An aquarium of length , width (front to back) , and depth is filled to the top with liquid of density
a. Find an expression for the force of the liquid on the bottom of the aquarium.
b. Find an expression for the force of the liquid on the front window of the aquarium.
c. Evaluate the forces for a-long, deep aquarium filled with water.
Force acting on the aquarium in plane is
Expression for hydrostatic pressure at depth is given by
Here is the surface pressure where is the density of the liquid.
Let force acted upon by the liquid on bottom of the aquarium, Force acted upon by the liquid on the bottom of the aquarium equals to the gravity of the liquid because the bottom of the aquarium is the only support in vertical direction:
The volume equals to length times width times depth,
Density of the liquid is given by,
Re arrange equation for
Substitute in the equation
Substitute in the equation
Therefore force acted upon by the liquid on bottom of the aquarium is
The hydrostatic pressure at depth is given by
So in the front window, at spot depth , there is the pressure indicated above pushing the windows outward. There is also a pressure due the atmosphere that pushes the window inward, which cancels the first term in the above equation. So the net pressure is the one only related to the liquid:
Here is the density of the liquid; is the distance from the point of interest to the surface of the liquid. And is the gravitational acceleration. The total force on the window is to integrate the pressure in the plane,
At depth in our coordinates is actually so we have,
The function does not depend on .So we can integrate the above equation
We can continue to do the integration over
The result of inetgration of
Conversion of units of length of the aquarium from cm to
Conversion of units of breadth of aquarium from to
Conversion of units of depth of the aquarium from
for in equation
Therefore the force acting on aquarium is
Using part (b) expression for force acting on aquarium in plane
A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends , but it's been a while since you've checked. You can't find a tire gauge in the car, but you do find the owner's manual and a ruler. Fortunately, you've just finished taking physics, so you tell your friend, "I don't know, but I can figure it out." From the owner's manual you find that the car's mass is. It seems reasonable to assume that each tire supports one-fourth of the weight. With the ruler you find that the tires are wide and the flattened segment of the tire in contact with the road is long. What answer-in psi-will you give your friend?
An object has density ρ.
a. Suppose each of the object’s three dimensions is increased by a factor of 2 without changing the material of which the object is made. Will the density change? If so, by what factor? Explain. b. Suppose each of the object’s three dimensions is increased by a factor of 2 without changing the object’s mass. Will the density change? If so, by what factor? Explain.
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