In the situation shown above, Point A is just near the wall at negligible distance. But how is pressure at A is due to weight of water and atmospheric pressure only,why does it not have any pressure due to horizontal net force (point would be accelerating with container horizontally therefore it would have some net force horizontally) even though. If it is due to it being point mass then how does pressure at B have additional pressure due to horizontal net force even though it is also a point mass
$\begingroup$
$\endgroup$
0
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
Let us start with a little cube-shaped drop of water floating in space. Squeeze it on two sides. It squishes into a flatter shape. To prevent this, you would have to squeeze with equal force on left/right, front/back, and top/bottom. The forces on all sides have to be equal.
Consider a similar drop of water sitting quietly in a container of water. Forces are acting on the drop. First, there is the weight of all the water above it plus the weight of all the air above that. Because the drop is sitting quietly, not being accelerated, we know the total force is $0$. The force on the front equals the force on the back. Likewise left and right.
The drop has a little weight, so top and bottom are slightly different. The force on the top is the weight of the water above. The force on the bottom is that weight plus the weight of the drop. The drop below our drop pushes back just as hard as our drop pushes on it. So the total vertical force is also $0$. The downward forces are the weight of the water above plus the weight of the drop. And that is the same as the upward force from below.
So the horizontal forces do matter. Without the walls, there would be no horizontal forces, and the water would flow all over the floor. Again, to keep the drop from squishing out of shape, the sideways forces must be equal to the vertical forces. The sideways forces must be a little bigger at the bottom of the drop.
Your question is probably in response to only needing atmospheric pressure and depth to calculate the pressure at A. This is the nature of reaction forces.
If you push on an object, it accelerates. If the object is against a wall, it doesn't accelerate because the wall pushes back just a hard. The force from the wall adjusts to be just as hard as you push on it.
So if you calculate the vertical forces, you know the horizontal forces are just as big. You don't need a separate calculation for them.
answered Jan 26, 2023 at 16:21
-
$\begingroup$ So sir do you mean to say that if a drop is against a wall then it won't accelerate even though the wall is accelerating because the wall will apply equal force? So the drop will remain in its position and when after some instance the drop is no longer against the wall and there is some fluid between our drop and wall then the drop will start accelerating because the fluid(between drop and wall) won't balance out forces on both sides? $\endgroup$– RaghavCommented Jan 26, 2023 at 17:00
-
$\begingroup$ I was assuming the container was not accelerating. In general, the wall pushes back just hard enough to prevent an object from penetrating the wall. This will happen if the object has the same acceleration as the wall. In your drawing, if you push the object against the wall, the wall will push back less hard so the object has the same acceleration as the wall. In this case, pressure changes horizontally, like weight makes pressure change vertically. $\endgroup$ Commented Jan 26, 2023 at 17:07
-
$\begingroup$ So then, if drop at A is accelerating along with wall then it means that there is some net force along horizontal, then shouldn't that net force also contribute to pressure at A and therefore pressure at A should be more than Patm+dHg(d=density)? $\endgroup$– RaghavCommented Jan 26, 2023 at 17:20
$\begingroup$
$\endgroup$
2
If the container is accelerating to the right , there is a force in the opposite direction due to inertia. So the pressure at point B would in fact be greater than at point A. The acceleration has the same effect in a horizontal direction as earth's gravity has in the vertical direction. Pressure $p = p_{atm} + \rho \times (a \times x + g \times y)$, where $p_{atm}$ is atmospheric pressure, $\rho$ is the density of water, a is the acceleration of the container, x is the distance of the point from the right wall, g is the local g factor, and y is the distance from the surface
-
$\begingroup$ But why would pressure at A would only be due to weight of fluid above it and Atmospheric pressure? If the point A is accelerating right along with container then there should be some net force along horizontal to provide it and that net force should also provide additional pressure at A right? $\endgroup$– RaghavCommented Jan 26, 2023 at 17:03
-
$\begingroup$ The additional pressure at point B in the horizontal direction results from the sum of the forces acting along the distance AB. Since there are no forces acting on the fluid at point A from the right side, there is no additional pressure here. It may be helpful to consider that acceleration and gravitational attraction have an equivalent effect: You are not surprised that the pressure is only the atmospheric pressure at the surface of a stationary container in gravity, but that container is in the same situation as a container being accelerated upwards. $\endgroup$ Commented Jan 28, 2023 at 11:43