At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle \alpha to the horizon if the friction coefficient is \mu?

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    <h3 id="back-link"><a href="/#4.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $4.1.22.$ At what angle to the horizon will the surface of the liquid in the vessel slide on an inclined plane that makes an angle $\alpha$ to the horizon if the friction coefficient is $\mu$?
    </p>

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

The vessel with liquid goes downward with acceleration $a$ as shown in the figure

Given the reaction force $N=mg\cos\alpha$, the sliding friction force is found as
$$F_\text{fr} = \mu N = \mu mg\cos\alpha$$
Newton's second law for the vessel
$$ma = mg\sin\alpha-F_\text{fr}$$
$$\boxed{a=g(\sin\alpha-\mu\cos\alpha)}\quad(1)$$
Let us introduce the effective acceleration of free fall $\vec{g^*}$ in a non-inertial frame of reference associated with the vessel, then vectorially, it is found as
$$\vec{g^*}=\vec{g}-\vec{a}$$

From the drawing it could be found that
$$\tan\beta = \frac{a - g\sin\alpha}{g\cos\alpha} = -\mu$$
The angle between the water level and the horizon will be
$$\boxed{\gamma= \alpha-\operatorname{arctg}\mu}$$

    <h4>Answer</h4>
    <p>
        $$\gamma=\alpha-\operatorname{arctg}\mu $$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
        Aliaksandr Melnichenka<br>
    </p>


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