Finding the force using calculus

In summary, the Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curve y=0.5 x^2 and below the line y= 160. The water level is 22 meters below the top of the dam. Find the force (in Newtons) exerted on the dam by water pressure.
  • #1
blumfeld0
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The Deligne Dam on the Cayley River is built so that the wall facing the water is shaped like the region above the curve y=0.5 x^2 and below the line y= 160. (Here, distances are measured in meters.) The water level is 22 meters below the top of the dam. Find the force (in Newtons) exerted on the dam by water pressure. (Water has a density of 1000 kg/m^3, and the acceleration of gravity is 9.8 m/sec^2 .)
F(x) = density of fluid* Volume displaced *g = 1000*9.8*volume displaced

so what is my volume displaced? is it simply .5 *x^2 -160 ??

and what integral do I do? are my limits of integration from 0 to 22?thank you!

blumfeld0
 
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  • #2
blumfeld0 said:
so what is my volume displaced? is it simply .5 *x^2 -160 ??

Woah, woah. This isn't a buoyancy problem. I don't think any fluid is in fact being displaced here. By the way, you didn't mention the overall depth of the water.

Go back to your definitions. Pressure is Force per unit Area, P=F/A. You're being asked to compute the total force on the dam due to fluid pressure. At the simplest level, what you need to do is find the area of the face of the dam touching the water, then multiply that area by the "average" water pressure on the area. That will give you a very, very crude "guess"timate of the answer.

OK, the trouble here is that not only does the water pressure vary with depth, increasing as you go downwards, the direction the dam faces in is also changing, meaning you have to compute the direction of the force as well as magnitude. To top it all off, the surface area may also be dependant on position, by which I mean the amount of surface area as you look "head on" at the dam really depends on how sloped it is at the point you're looking at.

This is where your integral calculus comes in. You will need to find two things;

1) A representation of the pressure as a function of the depth d. P(d)
2) A representation of the surface area vector for the dam. dS(x,d), where d is the depth. This will give you the direction the surface faces in, and the infinitesimal area of the surface at a point.

You then need to compute the two dimensional integral
[tex]\iint_A P(d) d\mathbf{S}(x,d) dd dx[/tex]

Integration here is really the advanced "multiplication" we need to compute the total force on the surface face. It's essentially analagous to the basic F= P*A in the simple case.
 
  • #3
No, you do not integrate "from 0 to 22" and, more importantly, you do not integrate with respect to x!

x is measured horizontally and pressure does not change horizontally. You need to integrate vertically, with respect to y. The height of the water is 160- 22= 158 m. You will need to integrate, with respect to y, from 0 to 158.
Of course, since y= (1/2)x2, [itex]x= \pm \sqrt{2y}[/itex] so the length of a thin horizontal strip is is [itex]2\sqrt{2y}[/itex] and its area is [itex]2\sqrt{2y}dy[/itex]. The force on that strip is the pressure due to a weight of water above that height, y, in a 1m by 1m column, times that area. The weight of that water is mgh= (9.81)(1000)(158- y) so your integral is
[tex](2)9810\int_0^{158}(158-y)\sqrt{2y}dy[/tex]
 
  • #4
omg! I am in calculus 2. we haven't even learned 2-D integrals. that's not til next semester. ok thank you! i will need to sit down and understand this. i have a feeling it will on the final!


blumfeld0
 

Related to Finding the force using calculus

What is the concept of finding the force using calculus?

The concept of finding the force using calculus involves using mathematical techniques to determine the force acting on an object at a given point in time. This can be done by calculating the derivative of the object's position or velocity, or by integrating the object's acceleration over a given time interval.

Why is calculus necessary for finding force?

Calculus is necessary for finding force because it allows us to analyze the relationship between an object's position, velocity, and acceleration. By taking derivatives and integrals, we can determine the instantaneous force acting on an object at any given time.

What is the difference between using calculus and using other mathematical methods to find force?

Using calculus allows for a more precise and accurate determination of force compared to other mathematical methods. This is because calculus takes into account the changing nature of an object's position, velocity, and acceleration over time, rather than just looking at a single moment in time.

What are some real-world applications of finding force using calculus?

Finding force using calculus has many practical applications, such as determining the forces acting on a moving object in physics, analyzing the stress and strain on structures in engineering, and predicting the motion of objects in space in astronomy.

Are there any limitations to using calculus for finding force?

One limitation of using calculus for finding force is that it assumes the object in question is moving in a continuous and smooth manner. This may not always be the case in real-world situations, such as when an object experiences sudden changes in acceleration or encounters external forces.

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