. It leapt off 1.5 meters. In part, they were us to find the pressure at the bottom off the stank. We know that pressure is density times G times. That sedan city for this fluid is 820 g is 9.8, and that is given that is 1.5 a few months old. We see that then the pressure is 12 1000 future for Paschal's, and we can write this one us about 12. Kill Paschal's in part B. Where has to find a force acting at the bottom of the cylinder. It means that we would need to talk plate the area that this pressure's acting on, and that would be this shaded area. Force will be pressure times area where area in this case as eight times for that is 32 meter skirt, so force would then be 12 1054 times 32. That would be three times 86 10 to the fifth in evidence and in party where has to find the force acting on Ah, one end off this tank's. Imagine that we're looking at this time from this side and this is the side of this Think so He had this one. And, um, certain this is four meters, since that is for me, it is white and the oil that is two meters of everyday this field. What if would up to 1.5 meters? So we're interested in this tin strip right now. Let's assume that we have this sense stripped and the thickness as D X sort of totaling as four area off. This would be four times DX. All right, now, let's assume not be having origin located here. And the positive why is in the downwards directions as we go down? What is positive? You know, that pressure is ro times G times, uh, that hear that is the distance measure for Miss Surface. Unless said, I bet his ex Sadat would be X and pressure then would be, um a 20 times 9.8 times. Thanks force would then be pressured times area forever. This is a 10 strip and if he some all those 10 strips up, we would find the total force acting on one answer I need It means that mean it Intergrated starting from zero to a 1.5. Since it is filled with water up to 1.5 meters, you would have a 20 times 9.8 times sex. That is pressure multiplied by area. And we found area to be four times the X from the species that we have a 20 times 9.8 times. For this is a constant and form this integral be would got excrete over to Now we need to let this between zero and 1.5. If you do so, we find a force to be 36,000 and 162 engines.

#### Topics

Applications of Integration

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp