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| <meta name="author" content="Aliaksandr Melnichenka"> | | <meta name="author" content="Aliaksandr Melnichenka"> |
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| <title>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.</title> | | <title>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.</title> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.1.22.$ After what time will the velocity of a body that has been sent up an inclined plane with velocity $v$ be equal to $v$ again? Coefficient of friction $\mu$, the angle between the plane and the horizon $\alpha$, $tg \alpha > \mu$. | | $2.1.22.$ After what time will the velocity of a body that has been sent up an inclined plane with velocity $v$ be equal to $v$ again? Coefficient of friction $\mu$, the angle between the plane and the horizon $\alpha$, $tg \alpha > \mu$. |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="https://savchenkosolutions.com/2/2.1.22/sol.png" | | <img src="https://savchenkosolutions.com/2/2.1.22/sol.png" |
| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
| <figcaption> | | <figcaption> |
| Forces acting on the body | | Forces acting on the body |
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| <p> | | <p> |
| The body on the inclined plane is under the action of three forces: the force of gravity $\vec{F}_{1} = m \vec{g}$, the force of friction $\vec{F}_{2} = - \mu mg$ and the normal reaction of the constraint $\vec{N}$, however, when further considering the motion, the normal reaction can be ignored, since its projection onto the horizontal axis along which the body moves is equal to zero. | | The body on the inclined plane is under the action of three forces: the force of gravity $\vec{F}_{1} = m \vec{g}$, the force of friction $\vec{F}_{2} = - \mu mg$ and the normal reaction of the constraint $\vec{N}$, however, when further considering the motion, the normal reaction can be ignored, since its projection onto the horizontal axis along which the body moves is equal to zero. |
| </p> | | </p> |
| <p> | | <p> |
| The equation of Newton's second law in projection onto the OX axis when the body moves upward in vector form is written as follows | | The equation of Newton's second law in projection onto the OX axis when the body moves upward in vector form is written as follows |
| $$ \vec{F}_{1} + \vec{F}_{2} = m \vec{a} $$ | | $$ \vec{F}_{1} + \vec{F}_{2} = m \vec{a} $$ |
| Let us determine the projections of the acting forces on the OX axis and write down the equation of Newton's second law | | Let us determine the projections of the acting forces on the OX axis and write down the equation of Newton's second law |
| $$ F_{1x} = mg \sin \alpha, F_{2x} = \mu mg \cos \alpha $$ | | $$ F_{1x} = mg \sin \alpha, F_{2x} = \mu mg \cos \alpha $$ |
| | | |
| $$ \mu mg \cos \alpha + mg \sin \alpha = ma $$ | | $$ \mu mg \cos \alpha + mg \sin \alpha = ma $$ |
| We divide both parts of the last equation by the mass $m$ and express the magnitude of the acceleration | | We divide both parts of the last equation by the mass $m$ and express the magnitude of the acceleration |
| $$ a = g( \sin \alpha + \mu \cos \alpha ) $$ | | $$ a = g( \sin \alpha + \mu \cos \alpha ) $$ |
| The time of the body's upward movement $t_1$ is determined from the condition that the speed at the end of the ascent is equal to zero | | The time of the body's upward movement $t_1$ is determined from the condition that the speed at the end of the ascent is equal to zero |
| $$ 0 = v_{0} - at \Rightarrow t_{1} = \frac{v_{0} }{a} = \frac{v_{0} }{g ( \sin \alpha + \mu \cos \alpha ) } $$ | | $$ 0 = v_{0} - at \Rightarrow t_{1} = \frac{v_{0} }{a} = \frac{v_{0} }{g ( \sin \alpha + \mu \cos \alpha ) } $$ |
| The downward motion of the body corresponds to the equation | | The downward motion of the body corresponds to the equation |
| $$ mg \sin \alpha - \mu mg \cos \alpha = ma $$ | | $$ mg \sin \alpha - \mu mg \cos \alpha = ma $$ |
| | | |
| $$ a = g( \sin \alpha - \mu \cos \alpha ) $$ | | $$ a = g( \sin \alpha - \mu \cos \alpha ) $$ |
| The speed will become equal to $v_0$ only at the end of the descent, because the law of conservation of energy has not been repealed, therefore | | The speed will become equal to $v_0$ only at the end of the descent, because the law of conservation of energy has not been repealed, therefore |
| $$ \frac{dv}{dt} = g( \sin \alpha + \mu \cos \alpha ) $$ | | $$ \frac{dv}{dt} = g( \sin \alpha + \mu \cos \alpha ) $$ |
| | | |
| $$ \int_{0}^{v_{0} } dv = g( \sin \alpha + \mu \cos \alpha ) = \int_{0}^{t_{2} } dt $$ | | $$ \int_{0}^{v_{0} } dv = g( \sin \alpha + \mu \cos \alpha ) = \int_{0}^{t_{2} } dt $$ |
| | | |
| $$ t_{2} = \frac{v_{0} }{g ( \sin \alpha + \mu \cos \alpha ) } $$ | | $$ t_{2} = \frac{v_{0} }{g ( \sin \alpha + \mu \cos \alpha ) } $$ |
| The required time will be determined as the sum $t = t_{1} + t_{2}$ | | The required time will be determined as the sum $t = t_{1} + t_{2}$ |
| $$ \boxed{t = \frac{2v_{0} \sin \alpha }{g( \sin^{2} \alpha - \mu^{2} \cos^{2} \alpha )}} $$ | | $$ \boxed{t = \frac{2v_{0} \sin \alpha }{g( \sin^{2} \alpha - \mu^{2} \cos^{2} \alpha )}} $$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$t = \frac{2v_{0} \sin \alpha }{g( \sin^{2} \alpha - \mu^{2} \cos^{2} \alpha )}$$ | | $$t = \frac{2v_{0} \sin \alpha }{g( \sin^{2} \alpha - \mu^{2} \cos^{2} \alpha )}$$ |
| </p> | | </p> |
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