Question

A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 1.80 cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 18.0 cm diameter) breaks away. Calculate the extra force (in N) exerted against the bottom if he pounded the cork with a 120 N force.

Answer 1

We would consider the hydraulic lift where the pressure is the same on the

master ad slave cylinder but force changes with the change in cross section area. We would apply the Pascal's princile which is expressed as

F1/A1 = 2/A2

where

F1 is the force applied to the master cylinder and A1 is thecross sectional area of the master cylinder

F2 is the force created at each of the slave cylinder and A2 is the cross sectional area of slave cylinder.

In this scenario, the cork works as master cylinder and the jug as slave cylinder. From the information given,

F1 = 120

Cross sectional area of cork is

A1 = πr1^2

where r1 is the radius of the cork

Given that diameter of cork = 1.8,

radius = diameter/2 = 1.8/2

r1 = 0.9

A1 = π * 0.9^2 = 0.81π

Cross sectional area of bottle, A2 = πr2^2

where r2 is the radius of the bottom of the jug.

Given that diameter of the bottom of the jug = 18,

r2 = 18/2 = 9

A2 = π * 9^2 = 81π

From the first equation.

F2 = A2F1/A1

By substituting the given values, we have

F2 = 81π * 120/ 0.81π

F2 = 12000 N

Extra force = F2 - F1 = 12000 - 120

Extra force = 11880 N

The extra force against the bottom is 11880 N