Hydrostatic Force on a Submerged Surface, Exams of Physical education

. Objective 1. The purpose of this experiment is to experimentally locate the center of pressure of a vertical submerged plane surface. 2. The experimental measurement is compared with a theoretical prediction. 2. Apparatus Figure 1.1 is a sketch of the device used to measure the center of pressure on a submerged vertical surface. It consists of an annular sector of solid material attached to a balance beam. When the device is properly balanced the face of the sector that is not attached to the beam is directly below (coplanar) with the pivot axis. The solid sector and the balance beam are supported above a tank of water.

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2023/2024

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Hydraulics Lab. Civil Eng. Dept., Taif University Dr. Tarek Tahawy Page 1
Experiment No. 1
Hydrostatic Force on
a Submerged Surface.
1. Objective
1. The purpose of this experiment is to experimentally locate the center of
pressure of a vertical submerged plane surface.
2. The experimental measurement is compared with a theoretical prediction.
2. Apparatus
Figure 1.1 is a sketch of the device used to measure the center of pressure on a
submerged vertical surface. It consists of an annular sector of solid material
attached to a balance beam. When the device is properly balanced the face of
the sector that is not attached to the beam is directly below (coplanar) with the
pivot axis. The solid sector and the balance beam are supported above a tank of
water.
Figure 1.1 Apparatus for measuring the location of the center of pressure
Figure 1.2 Detailed nomenclature for locating the center of
pressure
pf3

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Experiment No. 1

Hydrostatic Force on

a Submerged Surface.

1. Objective

  1. The purpose of this experiment is to experimentally locate the center of pressure of a vertical submerged plane surface.
  2. The experimental measurement is compared with a theoretical prediction.

2. Apparatus

Figure 1.1 is a sketch of the device used to measure the center of pressure on a submerged vertical surface. It consists of an annular sector of solid material attached to a balance beam. When the device is properly balanced the face of the sector that is not attached to the beam is directly below (coplanar) with the pivot axis. The solid sector and the balance beam are supported above a tank of water. Figure 1.1 Apparatus for measuring the location of the center of pressure Figure 1. 2 Detailed nomenclature for locating the center of pressure

3. Related Theory

Figure 1.2 shows the submerged surface viewed from the left side of the tank in Figure 1.1. The depth of the centroid below the surface of the water is. The x-y coordinate system has its origin at the centroid. The y-direction position of the center of pressure, yR,th , is given by: where Ixc is the moment of inertia of the surface about the x-axis, and A is the vertical projection surface area. The location of the center of pressure can be measured using the apparatus sketched in Figure 1.1. The counterweight is adjusted so that the beam is horizontal when there is no water in the tank and no weight in the pan. When the tank is filled with water the unbalanced hydrostatic force causes the beam to tilt. Adding weight W to the pan at a distance L from the pivot O exerts a moment WL that counterbalances the resultant moment due to the hydrostatic forces on the quarter-annulus-shaped body ABPQ. When the water level is as shown in the Figure 1.1 , there are hydrostatic forces on surfaces AB , BS and AT. Since BS and AT are concentric cylindrical surfaces with the common axis passing through O, the hydrostatic forces on BS and AT do not exert any moment about O. As a result, WL is equal to the moment due to the hydrostatic force F acting on the vertical plane surface AB. In this experiment the force F is not measured. Instead, the theoretical value

F =  ghA is assumed, where h is the depth of the centroid of the surface. The

moment due to F is measured and the theoretical value of F is used to compute the location of the center of pressure. Balancing the moments about O gives WL = F(H + yR)

Substituting F=  ghA and solving for yRm yields

Equation 1. Equation 1. 2 th m