Edit Problem

Year

Paper

Question Number

Course

Section

Worksheet Citation (for copying)

Click the copy button or select the text to copy this citation for use in worksheets.

Problem Text

A plank $AB$ of length $L$ initially lies horizontally at rest along the $x$-axis on a flat surface, with $A$ at the origin.
Point $C$ on the plank is such that $AC$ has length $sL$, where $0 < s < 1$.
End $A$ is then raised vertically along the $y$-axis so that its height above the horizontal surface at time $t$ is $h(t)$, while end $B$ remains in contact with the flat surface and on the $x$-axis.
The function $h(t)$ satisfies the differential equation 
$$\frac{d^2h}{dt^2} = -\omega^2 h, \text{ with } h(0) = 0 \text{ and } \frac{dh}{dt} = \omega L \text{ at } t = 0$$
where $\omega$ is a positive constant.
A particle $P$ of mass $m$ remains in contact with the plank at point $C$.
\begin{questionparts}
\item Show that the $x$-coordinate of $P$ is $sL\cos\omega t$, and find a similar expression for its $y$-coordinate.
\item Find expressions for the $x$- and $y$-components of the acceleration of the particle.
\item $N$ and $F$ are the upward normal and frictional components, respectively, of the force of the plank on the particle. Show that 
$$N = mg(1 - k\sin\omega t)\cos\omega t$$
and that 
$$F = mgsk\frac{\omega^2}{g}\tan\omega t$$
where $k = \frac{L\omega^2}{g}$.
\item The coefficient of friction between the particle and the plank is $\tan\alpha$, where $\alpha$ is an acute angle.
Show that the particle will not slip initially, provided $sk < \tan\alpha$.
Show further that, in this case, the particle will slip
\begin{itemize}
\item while $N$ is still positive,
\item when the plank makes an angle less than $\alpha$ to the horizontal.
\end{itemize}
\end{questionparts}

Solution (Optional)

\begin{questionparts}
\item Since we have $h'' + \omega^2 h = 0$ we must have that $h(t) = A \cos \omega t + B \sin \omega t$. The initial conditions tell us that $A = 0$ and $B = L$, so $h(t) = L \sin \omega t$.

\begin{center}
    \begin{tikzpicture}[scale=2]
        \draw[->] (0, -0.5) -- (0, 1);
        \draw[->] (-0.5,0 ) -- (1,0);

\draw (0, 0.5) -- (0.7, 0);

\node[left] at (0, 0.25) {$h = L \sin \omega t$};
        \node[above] at (0.35, 0.25) {$L$};
        
    \end{tikzpicture}
\end{center}

Therefore we can see the angle at $B$ is $\omega t$ and so $P$ has $y$-coordinate $(1-s)L \sin \omega t$ and $x$-coordinate $sL \cos \omega t$

\item If the position is $\binom{sL \cos \omega t}{(1-s) L \sin \omega t}$ then the acceleration is $-\omega^2 \binom{sL \cos \omega t}{(1-s) L \sin \omega t}$

\item
\begin{center}
    \begin{tikzpicture}[scale=4]

\coordinate (A) at (0, 0.5);
        \coordinate (B) at (0.7, 0);
        \coordinate (P) at ($0.6*(A)+0.4*(B)$);
    
        \draw[->] (0, -0.5) -- (0, 1);
        \draw[->] (-0.5,0 ) -- (1,0);

\draw (A) -- (B);
        \draw[dashed] (A) -- ++(0.5, 0);

\node[left] at ($0.5*(A)$) {$h = L \sin \omega t$};
        \node[above] at ($0.25*(A)+0.75*(B)$) {$L$};
        \filldraw (P) circle (0.3pt);

\draw[-latex, blue, ultra thick] (P) -- ++(0,-0.2) node[right] {$mg$};
        \draw[-latex, blue, ultra thick] (P) -- ($0.8*(A)+0.2*(B)$) node[left] {$F$};
        \draw[-latex, blue, ultra thick] (P) -- ++({0.5/5}, {0.7/5}) node[right] {$N$};
    \end{tikzpicture}
\end{center}

\begin{align*}
\text{N2}(\rightarrow): && - F\cos \omega t + N \sin \omega t &= -m\omega^2 sL \cos \omega t\\
\text{N2}(\uparrow): && -mg + F\sin \omega t + N \cos \omega t &= -m\omega^2 (1-s) L \sin \omega t \\
\Rightarrow && \begin{pmatrix} \cos \omega t & -\sin \omega t \\ \sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} F \\ N \end{pmatrix} &= \begin{pmatrix} m\omega^2 s L \cos \omega t \\
mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\
\Rightarrow && \begin{pmatrix}F \\ N \end{pmatrix}
 &= \begin{pmatrix} \cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t \end{pmatrix} \begin{pmatrix} m\omega^2 s L \cos \omega t \\
mg - m\omega^2(1-s)L \sin \omega t \end{pmatrix} \\
\Rightarrow && N &= m \omega^2 s L (-\sin \omega t \cos \omega t) + mg \cos \omega t - m \omega^2 (1-s)L \sin \omega t \cos \omega t \\
&&&=mg \cos \omega t - m \omega^2 L \sin \omega t \cos \omega t \\
&&&= mg \cos \omega t \left (1 - \frac{L \omega^2}{g} \sin \omega t \right) \\
&&&= mg (1 - k \sin \omega t) \cos \omega t \\
\Rightarrow && F &= m \omega^2 s L \cos^2 \omega t + mg \sin \omega t - m \omega^2 (1-s) L \sin ^2 \omega t \\
&&&= m \omega^2 s L + mg \sin \omega t - m \omega^2 L \sin^2 \omega t \\
&&&= mg \frac{\omega^2 L}{g} s + mg(1-\frac{\omega^2 L}{g} \sin \omega t)\sin \omega t \\
&&&= mg sk + mg(1-k \sin \omega t) \cos \omega t \tan \omega t \\
&&&= mgsk + N \tan \omega t
\end{align*}

\item The particle will not slip if $F < \tan \alpha N$. When $t = 0$, $N = mg, F = mgsk$, but clearly $sk < \tan \alpha \Rightarrow mgsk = F < \tan \alpha mg = \tan \alpha N$.

The particle will slip when: $F > \tan \alpha N$, but we have $F = mgsk + N \tan \omega t$. Clearly when $\omega t = \alpha$ we have reached a point where $F > \tan \alpha N$. Therefore we must slip before we reach this point, ie at a point where the plank makes an angle of less than $\alpha$ to the horizontal. Notice also that $N$ changes sign when $1-k \sin \omega t = 0$, however, to do this $N$ must become very small, smaller than $mgsk$, therefore we must slip before this point too. Since we slip before either condition occurs, we must be in a position when $N$ is positive AND the plank still makes a shallow angle.
\end{questionparts}

Preview

Problem

Solution

Cancel

Current Ratings

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0