Variation of Pressure with Depth problem

In summary, a rectangular tank filled with water 2 m deep has a rectangular hatch on one side that is 1 m high and 2 m wide, hinged at the top. For part (a), the force of the water on the hatch was found to be 2.943 * 10^-4 N. To find the torque caused by the water about the hinges, the distance to the top of the hatch must be integrated with the force on a horizontal strip of the hatch. The correct expression for dT is pg(2-y)*2dy*2, giving a final answer of 16.3 kN.
  • #1
Fantasmickat
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Homework Statement



A rectangular tank is filled with water 2 m deep. At the bottom of one side wall is a rectangular hatch 1 m high and 2 m wide that is hinged at the top of the hatch. a) Determine the force the water causes on the hatch. b) Find the torque caused by the water about the hinges.

I only need help with part (b)

Homework Equations



T = Fd

The Attempt at a Solution



For part a)
P = pg(2-y) ; dA = w*dy
dF = PdA = (integral from 0 to 1) pg(2-y)(wdy)
Plugging in numbers, (integral from 0 to 1) (1000)(9.81)(2-y)(2)dy
=1000(9.81)(2) [2y - y^2/2]
= 2.943 * 10^-4 N

I found out how to get force. With this answer I need to find the torque about the hinges. I know T = Fd but I'm confused as to how to approach this problem because the force varies with depth. What is the value of d?

dT = dF*r = pg(2-y)*2dy*2

This was my attempt but it's definitely way off. The answer is 16.3 kN.

Thanks!
 
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  • #2
you have to integrate the distance to the top of the hatch multiplied by the force on a horizontal strip of the hatch. If your y-coordinates start with 0 at the bottom of the tank and positivie is up, than the distance to the top of the hatch is NOT (2-y).
I also don't understand the dy^2 in your expression for dT.
 

1. What is the Variation of Pressure with Depth problem?

The Variation of Pressure with Depth problem is a concept in fluid mechanics that describes how the pressure of a fluid changes as the depth increases. It is based on the principle of hydrostatics, which states that the pressure in a fluid at rest increases with depth.

2. What factors affect the variation of pressure with depth?

The main factors that affect the variation of pressure with depth are the density of the fluid, the acceleration due to gravity, and the depth itself. As the density and depth increase, the pressure also increases. However, as the acceleration due to gravity decreases, the pressure decreases.

3. How is the variation of pressure with depth calculated?

The variation of pressure with depth can be calculated using the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth. This equation is known as the hydrostatic equation and is used to determine the pressure at any given depth in a fluid at rest.

4. How does the variation of pressure with depth affect objects submerged in a fluid?

The variation of pressure with depth can have a significant impact on objects submerged in a fluid. As the depth increases, the pressure exerted on the object also increases, causing it to experience a greater force in the upward direction. This is known as buoyancy and is the reason why objects float in fluids.

5. What are some real-world applications of the variation of pressure with depth problem?

The variation of pressure with depth problem has many real-world applications, including in the design of submarines, scuba diving, and deep-sea exploration. It is also important in understanding weather patterns and ocean currents, as well as in the study of the Earth's crust and the behavior of volcanoes.

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