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| The main feature of this problem is that the compression force of the column is non-uniform: the upper part is not loaded with anything, while the very base bears the entire huge weight of the column. Let us consider two methods for finding the total deformation. | | The main feature of this problem is that the compression force of the column is non-uniform: the upper part is not loaded with anything, while the very base bears the entire huge weight of the column. Let us consider two methods for finding the total deformation. |
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| <b>Method 1. Using average force </b> | | <b>Method 1. Using average force </b> |
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| Since the column is uniform and has a constant cross-section, the pressure force (weight) increases linearly as we move from the top to the base. At the very top, the force is zero ($F_{\text{top}} = 0$), and at the base, it is maximal and equal to the total weight of the column ($F_{\text{bottom}} = Mg = \rho S h g$). | | Since the column is uniform and has a constant cross-section, the pressure force (weight) increases linearly as we move from the top to the base. At the very top, the force is zero ($F_{\text{top}} = 0$), and at the base, it is maximal and equal to the total weight of the column ($F_{\text{bottom}} = Mg = \rho S h g$). |
| Due to this strictly linear dependence, we can use the average force acting along the entire length of the column: | | Due to this strictly linear dependence, we can use the average force acting along the entire length of the column: |
| $$F_{\text{avg}} = \frac{F_{\text{top}} + F_{\text{bottom}}}{2} = \frac{0 + \rho S h g}{2} = \frac{\rho S h g}{2}$$ | | $$F_{\text{avg}} = \frac{F_{\text{top}} + F_{\text{bottom}}}{2} = \frac{0 + \rho S h g}{2} = \frac{\rho S h g}{2}$$ |
| Let us write Hooke's law for the entire column as a whole, substituting the average force into it: | | Let us write Hooke's law for the entire column as a whole, substituting the average force into it: |
| $$\Delta h = \frac{F_{\text{avg}} h}{E S} = \frac{\frac{\rho S h g}{2} \cdot h}{E S} = \frac{\rho g h^2}{2E}$$ | | $$\Delta h = \frac{F_{\text{avg}} h}{E S} = \frac{\frac{\rho S h g}{2} \cdot h}{E S} = \frac{\rho g h^2}{2E}$$ |
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| <b>Method 2. Using integration </b> | | <b>Method 2. Using integration </b> |
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| Let us divide the column into infinitely small horizontal layers of thickness $dy$. Let the coordinate $y$ be measured from the top of the column downwards (from $0$ to $h$). | | Let us divide the column into infinitely small horizontal layers of thickness $dy$. Let the coordinate $y$ be measured from the top of the column downwards (from $0$ to $h$). |
| An isolated micro-layer located at depth $y$ is pressed by the weight of the granite located strictly above it. The mass of this part is $m(y) = \rho S y$, and the pressure force is: | | An isolated micro-layer located at depth $y$ is pressed by the weight of the granite located strictly above it. The mass of this part is $m(y) = \rho S y$, and the pressure force is: |
| $$F(y) = m(y)g = \rho S g y$$ | | $$F(y) = m(y)g = \rho S g y$$ |
| According to Hooke's law, the absolute compression $d(\Delta y)$ of this infinitely small layer is equal to: | | According to Hooke's law, the absolute compression $d(\Delta y)$ of this infinitely small layer is equal to: |
| $$d(\Delta y) = \frac{F(y) dy}{E S} = \frac{\rho S g y dy}{E S} = \frac{\rho g y}{E} dy$$ | | $$d(\Delta y) = \frac{F(y) dy}{E S} = \frac{\rho S g y dy}{E S} = \frac{\rho g y}{E} dy$$ |
| To find the total compression of the entire column, let us integrate (sum) the deformations of all such layers from the top ($y = 0$) to the base ($y = h$): | | To find the total compression of the entire column, let us integrate (sum) the deformations of all such layers from the top ($y = 0$) to the base ($y = h$): |
| $$\Delta h = \int_{0}^{h} \frac{\rho g y}{E} dy = \frac{\rho g}{E} \int_{0}^{h} y dy = \frac{\rho g}{E} \left[ \frac{y^2}{2} \right]_{0}^{h} = \frac{\rho g h^2}{2E}$$ | | $$\Delta h = \int_{0}^{h} \frac{\rho g y}{E} dy = \frac{\rho g}{E} \int_{0}^{h} y dy = \frac{\rho g}{E} \left[ \frac{y^2}{2} \right]_{0}^{h} = \frac{\rho g h^2}{2E}$$ |
| As we can see, both methods naturally lead to exactly the same result. | | As we can see, both methods naturally lead to exactly the same result. |
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| <b>Calculation:</b> | | <b>Calculation:</b> |
| Let us substitute the given values, assuming the acceleration of gravity is $g \approx 9.8$ m/s$^2$: | | Let us substitute the given values, assuming the acceleration of gravity is $g \approx 9.8$ m/s$^2$: |
| $$\Delta h = \frac{2700 \cdot 9.8 \cdot 30^2}{2 \cdot 10^{11}} = \frac{2700 \cdot 9.8 \cdot 900}{2 \cdot 10^{11}} \approx 1.19 \cdot 10^{-4} \text{ m}$$ | | $$\Delta h = \frac{2700 \cdot 9.8 \cdot 30^2}{2 \cdot 10^{11}} = \frac{2700 \cdot 9.8 \cdot 900}{2 \cdot 10^{11}} \approx 1.19 \cdot 10^{-4} \text{ m}$$ |
| *(When using the classical value for this problem book, $g = 10$ m/s$^2$, the calculated value will be $\approx 1.215 \cdot 10^{-4}$ m, which, when rounded to two significant figures, gives the final answer).* | | *(When using the classical value for this problem book, $g = 10$ m/s$^2$, the calculated value will be $\approx 1.215 \cdot 10^{-4}$ m, which, when rounded to two significant figures, gives the final answer).* |
