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| <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.1">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.21.$ The belt lift forms an angle $\alpha$ with the horizon. With what maximum acceleration can a box be lifted on such a lift, if the coefficient of friction is equal to $\mu$? The tape doesn't bend. | | $2.1.21.$ The belt lift forms an angle $\alpha$ with the horizon. With what maximum acceleration can a box be lifted on such a lift, if the coefficient of friction is equal to $\mu$? The tape doesn't bend. |
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| For problem $2.1.21$ | | For problem $2.1.21$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| Forces acting on the box | | Forces acting on the box |
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| <p> | | <p> |
| 1. By discarding the constraints imposed on the box and replacing them with reactions, we can consider it as a free body capable of moving along the $OX$ axis. The friction force in this case is directed toward acceleration, i.e. against the possible movement of the box. | | 1. By discarding the constraints imposed on the box and replacing them with reactions, we can consider it as a free body capable of moving along the $OX$ axis. The friction force in this case is directed toward acceleration, i.e. against the possible movement of the box. |
| </p> | | </p> |
| <p> | | <p> |
| 2. The equation of Newton's second law allows us to determine the maximum value of acceleration | | 2. The equation of Newton's second law allows us to determine the maximum value of acceleration |
| $$ \mu mg \cos \alpha-mg \sin \alpha \leq ma $$ | | $$ \mu mg \cos \alpha-mg \sin \alpha \leq ma $$ |
| | | |
| $$ \boxed{a \leq g (\mu \cos \alpha-\sin \alpha)} $$ | | $$ \boxed{a \leq g (\mu \cos \alpha-\sin \alpha)} $$ |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$a_{max} = g(\mu \cos \alpha − \sin \alpha)$$ | | $$a_{max} = g(\mu \cos \alpha − \sin \alpha)$$ |
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