Why must we use integral to find the resultant force?

In summary, the homework statement is that Hydrostatic force on a plane surface ex: and Hydrostatic force on a gate: are different. The Attempt at a Solution says that the homework equation for problem 3.57 can be used, but there are adjustments to be made because the width (the b variable in dbdh) is constant.
  • #1
EastWindBreaks
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Homework Statement


Hydrostatic force on a plane surface ex:
upload_2017-11-13_3-15-19.png


Hydrostatic force on a gate:
upload_2017-11-13_3-12-35.png


Homework Equations



The Attempt at a Solution


why can't we just use the formula in the red box above for problem 3.57?
instead I have to use integral
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I am confused, how does this gate different from a slanted plane. if I look at the gate from front view, i feel like they are just the same thing. How does
upload_2017-11-13_3-38-24.png
a double integral with respect to area (dhdb) became a single integral with respect to dh?
 

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  • #2
oh! I can use it but because its the top view, we have to use 5 meter for h_c, which is the depth that affects the pressure.
 
  • #3
ok, I have one question left, how does
upload_2017-11-13_3-38-24-png.png
a double integral with respect to area (dhdb) became a single integral with respect to dh?
 

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  • #4
EastWindBreaks said:
double integral with respect to area (dhdb) became a single integral with respect to dh
Because the width (the b variable in dbdh) is constant.
EastWindBreaks said:
how does this gate different from a slanted plane.
It doesn't really. Just bear in mind that the gates are vertical, so in the equation in the red box you would put θ=π/2. On the other hand, the area is not WD. To express the answer in terms of W and D you need to make reference to the 15° angle.
With those adjustments, I think you will find the two expressions are the same.
 
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  • #5
haruspex said:
Because the width (the b variable in dbdh) is constant.

It doesn't really. Just bear in mind that the gates are vertical, so in the equation in the red box you would put θ=π/2. On the other hand, the area is not WD. To express the answer in terms of W and D you need to make reference to the 15° angle.
With those adjustments, I think you will find the two expressions are the same.
oh yeah, I got it now, thank you!
 

FAQ: Why must we use integral to find the resultant force?

Why is the use of integrals necessary in finding resultant force?

Integrals are necessary because they allow us to calculate the total effect of continuous and varying forces on an object. This is important because forces can vary in magnitude and direction, and integrals allow us to consider the effect of each infinitesimal force on the object.

Can't we just use simple addition to find resultant force?

No, simple addition is only applicable to forces that are constant and acting in the same direction. When forces are varying and/or acting in different directions, simple addition will not give an accurate result.

How does an integral help in finding the direction of resultant force?

An integral helps in finding the direction of resultant force by breaking down the varying forces into smaller components and considering their effects in different directions. By summing up these components using an integral, we can determine the direction of the resultant force.

Is using an integral the only way to find resultant force?

No, there are other methods such as vector addition and trigonometric methods that can be used to find resultant force. However, integrals are often the most accurate and efficient method when dealing with continuous and varying forces.

Are there any real-life applications of using integrals to find resultant force?

Yes, there are many real-life applications such as calculating the total force exerted on a structure by wind or water, finding the center of mass of irregularly shaped objects, and determining the torque on a rotating object.

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