Question

A host pours the remnants of several bottles of wine into a jug after a party. The host then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. The host is amazed when the host pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

Answer 1

To solve this problem it is necessary to apply the Pascal principle for a hydraulic system. Of which you have to

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

Where

$F_{1,2}$= Force applied to master cylinder and created at each of the slave cylinder

$A_{1,2}$ = Cross sectional area of master cylinder and slave cylinder.

The cork works as master cylinder and jug as slave cylinder:

$F_1 = 120N$

Each sectional area would be

$A_1 = \pi r_1^2 = \pi (1)^2 = \pi$

$A_2 = \pi r_2^2 = \pi 7^2 = 49\pi$

From this Pascal law we have

$F_2 = \frac{A_2}{A_1} F_1$

$F_2 = \frac{49\pi}{\pi}(120)$

$F_2 = 5880N$

Therefore the extra force is

$F = F_2 - F_1$

$F = 5880-120$

$F = 5760N$

Therefore the extra force against the bottom is 5760N