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<meta name="description" content="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
<meta property="og:title" content="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
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@@ -14,9 +14,9 @@
<meta property="og:description" content="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
<title>On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.</title>
$2.3.48^*.$ On the horizontal plane lie two bars of mass $m_1$ and $m_2$, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is $\mu$.
</p>
<center>
<figure>
<img src="2.3.48.png"
loading="lazy" width="200" />
<figcaption>
For problem $2.3.48^*$
</figcaption>
</figure>
</center>
<h3>Solution</h3>
<p>
</p>
<center>
<figure>
<img src="2.3.48_1.png"
loading="lazy" width="300" />
<figcaption>
Forces acting on the system
</figcaption>
</figure>
</center>
<p>
States of equilibrium for horizontal axes
$$F_\text{fr1} = \mu N_1 \quad (1)$$
$$F_\text{fr2} = \mu N_2 \quad (2)$$
Alternatively for a vertical axes
$$N_1 = m_1g \quad (3)$$
$$N_2 = m_2g \quad (4)$$
Substituting $(3)$ and $(4)$ to the $(1)$ and $(2)$ equations
$$F_\text{fr1} = \mu m_1g$$
$$F_\text{fr2} = \mu m_2g$$
By the law of energy conservation, force $\vec{F}$ acts against the tension of the spring and against the friction force
$$A=Fx=\Delta E$$
Substituting the potential energy of the spring
$$Fx = \frac{kx^2}{2} + F_\text{fr1}x$$
$$F= \frac{kx}{2} + \mu m_1 g \quad (5)$$
From the state equilibrium in the steady state
$$kx = \mu m_2 g \quad (6)$$
After substituting $(6)$ into $(5)$, we finally get
$$\boxed{ F = \mu g \left( m_1 +\frac{m_2}{2}\right)}$$
<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="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
<meta name="description" content="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
<meta property="og:title" content="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
<meta property="og:title" content="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
<meta property="og:image" content="img/logo.png">
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@@ -14,9 +14,9 @@
<meta property="og:description" content="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
<meta property="og:description" content="On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.">
<title>On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.</title>
<title>On the horizontal plane lie two bars of mass m_1 and m_2, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is \mu.</title>
$2.3.48^*.$ On the horizontal plane lie two bars of mass $m_1$ and $m_2$, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is $\mu$.
$2.3.48^*.$ On the horizontal plane lie two bars of mass $m_1$ and $m_2$, connected by an undeformed spring. Determine what is the smallest constant force to be applied to the left bar to move the right one, if the coefficient of friction of loads on the plane is $\mu$.
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="2.3.48.png"
<img src="2.3.48.png"
loading="lazy" width="200" />
loading="lazy" width="200" />
<figcaption>
<figcaption>
For problem $2.3.48^*$
For problem $2.3.48^*$
</figcaption>
</figcaption>
</figure>
</figure>
</center>
</center>
<h3>Solution</h3>
<h3>Solution</h3>
<p>
<p>
</p>
</p>
<center>
<center>
<figure>
<figure>
<img src="2.3.48_1.png"
<img src="2.3.48_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>
States of equilibrium for horizontal axes
States of equilibrium for horizontal axes
$$F_\text{fr1} = \mu N_1 \quad (1)$$
$$F_\text{fr1} = \mu N_1 \quad (1)$$
$$F_\text{fr2} = \mu N_2 \quad (2)$$
$$F_\text{fr2} = \mu N_2 \quad (2)$$
Alternatively for a vertical axes
Alternatively for a vertical axes
$$N_1 = m_1g \quad (3)$$
$$N_1 = m_1g \quad (3)$$
$$N_2 = m_2g \quad (4)$$
$$N_2 = m_2g \quad (4)$$
Substituting $(3)$ and $(4)$ to the $(1)$ and $(2)$ equations
Substituting $(3)$ and $(4)$ to the $(1)$ and $(2)$ equations
$$F_\text{fr1} = \mu m_1g$$
$$F_\text{fr1} = \mu m_1g$$
$$F_\text{fr2} = \mu m_2g$$
$$F_\text{fr2} = \mu m_2g$$
By the law of energy conservation, force $\vec{F}$ acts against the tension of the spring and against the friction force
By the law of energy conservation, force $\vec{F}$ acts against the tension of the spring and against the friction force
$$A=Fx=\Delta E$$
$$A=Fx=\Delta E$$
Substituting the potential energy of the spring
Substituting the potential energy of the spring
$$Fx = \frac{kx^2}{2} + F_\text{fr1}x$$
$$Fx = \frac{kx^2}{2} + F_\text{fr1}x$$
$$F= \frac{kx}{2} + \mu m_1 g \quad (5)$$
$$F= \frac{kx}{2} + \mu m_1 g \quad (5)$$
From the state equilibrium in the steady state
From the state equilibrium in the steady state
$$kx = \mu m_2 g \quad (6)$$
$$kx = \mu m_2 g \quad (6)$$
After substituting $(6)$ into $(5)$, we finally get
After substituting $(6)$ into $(5)$, we finally get
$$\boxed{ F = \mu g \left( m_1 +\frac{m_2}{2}\right)}$$
$$\boxed{ F = \mu g \left( m_1 +\frac{m_2}{2}\right)}$$
<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>
