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<meta name="description" content="A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.">
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<meta property="og:description" content="A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.">
<title>A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.</title>
$2.1.21.$ 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.
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.
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2. The equation of Newton's second law allows us to determine the maximum value of acceleration
<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">
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<meta name="description" content="A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.">
<meta name="description" content="A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.">
<meta property="og:title" content="A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.">
<meta property="og:title" content="A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.">
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@@ -14,9 +14,9 @@
<meta property="og:description" content="A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.">
<meta property="og:description" content="A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.">
<title>A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.</title>
<title>A bar of mass m is placed on the horizontal board. The board is slowly tilted. Determine the dependence of the friction force acting on the bar on the angle of inclination of the board \alpha. Coefficient of friction \mu.</title>
$2.1.21.$ 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.
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
<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>
