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## Homework Statement

1)Calculate the necessary force to open a 0.3 m lenght square metallic container with a 410 mmHg inside pressure knowing external pressure is 0,980 atm

2)There is a hollow cilinder with a 5 kg mass and 30 cm diameter cap in a side and hanging on the ceiling in the other side. By a pump you reduce the internal pressure in the way of keeping the cap adherent to cilinder. Knowing external pressure is the normal atmospheric one and internal pressure is 12.6 mmHg, calculate the mass you need to separare cap.

## Homework Equations

p = F/A

P = m*g

## The Attempt at a Solution

*first problem)*

I converted all into pascal, so 54662 and 99299.

Then:

A = 0.09 m

^{2}

99299 - 54662 = 44637 Pa

F = 44637 Pa * 0.09 m

^{2}= 4017 N but my book says 4020 N, i forgot somewhat or is my procedure correct?

*second problem)*

Converted internal pressure in 1680 Pa.

A = 0.282743 m

^{2}

Atm. force = 101300 *0.282743 = 28642 N

Internal force = 1680 * 0.282743 = 475 N

Cop force = 49 N

Total acting force = 28642-49-475 = 28118 N

To win a 28118 N force I need a 28118/9.8 =

**2870 Kg**but my book says only

**470 Kg**how is it possible? where did i wrong?

Thanks!