# Amount of force on a trapezoidal surface following solution

• IntegrateMe
In summary, the conversation discusses the calculation of the force on one trapezoidal side of a container that is exposed to rainwater. The container can only support 6000 Newtons of force and eventually breaks due to water pressure. The solution involves using an integral to calculate the force, with the depth being (H-h) and the area being (1+h)ΔH. However, there is confusion about the correct values for depth and area.
IntegrateMe
A trash container is located outside a building. It starts to rain, causing water to enter the vessel. The trapezoidal side of the vessel is cracked and can only support 6000 Newtons of force; eventually the water pressure causes the wall to break.

Container

Write an integral expressing the amount of force on one trapezoidal side of the vessel when the height of the water is H meters. The density of rainwater is 1000 kg/m3 and the force of gravity is 9.8 N/kg.

The solution says:

F = δ*g*depth*A
depth = (H-h)
A = (1+h)ΔH

And so F = ∫1000(9.8)(H-h)(1+h)dh ; from 0 to H

I don't understand how they got (H-h) to be the depth nor how the area = (1+h)ΔH. Anyone care to explain?

Thank you!

Anyone?

IntegrateMe said:
A trash container is located outside a building. It starts to rain, causing water to enter the vessel. The trapezoidal side of the vessel is cracked and can only support 6000 Newtons of force; eventually the water pressure causes the wall to break.

Container

Write an integral expressing the amount of force on one trapezoidal side of the vessel when the height of the water is H meters. The density of rainwater is 1000 kg/m3 and the force of gravity is 9.8 N/kg.

The solution says:

F = δ*g*depth*A
depth = (H-h)
A = (1+h)ΔH

And so F = ∫1000(9.8)(H-h)(1+h)dh ; from 0 to H

I don't understand how they got (H-h) to be the depth nor how the area = (1+h)ΔH. Anyone care to explain?

Thank you!

H-h is not the depth of the water. It is the distance below the surface that the horizontal dh element where you are calculating the force is. And I also think that (1+h) is not the correct length of that horizontal dh element of area and dH should be dh in his expression for A.

## 1. What is the equation for calculating the amount of force on a trapezoidal surface?

The equation for calculating the amount of force on a trapezoidal surface is the product of the pressure and the area of the surface, multiplied by the cosine of the angle between the force vector and the surface. This can be represented as F = P x A x cos(theta).

## 2. How do you determine the pressure on a trapezoidal surface?

To determine the pressure on a trapezoidal surface, you will need to know the force applied to the surface, as well as the area of the surface. The pressure can then be calculated by dividing the force by the surface area.

## 3. What factors can affect the amount of force on a trapezoidal surface?

The amount of force on a trapezoidal surface can be affected by several factors, such as the angle of the force vector, the surface area, and the pressure applied to the surface. Additionally, any external forces acting on the surface, such as friction or gravity, can also impact the amount of force.

## 4. Can the amount of force on a trapezoidal surface be negative?

Yes, the amount of force on a trapezoidal surface can be negative. This would occur if the force vector is acting in the opposite direction of the surface, resulting in a negative value for the force.

## 5. How can the amount of force on a trapezoidal surface be measured experimentally?

The amount of force on a trapezoidal surface can be measured experimentally using a variety of methods, such as using a force sensor or a load cell. These instruments can accurately measure the force applied to a surface and provide numerical data for analysis.

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