Suppose I have a glass full of water, upside down, so that the open end is at the bottom, resting on top of a weight measuring scale. Now from what I've learnt scales show the weight as the external force acting upon it excluding the reaction force from the ground(or so my professor told me while I was solving those elevator problems). So if I put the glass full of water like I said, inverted on top of the scale, shouldn't the hydrostatic pressure of the water exert some force on the scale too(just like a did exerts pressure on the walls of the container)? Wouldn't that cause the measured weight of glass+water to be higher than the actual weight?
$\begingroup$
$\endgroup$
asked Nov 30, 2017 at 20:37
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
Weight of water causes pressure. The scale measures weight (force), not weight-per-unit-area (pressure), and makes the correct summation. To confuse the scale, one could rest a thumb on the pan, or attach a string and lift: there would then be a force OTHER than weight and the structural support of the scale pan that counters the weight.
-
$\begingroup$ I meant the force applied due to the hydrostatic pressure. Like if the height of the glass is h and the contact area of the water and scale is A, then the extra force applied on the scale due to pressure would be h(rho)gA, right? $\endgroup$ Dec 1, 2017 at 5:42
-
$\begingroup$ That 'hydrostatic pressure' is weight, reported on a per-square-centimeter basis. It isn't an extra term. Earth's gravity acting on the water is no different from any other weight force, and if the scale pan supports it, it is measured. $\endgroup$– Whit3rdDec 1, 2017 at 6:53
-
$\begingroup$ Hmm but in that case wouldn't the pressure at the bottom be different for different container shapes? Like suppose there is a cylindrical container and an inverted cone as a container. If both of them has the same height, and A represents the cross sectional area of the cylinder AND the area of the circular base of the cone, then the weight per unit area at the bottom comes out h(rho)g and h(rho)g(1/3) due the cylinder and cone respectively. But I was told that the hydrostatic pressure didn't depend on the shape of the container. What did I do wrong? $\endgroup$ Dec 1, 2017 at 8:45
-
$\begingroup$ The weight is the up/down component only (the sideways components of the pressure forces all cancel) of the pressure, integrated over all the container area, which is a rather complicated way to weigh a quantity of liquid. But, it is effective (and gives the same result as the scale). If the answer weren't the same, one could make a perpetual motion machine. If the sideways forces didn't cancel, that could propel a car.... $\endgroup$– Whit3rdDec 1, 2017 at 9:23
-
$\begingroup$ I never said anything about there being a net sideways force... $\endgroup$ Dec 1, 2017 at 9:35