Pascal's Law & Hydraulic Lift Video Lessons

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Problem: A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-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 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

FREE Expert Solution

Let's consider the situation of a hydraulic lift.

In a hydraulic lift, the pressure on the master cylinder is equal to the pressure on the slave cylinder.

However, force changes with the change in cross-section. 

Pressure:

P=FA

We now have:

F1A1=F2A2, where F1 is the force applied to the master cylinder, A1 is the cross-sectional area of the master cylinder, F2 is the force created at each slave cylinder, and A2 is the cross-sectional area of slave cylinder.

We consider the cork as the master cylinder and the jug as the slave cylinder.

F1 = 120 N

A1 = πr12 

r1 = 2.00/2 = 1.00 × 10-2m

A1 = π(1.00 × 10-2)2 = 3.142 × 10-4 m2 

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Problem Details

A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-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 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

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