2.1.23 | revision #7835 | Savchenko Solutions

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.

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    <h3> Statement </h3>
    <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.
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    <h3>Solution</h3>
    <p>
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1. The normal reaction of the connection in this case will be determined both by the force of gravity $$mg$$ and the projection on the axis $$OY$$ of the applied force:

$N=mg-F \\, sin \\, \alpha$

The friction force is determined as:

$тре$F_{тре} = (mg-F \\,sin\\, \alpha)$$

2. The basic law of dynamics, therefore. will be written as follows:

$F \\, cos \\,\alpha = \mu (mg-F \\,sin\\, \alpha)$

3. From the equation of Newton's second law it is easy to determine the desired acceleration

$a = \frac{1}{m}(F\\,cos\\,\alpha-\mu mg+F\\,sin\\,\alpha)$

$a = \frac{F}{m}(cos\\,\alpha-\mu \\,sin\\,\alpha)$

    <h4>Answer</h4>
    <p>
        $a = (F/m)(cos \\,α + \mu \\,sin \\, \varphi)$$ − \mu g$ if this expression is greater than zero, otherwise $a = 0$.
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