Video: Using Pascal’s Principle to Determine the Force Applied to a Surface

A host pours the remnants of several bottles of wine into a jug after a party. The jug has a 14.0 cm diameter. The host inserts a cork with a diameter of 2.00 cm into the bottle, placing it in direct contact with the wine. The host is amazed when he pounds the cork into place and the bottom of the jug breaks away. Calculate how much the magnitude of the force exerted against the bottom of the jug exceeds the 1.20 × 10² N magnitude force applied to the cork when it is struck by the host.

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Video Transcript

A host pours the remnants of several bottles of wine into a jug after a party. The jug has a 14.0-centimeter diameter. The host inserts a cork with a diameter of 2.00 cm into the bottle, placing it in direct contact with the wine. The host is amazed when he pounds the cork into place and the bottom of the jug breaks away. Calculate how much the magnitude of the force exerted against the bottom of the jug exceeds the 1.20-times-10-to-the-two-newton-magnitude force applied to the cork when it is struck by the host.

This is a problem having to do with pressure, force, and area and we’re given three important pieces of information in this problem statement. First we’re told that the jug overall has a 14.00-centimeter diameter. And second, we’re told that the diameter of the cork is 2.00 cm. And finally, we find out that the force applied to the cork when it is struck by the host is 1.20 times 10 to the two newtons.

Let’s carry this critical information over to another screen and draw a diagram of the situation. Now here’s a diagram of the jug that is filled up. And we’re told that the bottom of the jug, its diameter, is 14.0 cm and that we’re putting a cork into the neck of the jug and the diameter of the cork is 2.00 cm.

Now after all this, after putting the cork snugly into the neck of our jug, the host takes a hammer or a mallet and applies a force to the cork to fasten it in place. And we’re told that force — we’ll refer to it as capital 𝐹 sub 𝑚 — is equal to 1.20 times 10 to the two newtons. Now as we said before, this is a problem involving pressure, the pressure that supplied to the wine at the top of the jug and the pressure that the wine applies to the bottom of the jug that eventually causes it to burst.

Now for notation sake, for the rest of our work on this problem, let’s refer to the top of the jug where the cork is as 𝑡 and the bottom of the jug where the glass breaks out under pressure as 𝑏, so we have a top and a bottom location. Now it’s true that if we take any horizontal cross section of this jug, any cross section whatsoever, that if we were to measure the pressure exerted by the force of our mallet striking the cork, the pressure along this cross section is constant.

That means that if we were to measure the pressure at the top or the pressure at the bottom, those values would be the same. Now let’s recall at this point what is the equation that tells us pressure. Well pressure relates force and area. And in fact, pressure, which we can represent as a capital 𝑃, is equal to force divided by area, which we’ll abbreviate 𝐹 divided by capital 𝐴.

So if we know the force exerted and we know the area over which that force is exerted, then we can solve for the pressure. Now with this as background, let’s refer back to this top and bottom or 𝑡 and 𝑏 locations on our jug. Now if what we’ve said about pressure and cross sections is true, then that means we can write an equation in terms of the pressure at the top and the pressure at the bottom of the jug.

That equation would be the pressure at the top, which we can symbolize as 𝑃 sub 𝑡, is equal to the pressure at the bottom, 𝑃 sub 𝑏. Great! Now let’s use this relationship we just explained for pressure, that pressure is equal to force over area, to substitute in for 𝑃 sub 𝑡 and 𝑃 sub 𝑏. We’ll substitute in the force at the top, 𝐹 sub 𝑡, the area at the top, 𝐴 sub 𝑡, and the force and the area at the bottom, 𝐹 sub 𝑏 and 𝐴 sub 𝑏.

So our pressure relationship implies that 𝐹 sub 𝑡 divided by 𝐴 sub 𝑡 equals 𝐹 sub 𝑏 divided by 𝐴 sub 𝑏. Now in terms of the variables we defined, what we are after is 𝐹 sub 𝑏, the force at the bottom of the jug, minus 𝐹 sub 𝑡; the problem asked for what is the difference between these two forces; or specifically, how much did the force of the bottom on the jug exceed the force exerted at the top applied by the host.

So this is what we wanna solve for. Our course of action will be to solve for 𝐹 sub 𝑏, the force exerted on the bottom of the jug, in terms of 𝐹 sub 𝑡, the known force at the top of the jug. So let’s do that using this ratio equation we just derived.

To isolate 𝐹 sub 𝑏, we can multiply both sides of the equation by 𝐴 sub 𝑏, the area of the bottom of the jug. When we do that, 𝐴 sub 𝑏 cancels out from the right-hand side of our equation, and we’re left with an expression that reads 𝐹 sub 𝑏 equals 𝐹 sub 𝑡 multiplied by 𝐴 sub 𝑏 divided by 𝐴 sub 𝑡.

Now that we have this expression for the force at the bottom of the jug in terms of the force of the top and the areas of the bottom and top, let’s focus on the ratio of areas and figure out what is that value. Now we know in general that the area of a circle is equal to 𝜋 times the radius of that circle squared, so we can write that out for 𝐴 sub 𝑏 and 𝐴 sub 𝑡.

Now you can see right away that the 𝜋 and the numerator and denominator cancel, so this ratio of areas will depend on the square of the radius of the bottom to the square of the radius of the top. Now what are those radii? Well we know the diameters, so we can solve for radii in terms of those. The diameter of the bottom of the jug is given as 14.0 cm, which means the radius is half of that or 7.00 cm.

And likewise, for the radius of the top, that’s half of 2.00 cm or 1.00 cm. Both these terms are then squared. And when they are, we’ve established that the ratio of the bottom of this jug to the top of the jug is equal to 7.00 squared divided by 1.00 squared, which equals 49. Note that this is a unitless number, the units in our fraction cancelled out, so we have a unitless value which is a ratio of these two areas.

Knowing this, we can plug this value in to our original equation 𝐹 sub 𝑏 equals 𝐹 sub 𝑡 times the ratio of these areas to solve for the force at the bottom of our jug in terms of the force we applied at the top. So the force on the bottom of the jug equals 49 times the force at the top of the jug. And what was that force of the top? Well we’ve labeled it the force applied by the mallet, 𝐹 sub 𝑚, which equals 1.20 times 10 to the two newtons.

So the force applied to the bottom of the jug is 49 times the force applied to the top of the jug. And again what we’re seeking is the difference between the force applied at the bottom and the force applied at the top; in other words, 𝐹 sub 𝑏 minus 𝐹 sub 𝑡. We found that we can write 𝐹 sub 𝑏 as 49 times 𝐹 sub 𝑡, which means that what we’re solving for can also be written as 𝐹 sub 𝑡 times 49 minus one or 𝐹 sub 𝑡 times 48.

When we plug in for 𝐹 sub 𝑡 1.20 times 10 to the two newtons and multiply that by 48, we get an answer of 5.76 times 10 to the third newtons. That is the difference between the force at the bottom and the force of the top; or specifically, it’s how much more force is applied on the bottom of the jug then is applied on the cork through the mallet. An all of this is thanks to the fact that pressure over any cross section of this jug is constant.

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