Question

A host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.30 cm diameter 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 (with a 13.0 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

Answer: 3.88 N

Step-by-step explanation:

From the question we have the following:

cork diameter (D1) = 2.3 cm = 0.023 m

bottom diameter (D2) = 13 cm = 0.13 m

force on the cork (F1) = 120 N

force at the bottom (F2) = ?

we can get the force at the bottom by applying the formula below

`(F1)/(A1)` =
`(F2)/(A2)`

therefore

F2 =
`(F1)/(A1)` × A2

where A1 and A2 are the areas of the cork and bottom respectively

A1 = π×
`r^(2)` = π×
`(0.023 ÷ 2 )^(2)` = 0.00042

A2 = π×
`r^(2)` = π×
`(0.13 ÷ 2 )^(2)` = 0.013

now substituting all values into F2 =
`(F1)/(A1)` × A2

F2 =
`(120)/(0.00042)` × 0.013

F2 = 3.88 N