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| ### Statement |
| ### Statement |
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| $2.2.42.$ With what force does the cobra press on the ground when it is preparing to jump, rises vertically upwards at a constant speed $v$? The mass of the snake is $m$, its length is $l$. |
| $2.2.42.$ With what force does the cobra press on the ground when it is preparing to jump, rises vertically upwards at a constant speed $v$? The mass of the snake is $m$, its length is $l$. |
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| ### Solution |
| ### Solution |
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| Let's consider small part of cobra with the length $\Delta x$ and mass $\Delta m$: \begin{equation*} \Delta P = \Delta m \cdot v \end{equation*} Since the snake is homogeneous \begin{equation*} \Delta m = \frac{m}{L} \cdot \Delta x \end{equation*} By newton's second law: \begin{equation*} F=\frac{\Delta P}{\Delta t} = \frac{\Delta m}{\Delta t} \cdot v \end{equation*} |
| Let's consider a small part of the cobra with length $\Delta x$ and mass $\Delta m$. When this part is lifted off the ground, it acquires a momentum: |
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| $$\Delta P = \Delta m \cdot v$$ |
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| Since the snake is homogeneous, the mass of this element is: |
| F = \frac{m}{L} \cdot \frac{\Delta x}{\Delta t} \cdot v = \frac{mv^2}{L} |
| $$\Delta m = \frac{m}{l} \Delta x$$ |
| $$ |
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| By definition of speed \begin{equation*} \frac{\Delta x}{\Delta t} = v \end{equation*} Whence the force associated with the acceleration of the centre of mass \begin{equation*} F = \frac{\Delta P}{\Delta t} = \frac{mv^2}{L} \end{equation*} Let's take everything together and find the support reaction force \begin{equation*} N = mg + \frac{mv^2}{L} = m\left(g + \frac{v^2}{L}\right) \end{equation*} \begin{equation*} \boxed{N = m\left(g + \frac{v^2}{L}\right)} \end{equation*} |
| By Newton's second law, the additional force associated with accelerating these mass elements from rest to speed $v$ is equal to the rate of change of momentum: |
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| $$F = \frac{\Delta P}{\Delta t} = \frac{\Delta m}{\Delta t} v = \frac{m}{l} \frac{\Delta x}{\Delta t} v$$ |
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| #### Answer |
| By the definition of speed, $\frac{\Delta x}{\Delta t} = v$. Whence the force associated with the acceleration of the mass is: |
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| $$F = \frac{mv^2}{l}$$ |
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| $$ |
| Let's take everything together and find the total support reaction force $N$. The ground must support the entire gravitational weight of the snake ($mg$) plus provide the extra force $F$ to accelerate the lifting parts: |
| N = m\left(g + \frac{v^2}{L}\right) |
| $$N = mg + \frac{mv^2}{l} = m\left(g + \frac{v^2}{l}\right)$$ |
| $$ |
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| By Newton's third law, the force with which the cobra presses on the ground is equal in magnitude to the support reaction force $N$. |
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| #### Answer |
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| $$N = m\left(g + \frac{v^2}{l}\right)$$ |
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