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<title>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?</title>
+
<title>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 ?</title>
$4.1.22.$ 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>
<center>
<figure>
<img src="4.1.22_1.png"
loading="lazy" width="300" />
<figcaption>
Forces acting on the system
</figcaption>
</figure>
</center>
<p>
The vessel with liquid goes downward with acceleration $a$ as shown in the figure</p><p>
@@ -71,8 +71,8 @@
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
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$$ma = mg\sin\alpha-F_\text{fr}$$
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$$\boxed{a=g(\sin\alpha-\mu\cos\alpha)}\quad(1)$$
+
$$ma = mg\sin\alpha-F_\text{fr}$$
+
$$\boxed{a=g(\sin\alpha-\mu\cos\alpha)}\quad(1)$$
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
<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>
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<meta name="description" content="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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<meta property="og:description" content="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?">
<meta property="og:description" content="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?">
<title>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?</title>
<title>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 ?</title>
$4.1.22.$ 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 $\mu$?
$4.1.22.$ 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>
</p>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="4.1.22_1.png"
<img src="4.1.22_1.png"
loading="lazy" width="300" />
loading="lazy" width="300" />
<figcaption>
<figcaption>
Forces acting on the system
Forces acting on the system
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<p>
<p>
The vessel with liquid goes downward with acceleration $a$ as shown in the figure</p><p>
The vessel with liquid goes downward with acceleration $a$ as shown in the figure</p><p>
@@ -71,8 +71,8 @@
Given the reaction force $N=mg\cos\alpha$, the sliding friction force is found as
Given the reaction force $N=mg\cos\alpha$, the sliding friction force is found as
$$F_\text{fr} = \mu N = \mu mg\cos\alpha$$
$$F_\text{fr} = \mu N = \mu mg\cos\alpha$$
Newton's second law for the vessel
Newton's second law for the vessel
$$ma = mg\sin\alpha-F_\text{fr}$$
$$ma = mg\sin\alpha-F_\text{fr}$$
$$\boxed{a=g(\sin\alpha-\mu\cos\alpha)}\quad(1)$$
$$\boxed{a=g(\sin\alpha-\mu\cos\alpha)}\quad(1)$$
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
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
<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>
