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<meta name="description" content="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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<meta property="og:description" content="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>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>
$2.1.22.$ 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 $\alpha$, $tg \alpha > \mu$.
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>
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
$$\vec{F}_{1} + \vec{F}_{2} = m \vec{a}$$
Let us determine the projections of the acting forces on the OX axis and write down the equation of Newton's second law
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
<meta name="description" content="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.">
<meta name="description" content="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.">
<meta property="og:title" content="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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<meta property="og:image" content="img/logo.png">
<meta property="og:image" content="img/logo.png">
@@ -14,9 +14,9 @@
<meta property="og:description" content="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.">
<meta property="og:description" content="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>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>
$2.1.22.$ 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 $\alpha$, $tg \alpha > \mu$.
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
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
