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| <title>A body of mass m falls from a height h onto a spring of stiffness k and length l standing vertically on the floor. Determine the maximum pressure on the floor. Explain why this force increases as the spring stiffness increases.</title> | | <title>A body of mass m falls from a height h onto a spring of stiffness k and length l standing vertically on the floor. Determine the maximum pressure on the floor. Explain why this force increases as the spring stiffness increases.</title> |
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| <h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.3.45.$ A body of mass $m$ falls from a height $h$ onto a spring of stiffness $k$ and length $l$ standing vertically on the floor. Determine the maximum pressure on the floor. Explain why this force increases as the spring stiffness increases. | | $2.3.45.$ A body of mass $m$ falls from a height $h$ onto a spring of stiffness $k$ and length $l$ standing vertically on the floor. Determine the maximum pressure on the floor. Explain why this force increases as the spring stiffness increases. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.3.45.png" | | <img src="2.3.45.png" |
| loading="lazy" width="170" /> | | loading="lazy" width="170" /> |
| <figcaption> | | <figcaption> |
| For problem $2.3.45$ | | For problem $2.3.45$ |
| </figcaption> | | </figcaption> |
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| <p> | | <p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.3.45_1.png" | | <img src="2.3.45_1.png" |
| loading="lazy" width="320" /> | | loading="lazy" width="320" /> |
| <figcaption> | | <figcaption> |
| Compression of the spring under the force of the weight's gravity | | Compression of the spring under the force of the weight's gravity |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| Conservation of kinetic and potential energies | | Conservation of kinetic and potential energies |
| $$mgh = \frac{mv^2}{2} + mgH + \frac{k(l-H)^2}{2}$$ | | $$mgh = \frac{mv^2}{2} + mgH + \frac{k(l-H)^2}{2}$$ |
| Since at a finite instant of time, the body stops $(v=0)$ | | Since at a finite instant of time, the body stops $(v=0)$ |
| $$mgh=mg(l-\Delta x)+\frac{k\Delta x^2}{2}\quad (1)$$ | | $$mgh=mg(l-\Delta x)+\frac{k\Delta x^2}{2}\quad (1)$$ |
| From the drawing | | From the drawing |
| $$H=l-\Delta x$$ | | $$H=l-\Delta x$$ |
| Let's substitute into $(1)$ | | Let's substitute into $(1)$ |
| $$\frac{k\Delta x^2}{2}-mg\Delta x-mg(h-l)=0$$ | | $$\frac{k\Delta x^2}{2}-mg\Delta x-mg(h-l)=0$$ |
| Solve as a quadratic equation | | Solve as a quadratic equation |
| $$\Delta x = \frac{mg}{k}\left(1\pm\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)$$ | | $$\Delta x = \frac{mg}{k}\left(1\pm\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)$$ |
| As $\Delta x$ must be positive, we keep only the plus solution | | As $\Delta x$ must be positive, we keep only the plus solution |
| $$\Delta x = \frac{mg}{k}\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)$$ | | $$\Delta x = \frac{mg}{k}\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)$$ |
| Since at the bottom point, there are no forces other than the support reaction force $N$ and spring elasticity force $k\Delta x$, and at the bottom position the acceleration is zero ($a=0$) and the forces are compensated, the condition of equilibrium on the vertical axis has the form | | Since at the bottom point, there are no forces other than the support reaction force $N$ and spring elasticity force $k\Delta x$, and at the bottom position the acceleration is zero ($a=0$) and the forces are compensated, the condition of equilibrium on the vertical axis has the form |
| $$F_y=N$$ | | $$F_y=N$$ |
| Whence we obtain the maximum force of pressure on the floor, which is equal to the force with which the support acts on the load, according to Newton's third law | | Whence we obtain the maximum force of pressure on the floor, which is equal to the force with which the support acts on the load, according to Newton's third law |
| $$F_y=k\cdot \Delta x\Rightarrow \boxed{N=mg\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)}$$ | | $$F_y=k\cdot \Delta x\Rightarrow \boxed{N=mg\left(1+\sqrt{1+\frac{2k\cdot(h-l)}{mg}}\right)}$$ |
| Analysing the obtained expression, we obtain that $F\sim \sqrt{k}\to$ Therefore, the greater the spring stiffness $k$, the greater the support reaction force $N$ | | Analysing the obtained expression, we obtain that $F\sim \sqrt{k}\to$ Therefore, the greater the spring stiffness $k$, the greater the support reaction force $N$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$F=mg\left(1+\sqrt{1+2k(h-l)/(mg)}\right)$$ | | $$F=mg\left(1+\sqrt{1+2k(h-l)/(mg)}\right)$$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
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