How to calculate pressure when two forces are exerted

[physics565]

Homework Statement

John exerts a force of 10 N on a side of a table. The cross-sectional area is $20 cm^2$. A friend of John exerts a force of 20 N on a different point of the same side of the table. The cross-sectional area is the same ($20 cm^2$). The two forces are of the same direction and perpendicular to the side of the table. Calculate the total pressure exerted on the table.

Relevant Equations

$P=\frac{F}{A}$

I have come up with two different solutions but i'm not sure which is the correct one.
Solution 1
Pressure exerted by John:
$P_1=\frac{F_1}{A_1}=\frac{10}{0.002} pa=5000 pa$
Pressure exerted by John's friend:
$P_2=\frac{F_2}{A_2}=\frac{20}{0.002} pa= 10000 pa$
Total pressure:
$P=P_1+P_2=15000 pa$

Solution 2
Total Force: $F=F_1+F_2=30 N$
Total area: $A=A_1+A_2=0.004 m^2$
Total pressure: $P=\frac{F}{A}=7500 pa$

 

[kuruman]

The force that each person exerts on the table is distributed over the same area. Which method do you think is appropriate when that is the case? Remember that force is a vector, but pressure is not.

 

[haruspex]

Calculate the total pressure exerted on the table.

Total pressure is only meaningful for pressures that do physically add, and that requires them to be exerted on the same actual areas, not merely areas of the same magnitude.
If we allow it to mean average pressure exerted on the table then we need to be told the total area of the table.

The question is daft.

 

[Lnewqban]

Welcome, @physics565 !

It seems to me that solution 1 is not correct, because the forces are acting in parallel on different areas of the table, rather than in series over a common area.
Solution 2 is only the average of both pressures.

The problem does not make much sense, unless I don't understand it properly.

 

[Steve4Physics]

Homework Statement: John exerts a force of 10 N on a side of a table. The cross-sectional area is $20 cm^2$. A friend of John exerts a force of 20 N on a different point of the same side of the table. The cross-sectional area is the same ($20 cm^2$). The two forces are of the same direction and perpendicular to the side of the table.

Not a great question, but it does make sense. (Unless I've misunderstood!)

From the wording in the question, there aren't two separate areas $A_1$ and $A_2$. There is just a single side, so just a single area, $A$.

Possibilities are:
$F_1$ alone is exerted on $A$ or
$F_2$ alone is exerted on $A$ or
$F_1+F_2$ is exerted on $A$.

Solution 2 is wrong because it uses the wrong value of area.

Also, note that the symbol for pascal is 'Pa' with upper case 'P'.

 

[haruspex]

From the wording in the question, there aren't two separate areas

"A friend of John exerts a force of 20 N on a different point of the same side of the table…​

The cross-sectional area is the same…"​

and

"Calculate the total pressure exerted on the table."​

not "the total pressure on that area".

 

[jbriggs444]

From the wording in the question, there aren't two separate areas $A_1$ and $A_2$. There is just a single side, so just a single area, $A$.

The wording makes it clear that there are indeed at least two separate areas on the side of the table in question. Each of those two areas having an area of 20 cm².

The question also indicates that an area of 20 cm² qualifiies as a "point". If the author were at all trustworthy, this would suggest that the side of the table has a total area well in excess of 40 cm². Indeed, the side of a typical table would exceed 40 cm².

 

[physics565]

Thank you all for your replies. The problem is probably wrong. Note that the wording might not be perfect. English is not my mother tongue.

 

[physics565]

I was actually trying to understand a different problem. The solution to the previous problem would give me a hint.
I have a hydraulic press system with three pistons A, B and C. Each piston has a cross sectional area of $1 m^2$. A force of 10 N is exerted on each of the pistons A and B at the same time. I need to calculate the pressure exerted on piston C. How does Pascal's principle work here? Should i add up the pressures? Excuse me if my question is stupid. It's not exactly a homework question but rather something i'm trying to understand.

 

[haruspex]

I was actually trying to understand a different problem. The solution to the previous problem would give me a hint.
I have a hydraulic press system with three pistons A, B and C. Each piston has a cross sectional area of $1 m^2$. A force of 10 N is exerted on each of the pistons A and B at the same time. I need to calculate the pressure exerted on piston C. How does Pascal's principle work here? Should i add up the pressures? Excuse me if my question is stupid. It's not exactly a homework question but rather something i'm trying to understand.

I assume all three pistons are at the same height.
A thought experiment… if the three pistons exert the same pressure, will two of them push the third out?

 

[Steve4Physics]

The wording makes it clear that there are indeed at least two separate areas on the side of the table in question. Each of those two areas having an area of 20 cm².

At the risk of side-tracking, I'd like to respond and defend my Post #5 by explaining my interpretation of the question.

The question says "... a force of 20 N on a different point of the same side of the table".

And it also says "The two forces are of the same direction and perpendicular to the side of the table."

These indicate that there is only a single side to consider.

