Month: November 2022

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202211031053 Statics Figures (Elementary) Q2

87. A homogeneous chain (i.e., \rho =\textrm{Const.}) with a length l lies on a table. What is the maximum length l_1 of the part of the chain hanging over the table if the coefficient of friction between the chain and the table is k?

Extracted from B. Bukhovtsev et al. (1978). Problems in Elementary Physics.

Roughwork.

\begin{aligned} f = kN & = W_1 \\ k\rho (l-l_1)g & = \rho l_1g \\ \end{aligned}

Answer. \displaystyle{l_1=l\frac{k}{k+1}}.

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202211021054 Dynamics Figures (Elementary) Q9

Referring to the figure below, the pulleys are assumed to be weightless and frictionless and the ropes massless and inextensible.

Prove that m_1=2m_2 if the system is in equilibrium. Find the equation of motion for each mass, i. if m_1>2m_2; and ii. if m_1<2m_2.

Roughwork.

In equilibrium,

\textrm{Net }\mathbf{F}=m\mathbf{a}=\mathbf{0}.

Thus

\begin{cases} W_1-2T =m_1(0) \\ W_2-T =m_2(0) \end{cases} \Longrightarrow\quad \begin{cases} m_1g =2T \\ m_2g =T \\ \end{cases}

\therefore m_1=2m_2.

Mass m_1 will move down and m_2 up if m_1>2m_2, and the reverse if m_1<2m_2. When m_1 makes displacement of \Delta h the vertical, m_2 makes 2\Delta h.

Without gain of speciality, one may obtain the equations of motion by either (a) Newton’s second law, or (b) conservation of mechanical energy, or (c) Euler-Lagrange method.

Take downward positive. When m_1>2m_2, in one’s mind one can draw two free-body diagrams giving two equations:

\begin{cases} W_1(=m_1g)-2T=m_1a_1\\ W_2(=m_2g)-T=-m_2a_2 \end{cases}

Hence,

\begin{cases} 2T = m_1(g-a_1) \\ T = m_2(g+a_2) \end{cases}

note that 2a_1=a_2. And so

\begin{aligned} a_1 & = \bigg( 1-\frac{6m_2}{m_1+4m_2}\bigg)\cdot g \\ a_2 & = \bigg( \frac{3m_1}{m_1+4m_2}-1\bigg)\cdot g \\ \end{aligned}

Note the inequalities 0\leqslant a_1\leqslant a_2<g. One can thus find tension in magnitude T, yet this is left the reader.

The Lagrangian \mathcal{L}=T-V of the system is

\mathcal{L}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+m_1g\Delta h-2m_2g\Delta h,

or,

\mathcal{L}=\frac{1}{2}m_1\dot{x}^2+\frac{1}{2}m_2(\dot{2x})^2+m_1gx-2m_2gx.

where v_1=v_1(t) and v_2=v_2(t) vary with time t as are subjected to acceleration, and v_2=2v_1.

Euler-Lagrange equation reads

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial \mathcal{L}}{\partial\dot{x}}\bigg)-\frac{\partial\mathcal{L}}{\partial x}=0}.

Computing term by term,

\begin{aligned} \frac{\partial \mathcal{L}}{\partial x} & = m_1g-2m_2g \\ \frac{\partial\mathcal{L}}{\partial \dot{x}} & = m_1\dot{x}+4m_2\dot{x} \\ \frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{\partial\mathcal{L}}{\partial\dot{x}}\bigg) & = m_1\ddot{x}+4m_2\ddot{x} \\ \end{aligned}

and the result follows.

(to be continued)