# Tag Info

## Hot answers tagged fluid-dynamics

4

When the ball is suspended, the weight of the ball is balanced by the tension in the string and the buoyancy force. Here buoyancy force on the ball has an equal and opposite reaction on the water itself thus this extra force must be accounted for. Why shouldn't it be weight of beaker + Weight of ball - Bouyant force on the ball This would be the case if ...

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No matter how much denser the ball is than the water, when held submerged but not sank to the bottom, the weight of the beaker on a scale will only increase by the weight of the volume of water the ball displaces, its buoyant force, per Archimedes principle of buoyancy. The tension on the string will support the extra amount of the ball's weight that is ...

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We do. The walls of the tube are pulled down. However, virtually in all meaningful situations, the tube is sufficiently anchored such that this small force is undetectable. It just gets transferred onto the ground/table/etc. However, were you to do this experiment in zero-gravity, and expose an open tube to a drop of water, you would indeed see that the ...

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I stumbled upon this question while searching for something else. You have probably found and answer yourself by now, but here's a quick attempt to answer your question anyway. $-\nabla\Phi$ represents gravity. $\Phi$ is the so called geopotential, which is a nice way to handle gravity. If we want the simple, high-school physics version of gravity where we ...

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Newton's Third Law: action begets reaction. As the water exerts a buoyant force on the ball, the water exerts an opposite force on the beaker, thus scales. It's a classic part of density determination for irregularly shaped bodies.

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[...] when it is clearly changing density [...] The shape the water is in has nothing to do with its density. It is changing shape, not density. Fill a cube of 1 m side length with water. Now put the same water in a cube with 10 m side length. The water will fill the larger cube only 1 cm high and is spread out over a much larger area. However, the volume ...

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In an incompressible fluid in a gravitational field, the pressure at specific point in the fluid depends on the gravitational field strength, $g$ (some call this acceleration due to gravity, unfortunately), the density of the the fluid, $\rho$, and the depth (in the direction of the grav. field $\vec{g}$) of the point, $D$, plus whatever the pressure is at ...

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In $h \rho g$ h is the height or rather depth of that point from free surface. To find pressure at lower face of cube you consider height $h_°$ . When you are taking the block down the depth of the point on lower surface increases that is which it is said that pressure increases with depth the force exerted by box increases with depth Yes you are right ...

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Your last line has some errors and should read (with two denominators changed) $$=\left(\frac{\partial\sigma_{xx}}{\partial x}+\frac{\partial\tau_{yx}}{\partial y}+\frac{\partial\tau_{zx}}{\partial z}\right)dx\,dy\,dz$$ By "there opposite terms", I think you're asking why $[\sigma_{xx}(dy\,dz)+\tau_{xy}(dx\,dz)+\tau_{zx}(dx\,dy)]$ is subtracted. ...

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The continuity equation reads $$\frac{\partial \rho}{\partial t}+v\centerdot \nabla \rho+\rho \nabla \centerdot v=0$$where $\rho$ is the fluid density. Dividing this by $\rho$ gives $$\frac{1}{\rho}\left(\frac{\partial \rho}{\partial t}+v\centerdot \nabla \rho\right)+\nabla \centerdot v=0$$But, since the density is the inverse of the specific volume V, we ...

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If the bottom section were horizontal and the datum for potential energy were taken as the bottom, then the total potential energy would be $$U=\rho gA \frac{(H+y)^2}{2}+\rho gA \frac{(H-y)^2}{2}=\rho gA(2H^2+y^2)$$where H is the equilibrium height; and the total kinetic energy would be $$K=\frac{M}{2}\left(\frac{dy}{dt}\right)^2$$So the total mechanical ...

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The potential energy $U$ is a function of the displacement $y$ from the liquid's equilibrium position. Although the final answer will be symmetric in $y$, for the sake of definiteness let's take $y$ to be positive when the liquid in the right hand side of the tube is raised by a distance $y$. If the absolute potential energy at the equilibrium position is \$...

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