The question also indicates that an area of 20 cm² qualifiies as a "point".

The question says "... exerts a force of 10 N on a side of a table. The cross-sectional area is $20cm^2$".

That indicates (IMO) that the area of the (entire) side is $20cm^2$.

If the author were at all trustworthy, this would suggest that the side of the table has a total area well in excess of 40 cm². Indeed, the side of a typical table would exceed 40 cm².

That occurred to me too. But the term 'side' wouldn't generally be used to mean the top surface. A 'side' means one of the vertical edge-faces. E.g. a small square table could have each side measuring 40cm x 0.5cm (admittedly that would require an unusually thin top).

There. I've got it off my chest!

 

[physics565]

I assume all three pistons are at the same height.
A thought experiment… if the three pistons exert the same pressure, will two of them push the third out?

Assuming that the fluid is incompressible, if any of the pistons move then at least one of them is pushed out and one of them is pushed inwards. But as long as the three pistons exert the same pressure at the same time and they all have the same surface area, shouldn't they all move in the same way?
So none of them moves. I guess...

 

[haruspex]

The question says "... exerts a force of 10 N on a side of a table. The cross-sectional area is $20cm^2$".

That indicates (IMO) that the area of the (entire) side is $20cm^2$.

If it intended that, why would it say " The cross-sectional area is the same"?

 

[haruspex]

Assuming that the fluid is incompressible, if any of the pistons move then at least one of them is pushed out and one of them is pushed inwards. But as long as the three pistons exert the same pressure at the same time and they all have the same surface area, shouldn't they all move in the same way?
So none of them moves. I guess...

Right.

 

[physics565]

Right.

So what happens when the forces are exerted on only two pistons? Will the third piston move?

 

[jbriggs444]

So what happens when the forces are exerted on only two pistons? Will the third piston move?

Yes.

To be clear, you are applying an external force of 10N on two of the pistons while the third piston is subject to no external force other than that from the fluid pressure?

This third piston... Is it massless?
The fluid... Is it massless?

If everything is massless then we have troubles. An unbalanced net force with nowhere for momentum to flow.

Note that there are practical applications for this sort of thing.

 

[haruspex]

So what happens when the forces are exerted on only two pistons? Will the third piston move?

If no pressure, or a lesser pressure, is exerted on the third then obviously it will move. The point is that if equal pressures are exerted from outside on all three they will not move. That is enough to answer the question in post #9.

 

[Steve4Physics]

If it intended that, why would it say " The cross-sectional area is the same"?

I take the point.

 

[physics565]

Perhaps i wasn't clear. No external force is applied to the third piston except the one from the fluid like jbriggs444 said. The question is how one calculates the pressure exerted on the the third piston. Should i add up the pressures that the first two pistons exert on the fluid?

 

[physics565]

Yes.

To be clear, you are applying an external force of 10N on two of the pistons while the third piston is subject to no external force other than that from the fluid pressure?

This third piston... Is it massless?
The fluid... Is it massless?

If everything is massless then we have troubles. An unbalanced net force with nowhere for momentum to flow.

Note that there are practical applications for this sort of thing.
View attachment 348723

Well...i have no reason to assume that either the fluid or the pistons are massless.

 

[jbriggs444]

The question is how one calculates the pressure exerted on the the third piston. Should i add up the pressures that the first two pistons exert on the fluid?

No. You should not. Pressure does not add like that.

Let us modify the setup a bit. Put the two pistons where force is applied side by side with a tissue-thin wall between them. Give the two pistons a square cross-section.

If you remove the wall and weld the two pistons together on the side facing that wall, do you expect anything to change with respect to the fluid pressure?

You now have one piston with twice the area and twice the external force. Would the newly merged piston be at equilibrium with the fluid at an unchanged fluid pressure?

 

[jbriggs444]

Well...i have no reason to assume that either the fluid or the pistons are massless.

If the fluid has mass and we have zero pressure at the third piston and non-zero pressure at the first two pistons then the fluid will flow.

We are trying to use the rules of hydrostatics here. Hydodynamics brings in more than a few complications.

It may be worth putting a barrier in the third cylinder with a force sensor that measures how strongly the third piston is pushing on the force sensor. Then we are back to a hydrostatic situation.

 

[physics565]

I got it. Thank you all for your time.

 

[kuruman]

To go back to the original problem, I think that the conceptual goals of the problem can be clarified by first asking part (a) of the problem below. Part (b) is essentially the same as the original problem.

A tabletop of trapezoidal shape (shown on the right) has uniform thickness. John pushes on the tabletop from the right and his friend Jim pushes to the left from the other side so that the tabletop does not accelerate. Assume that the forces are uniformly distributed over the faces and that there are no other horizontal forces acting on the tabletop.

(a) The pressure on left face of the tabletop (John's side) is 200 Pa. What is the pressure on the right face of the tabletop?​

(b) Jim goes over to John's side and pushes to the right with the same force as before. The tabletop accelerates to the right. What is the pressure on the right face of the tabletop now?​