3.16 A partitioned tank as shown contains water and mercury. What is the gage pressure in the air trapped in the left chamber? What pres- sure would the air on the left need to be pumped to in order to bring the water and mercury free surfaces level?

3.16 A partitioned tank as shown contains water and mercury. What is the gage pressure in the air trapped in the left chamber? What pres- sure would the air on the left need to be pumped to in order to bring the water and mercury free surfaces level?

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Description: 3.16 A partitioned tank as shown contains water and mercury. Whatis the gage pressure in the air trapped in the left chamber? What pres-sure would the air on the left need to be pumped to in order to bringthe water and mercury free surfaces level?0.

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter8: Centroids And Distributed Loads

Section: Chapter Questions

Problem 8.121P: One side of the container has a 03-m square door that is hinged at its top edge. If the container is...

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Question

3.16 A partitioned tank as shown contains water and mercury. What
is the gage pressure in the air trapped in the left chamber? What pres-
sure would the air on the left need to be pumped to in order to bring
the water and mercury free surfaces level?
0.75 m
- 3.75 m
Water,
3 m
1 m
Mercury
3 m
2.9 m
Р3.16

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That's great but another question is in case 2 shouldn't the water move down and the oil move up so we can't just make both of them equal to 1 because as the oil moves up the free surface of the water should move down

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Follow-up Question

In step two if we subtract 9810 from 132925.5 it shouldn't give 3482.55 can you please check the steps and the answers for step 2 or finding the gauge pressure of the air trapped.Also the same equation is used to solve both of tge questions which when calculated give the same answer so can someone clarify

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