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| | | ### Statement |
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| | − | $2.8.43.$ The ant decided to drag a straw to the anthill. What should he do if the force with which he can pull the straw is somewhat less than the maximum frictional force at rest? |
| | + | $2.8.43.$ An ant decided to drag a straw to its anthill. What should it do if the force with which it can pull the straw is slightly less than the maximum force of static friction? |
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| | | ### Solution |
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| | − | The ant should lift the straw from the ends alternatively turning it through a certain angle while doing so.In this way it can move the straw without actually needing to drag it and thus would be able to bypass the frictional force. We must note that the angle through which the straw has been turned through in the air must be less than 360 degrees otherwise lifting and turning the straw would be of no use. |
| | + | #### Method 1. |
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| | − | #### Answer |
| | + | When the ant pulls the straw by one end along its length, a maximum static friction force $F_{\text{fr}} = \mu mg$ arises, directed opposite to the motion. If the ant's force $F < F_{\text{fr}}$, the straw will not move when it tries to pull it translationally. |
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| | − | The ant should lift the straw from alternate ends and turn it through a certain angle in the air continuing the process until it reaches the anthill. |
| | + | However, if the ant lifts one end of the straw, the normal reaction force $N$ at the other end will become equal to $mg/2$. Accordingly, the friction force at the far end will decrease by half. In this case, the ant can easily turn the straw around the other end. |
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| | + | By doing this alternately from different ends and slightly turning the straw, the ant will gradually move it in the desired direction without overcoming the full maximum static friction force required for sliding along its entire length. |
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| | + | #### Method 2. |
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| | + | The ant can drag the straw translationally, without rotating it, if it pulls not horizontally, but at a certain angle $\alpha$ to the horizontal (upwards). |
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| | + | Let us write the conditions for the uniform motion of the straw (limiting case): |
| | + | Along the $x$-axis: $$F \cos\alpha = F_{\text{fr}}$$ |
| | + | Along the $y$-axis: $$N + F \sin\alpha = mg \implies N = mg - F \sin\alpha$$ |
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| | + | The sliding friction force is: |
| | + | $$F_{\text{fr}} = \mu N = \mu (mg - F \sin\alpha)$$ |
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| | + | Let us substitute this into the equation for the horizontal axis: |
| | + | $$F \cos\alpha = \mu mg - \mu F \sin\alpha$$ |
| | + | $$F (\cos\alpha + \mu \sin\alpha) = \mu mg$$ |
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| | + | From this, we express the force $F$ required for motion: |
| | + | $$F(\alpha) = \frac{\mu mg}{\cos\alpha + \mu \sin\alpha}$$ |
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| | + | For the ant to need to apply the minimum force, the denominator of the fraction must be maximal. Let us find the maximum of the function $f(\alpha) = \cos\alpha + \mu \sin\alpha$. Taking the derivative with respect to $\alpha$ and equating it to zero, we obtain: |
| | + | $$-\sin\alpha + \mu \cos\alpha = 0 \implies \tan\alpha = \mu$$ |
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| | + | At this angle, the denominator is equal to $\sqrt{1 + \mu^2}$. Then the minimum required force is: |
| | + | $$F_{\text{min}} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$$ |
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| | + | Since the condition $\sqrt{1 + \mu^2} > 1$ is satisfied for any $\mu > 0$, then $F_{\text{min}} < \mu mg$. |
| | + | Thus, by choosing the correct angle $\alpha = \arctan(\mu)$, the ant will be able to drag the straw straight to the anthill, even if its force is slightly less than $\mu mg$. |
