the pressure in a fluid at rest is the same at all points if they are at the same height. This can be be demonstrated in a simple way.
The figure below shows an element in the interior of a fluid at rest. This element ABC-DEF is in the form of a right-angled prism. In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of the gravity is the same at all these points.
Let us enlarged this element. The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element . Thus, the fluid exerts pressures Pa, Pb and Pc on this element of area corresponding to the normal forces Fa, Fb and Fc as shown in fig.ures on the faces BEFC, ADFC and ADEB denoted by Aa, Ab and Ac respectively.
Fb sinθ = Fc, Fb cosθ = Fa (by equilibrium)
Ab sinθ = Ac, Ab cosθ = Aa (by geometry)
Thus, ratio of force per area and hence pressure is same at all points A,B and C.
Hence, pressure exerted is same in all directions in a fluid at rest. Pressure is not a vector quantity. No direction can be assigned to it. The force against any area within (or bounding) a fluid at rest and under pressure is normal to the area, regardless of the orientation of the area.
In a a fluid element in the form of a horizontal bar of uniform cross-section. The bar is in equilibrium. The horizontal forces exerted at its two ends must be balanced or the pressure at the two ends should be equal. Hence for a liquid in equilibrium the pressure is same at all points in a horizontal plane.
If the pressure were not equal in different parts of the fluid, then there would be a flow as the fluid will have some net force acting on it. Thus in the absence of flow the pressure in the fluid must be same everywhere. Wind is flow of air due to pressure differences.
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The figure below shows an element in the interior of a fluid at rest. This element ABC-DEF is in the form of a right-angled prism. In principle, this prismatic element is very small so that every part of it can be considered at the same depth from the liquid surface and therefore, the effect of the gravity is the same at all these points.
Let us enlarged this element. The forces on this element are those exerted by the rest of the fluid and they must be normal to the surfaces of the element . Thus, the fluid exerts pressures Pa, Pb and Pc on this element of area corresponding to the normal forces Fa, Fb and Fc as shown in fig.ures on the faces BEFC, ADFC and ADEB denoted by Aa, Ab and Ac respectively.
Fb sinθ = Fc, Fb cosθ = Fa (by equilibrium)Ab sinθ = Ac, Ab cosθ = Aa (by geometry)
Thus, ratio of force per area and hence pressure is same at all points A,B and C.
Hence, pressure exerted is same in all directions in a fluid at rest. Pressure is not a vector quantity. No direction can be assigned to it. The force against any area within (or bounding) a fluid at rest and under pressure is normal to the area, regardless of the orientation of the area.
In a a fluid element in the form of a horizontal bar of uniform cross-section. The bar is in equilibrium. The horizontal forces exerted at its two ends must be balanced or the pressure at the two ends should be equal. Hence for a liquid in equilibrium the pressure is same at all points in a horizontal plane.
If the pressure were not equal in different parts of the fluid, then there would be a flow as the fluid will have some net force acting on it. Thus in the absence of flow the pressure in the fluid must be same everywhere. Wind is flow of air due to pressure differences.
Related posts :
Problems on Bernoulli's theorem and Its Applications
Bulk Modulus
Shear modulus
Elastic behavior of Solids
Stress and strain
Stress and Strain Curve
Determination of Young's modules
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