# Is pressure from surface on cylinder = weight of cylinder when in water?

Written here are net forces acting on a body. But I am having difficulty in understand them after reading from textbook and online material.

In the book, we have been introduced to thrust.

It clearly says that pressure exerted by a body on the surface = weight of the body.

Then, why is it not that $$mg$$ is nothing but $$P_1A$$ and $$P_2A$$?

Can we also say that it is $$F = ma$$ expression (the first equation as a whole) where $$a = 0$$?

• Thrust has the units of force not pressure. Please recheck Dec 26 '20 at 6:53
• @shahrozeshahab Please recheck my post.I Dec 26 '20 at 6:58
• Weight of a body is equal to the difference of thrust P1A and P2A ( thrust acting on the top and bottom respectively) only if its in equilibrium Dec 26 '20 at 7:02

It clearly says that pressure exerted by a body on the surface = weight of the body.

It is simply impossible — their units are different: $$\left[N \over m^2\right]$$ vs. $$[N]$$.

Your book says something other - thrust, not pressure.

The picture is about forces and their difference. There are

• 2 forces in the top-to-down direction (the gravitational force and the force caused by the water pressure)
• 1 force in the down-to-top (the force caused by the water pressure).

So the first two forces are added together and the last force is subtracted from their sum.

The resulting force must be a zero force, because the cylinder is not moving.

• Ok.I understood.I have also put the photo of what was there exactly in my book.I got very confused with that.Thank you for helping. Dec 26 '20 at 6:53
• You're welcome. What should I do when someone answers my question? Dec 26 '20 at 6:59
• Just one thing.how is it possible that my book has written wrong statement ?.That is a very big thing right Dec 26 '20 at 7:02
• No, the statement in your book is correct — see my (edited answer), it is about thrust, not about pressure. Dec 26 '20 at 7:05
• Oh.Ok thank you. Dec 26 '20 at 7:06