Determine the force acting on the vertical wall from the side of the wedge, if a weight of mass m is placed on it. Angle at the base of the wedge \alpha . Coefficient of friction between the load and the wedge surface \mu . There is no friction between the floor and the wedge.

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Solutions of Savchenko Problems in Physics
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    <h3> Statement </h3>
    <p>
        $2.1.32.$ Determine the force acting on the vertical wall from the side of the wedge, if a weight of mass $m$ is placed on it. Angle at the base of the wedge $\alpha$. Coefficient of friction between the load and the wedge surface $\mu$. There is no friction between the floor and the wedge.

    <h3>Solution</h3>
    <p>
        <center>

The occurrence of an action on the wedge from the chalk is due to the accelerated movement of the body along the wedge

The movement becomes possible under the condition $$mg \sin\alpha\geq \mu mg \cos\alpha$$ otherwise, the body will be at rest, and acceleration will not occur.

Using the principles of liberation, we will represent the body as a free material particle under the action of a system of forces $\{mg; F_\tau\}$

The equation of Newton's second law in projection onto the direction of motion of the body will be represented as follows: $$ma_x=mg \sin\alpha - \mu mg \cos\alpha$$

$$a_x=g \sin\alpha - \mu g \cos\alpha$$
Force acting on the wedge and the vertical wall
$$F = mg \cos\alpha (\sin\alpha - \mu\cos\alpha )$$

    <h4>Answer</h4>
    <p>
        <p>
      $F=mg\operatorname{\cos}\alpha (\operatorname{sin}\alpha -\mu\operatorname{\cos}\alpha )$ with $\mu\leqslant\operatorname{tg}\alpha$;
    </p>
    <p>
      $F=0$ with $\mu\geqslant\operatorname{tg}\alpha$
    </p>
    </p>


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