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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> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.23.$ A body of mass $m$ lying on a horizontal plane is affected by a force $F$ at an angle $\alpha$ to the horizon. Coefficient of friction $\mu$. Find the acceleration of the body if it does not detach from the plane. | | $2.1.23.$ A body of mass $m$ lying on a horizontal plane is affected by a force $F$ at an angle $\alpha$ to the horizon. Coefficient of friction $\mu$. Find the acceleration of the body if it does not detach from the plane. |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| loading="lazy" width="250" /> | | loading="lazy" width="250" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the body | | Forces acting on the body |
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| <div class="solutions"> | | <div class="solutions"> |
| <p> | | <p> |
| 1. The normal reaction of the connection in this case | | 1. The normal reaction of the connection in this case |
| will be determined both by the force of gravity $mg$ and | | will be determined both by the force of gravity $mg$ and |
| the projection on the axis $OY$ of the applied force: | | the projection on the axis $OY$ of the applied force: |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $N=mg-F \, sin \, \alpha$ | | $N=mg-F \, sin \, \alpha$ |
| </p> | | </p> |
| <p> | | <p> |
| The friction force is determined as: | | The friction force is determined as: |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $F_{тре} = (mg-F \,sin\, \alpha)$ | | $F_{тре} = (mg-F \,sin\, \alpha)$ |
| </p> | | </p> |
| <p> | | <p> |
| 2. The basic law of dynamics, therefore. will be written as follows: | | 2. The basic law of dynamics, therefore. will be written as follows: |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $F \, cos \,\alpha = \mu (mg-F \,sin\, \alpha)$ | | $F \, cos \,\alpha = \mu (mg-F \,sin\, \alpha)$ |
| </p> | | </p> |
| <p> | | <p> |
| 3. From the equation of Newton's second law it is easy to determine the desired acceleration | | 3. From the equation of Newton's second law it is easy to determine the desired acceleration |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $a = \frac{1}{m}(F\,cos\,\alpha-\mu mg+F\,sin\,\alpha)$ | | $a = \frac{1}{m}(F\,cos\,\alpha-\mu mg+F\,sin\,\alpha)$ |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $a = \frac{F}{m}(cos\,\alpha-\mu \,sin\,\alpha)$ | | $a = \frac{F}{m}(cos\,\alpha-\mu \,sin\,\alpha)$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $a = (F/m)(cos \,α + \mu \,sin \, \varphi)$$ − \mu g$ if this expression is greater than zero, otherwise $a = 0$. | | $a = (F/m)(cos \,α + \mu \,sin \, \varphi)$$ − \mu g$ if this expression is greater than zero, otherwise $a = 0$. |
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