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| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#2.8">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.8">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.8.13.$ Rolling mill rolls have radius $R$. Rotating, they retract the workpiece, if its thickness is small enough. The coefficient of friction between the rolls and the workpiece is $\mu$, and the gap between the rolls is $d_0$. Find the maximum thickness of the blank. The blank is not pushed. | | $2.8.13.$ Rolling mill rolls have radius $R$. Rotating, they retract the workpiece, if its thickness is small enough. The coefficient of friction between the rolls and the workpiece is $\mu$, and the gap between the rolls is $d_0$. Find the maximum thickness of the blank. The blank is not pushed. |
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| For problem $2.8.13$ | | For problem $2.8.13$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| <img src="2.8.13_1.png" | | <img src="2.8.13_1.png" |
| loading="lazy" width="280" /> | | loading="lazy" width="280" /> |
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| Forces acting on the mill rolls | | Forces acting on the mill rolls |
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| As the thickness of the workpiece increases, there will come a point where the mill rolls can no longer roll the workpiece.</p><p> | | As the thickness of the workpiece increases, there will come a point where the mill rolls can no longer roll the workpiece.</p><p> |
| It will happen until the vertical projection of friction force $\vec{F}_\text{fr}$ will exceed the vertical component of the support reaction force $\vec{N}$ | | It will happen until the vertical projection of friction force $\vec{F}_\text{fr}$ will exceed the vertical component of the support reaction force $\vec{N}$ |
| $$F_\text{fr}\cos\alpha\geq N\sin\alpha$$ | | $$F_\text{fr}\cos\alpha\geq N\sin\alpha$$ |
| Considering the value of friction force $F_\text{fr} = \mu N$: | | Considering the value of friction force $F_\text{fr} = \mu N$: |
| $$\mu = \tan\alpha\quad(1)$$ | | $$\mu = \tan\alpha\quad(1)$$ |
| From the drawing | | From the drawing |
| $$\frac{d-d_0}{2} = R(1-\cos\alpha) \Rightarrow \boxed{d = d_0 + 2R(1-\cos\alpha)}\quad(2)$$ | | $$\frac{d-d_0}{2} = R(1-\cos\alpha) \Rightarrow \boxed{d = d_0 + 2R(1-\cos\alpha)}\quad(2)$$ |
| From the expression $(1)$, | | From the expression $(1)$, |
| $$\cos\alpha = \frac{1}{\sqrt{1+\tan^2\alpha}}=\frac{1}{\sqrt{1+\mu^2}}\quad(3)$$ | | $$\cos\alpha = \frac{1}{\sqrt{1+\tan^2\alpha}}=\frac{1}{\sqrt{1+\mu^2}}\quad(3)$$ |
