fluid force on vertical side of tank the weight density of water is 62.4 trapazoid

Using formula from table of integrals for first part and u substitution for second part we get, Window is a circle of radius 1 ft, so we can use circle equation with radius r=1.

Lets assume a slice of width Δy which is 8-y ft below the water. If the tank is filled to the top with water, find the force on one end of the tank. lets calculate the force on one slice of width Δy which is y m deep down.Įxample3: The vertical ends of a tank are semicircles of radius 10 ft. Solution: a) As the bottom of pool act like a horizontal plate submerged in water, so pressure on bottom would be constant and can be calculated as,ī) Since 5×3 end of pool act like a vertical plate. If pool is completely filled with water find the force on Where h(y) is depth of fluid at y and L(y) is the horizontal length of region at y.Įxample2: Consider a swimming pool 5 m wide, 10 m long and 3m deep. So the force exerted by a fluid of constant weight density w, against a submerged vertical plane from y=a to y=b is given as, If plate is submerged vertically then pressure is not constant throughout, so force acting on it must be calculated through slicing which leads to integral. We don’t need calculus for this which is illustrated by the following example.Įxample1: Find the fluid force on a rectangular metal sheet measuring 3ft by 4ft, that is submerged horizontally in 6 ft. If the plate is submerged horizontally then pressure is constant and can be calculated using the equation F=P*A.

Weight density of water= 62.4 lb/cubic ft. Fluid force on a submerged horizontal surface of area A is , Pascal’s principle states that pressure exerted by a fluid at a depth h is transmitted equally in all directions. Where w= weight density of liquid per unit of volume. The pressure on an object at depth h in a liquid is given as,

Fluid force on vertical side of tank the weight density of water is 62.4 trapazoid