Solid body pressure — lesson. Science State Board, Class 8.

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எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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Before getting to know the solid body pressure, let us solve a problem.

Who puts more pressure on the surface - The elephant or the girl in high heels?


The elephant weighs \(4000\) \(kg\)	The girl weighs \(50\) \(kg\)

The answer to this question can be surprising.

The amount of pressure does not depend only on body weight!

Pressure:

Pressure is a physical quantity that describes how much force is applied to a unit of surface area perpendicular to the surface. The SI unit of pressure is measured in \(newtons\ per\ square\ meter\), or \(pascal\).

\begin{array}{l} pressure = \frac{pressure force}{Square area} \\ It is represented as \\ p = \frac{F}{S} \end{array}

1 Pa = 1 \frac{N}{m^{2}}

A force acting on a body perpendicular to its surface is called a compressive force.

With the same force, it is possible to obtain pressures of different magnitudes, as it depends on the dimensions of the support area. The smaller the area, the higher the pressure.

The unit of pressure is Pa (pascals).
\(1\) \(Pa\) is low pressure.

So use,

Hectopascals \(1\) \(hPa\) \(=\) \(100\) \(Pa\);

Kilopascals \(1\) \(kPa\) \(=\) \(1000\) \(Pa\);

Megapascals \(1\) \(MPa\) \(=\) \(1000,000\) \(Pa\);

Gigapascals \(1\) \(GPa\) \(=\) \(1000 000 000\) \(Pa\).

Now, Let us solve the problem about the elephant and the girl.

To determine the pressure, you still need to know the size of the support area.

The elephant's foot is about \(200\)

{cm}^{2}

, as the elephant stands on four legs, this figure must be multiplied by 4.

The ladies' shoes area that is in contact with the support area is approximately \(10\)

{cm}^{2}

(Look at the picture, the girl standing on one leg).

In this case, the weight of the bodies is the force pushing on the support area.

Let us assume,

\begin{array}{l} m_{z} - Mass of the elephant \\ s_{z} - Area of contact between the elephant feet and the land \\ m_{m} - Mass of the girl \\ s_{m} - Area of contact between the girl feet and the land \end{array}

\begin{array}{l} m_{z} = 4000 kg \\ S_{z} = 800 {cm}^{2} \\ m_{m} = 50 kg \\ S_{m} = 10 {cm}^{2} \\ g \approx 10 m / s^{2} \\ ___________ \\ Compare p \end{array}

\begin{array}{l} = 0, 08 m^{2} \\ = 0, 001 m^{2} \end{array}

\begin{array}{l} p = \frac{F}{A} \\ F = P = mg \end{array}

\begin{array}{l} p_{z} = \frac{4000 \cdot 10}{0, 08} = 500000 (Pa) \\ p_{m} = \frac{50 \cdot 10}{0, 001} = 500000 (Pa) \end{array}

Answer: The elephant and the girl put equal pressure on the support area.

Reference:

https://pxhere.com/en/photo/1126058
https://pxhere.com/en/photo/745664

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4. Solid body pressure

Theory: