Question

An airtight box, having a lid of area 80 $c{m^2}$ , is partially evacuated ( has lower pressure than outside atmosphere). Atmospheric pressure is $1.01{\text{ }} \times {10^5}$ Pa. A force of 600N is required to pull the lid of the box. Find the pressure in the box.

Answer 1

Hint: Here Atmospheric pressure exerts inward force on the lid, the applied force and the force due to pressure inside the box are outwards. Such that when the box is about to open ${F_{atm}} = {F_{applied}} + {F_{inside}}$
where ${F_{atm}}$ is force due to atmospheric pressure
${F_{applied}}$ is applied force
${F_{inside}}$ is force due to pressure inside the box

Complete step by step solution:
Given data: Area of the lid (A) = 80$c{m^2}$
Atmospheric pressure = $1.01{\text{ }} \times {10^5}$ Pa
Applied force (${F_{applied}}$ ) = 600N

We know that F=PA
where F Is force and P is pressure and A is area of the lid

Then ${F_{atm}}$ =${P_{atm}}A$
where ${P_{atm}}$ is atmospheric pressure, A is area of the lid

Similarly,
${F_{inside}}$ =${P_{inside}}$$A$
where ${P_{inside}}$ is pressure inside the box, A is area of the lid

Substituting in the equation ${F_{atm}} = {F_{applied}} + {F_{inside}}$
We have ${P_{atm}}A$ = ${F_{applied}}$ + ${P_{inside}}$$A$

Substituting the above values in the eqn
$1.01 \times {10^5} \times 80 \times {10^{ - 4}} = 600 + {P_{inside}}80 \times {10^{ - 4}}$
$1.01 \times {10^5} \times 80 \times {10^{ - 4}} - 600 = {P_{inside}}80 \times {10^{ - 4}}$ (substituting for ${P_{inside}}$)
${P_{inside}}$ \[ = \dfrac{{\left( {1.01 \times {{10}^5} \times 80 \times {{10}^{ - 4}} - 600} \right)}}{{80 \times {{10}^{ - 4}}}}\] (first multiplication and subtraction)
= ${{1.01 \times 80 \times 10 - 600}}{{80 \times {{10}^{ - 4}}}} ={{208 - 600}}{{80 \times {{10}^{ - 4}}}} = 2.6 \times {10^4}Pa$

Hence the pressure inside the box is $2.6 \times {10^4}Pa$

Note: This problem can be solved by taking ${P_{atm}} = {P_{applied}} + {P_{inside}}$