OCR H240/03 2019 June — Question 11 14 marks

Exam BoardOCR
ModuleH240/03 (Pure Mathematics and Mechanics)
Year2019
SessionJune
Marks14
PaperDownload PDF ↗
TopicMoments
TypeLadder against wall
DifficultyStandard +0.3 This is a standard A-level mechanics ladder problem requiring moments about a point, resolving forces, and using limiting friction. Part (a) is routine moment calculation, part (b) applies friction law, part (c) is simple algebra, and part (d) is recall. The problem follows a well-established template with clear scaffolding and no novel insight required.
Spec1.05c Area of triangle: using 1/2 ab sin(C)

The diagram shows a ladder \(AB\), of length \(2a\) and mass \(m\), resting in equilibrium on a vertical wall of height \(h\). The ladder is inclined at an angle of \(30°\) to the horizontal. The end \(A\) is in contact with horizontal ground. An object of mass \(2m\) is placed on the ladder at a point \(C\) where \(AC = d\). The ladder is modelled as uniform, the ground is modelled as being rough, and the vertical wall is modelled as being smooth.
  1. Show that the normal contact force between the ladder and the wall is \(\frac{mg(a + 2d)\sqrt{3}}{4h}\). [4]
It is given that the equilibrium is limiting and the coefficient of friction between the ladder and the ground is \(\frac{1}{3}\sqrt{3}\).
  1. Show that \(h = k(a + 2d)\), where \(k\) is a constant to be determined. [7]
  2. Hence find, in terms of \(a\), the greatest possible value of \(d\). [2]
  3. State one improvement that could be made to the model. [1]

\begin{tikzpicture}[scale=1.2, thick]
  % Compute positions
  \pgfmathsetmacro{\Wx}{2*3.5*cos(30)}
  \pgfmathsetmacro{\By}{2*3.5*sin(30)}

  \coordinate (A) at (0,0);
  \coordinate (W) at (\Wx,0);
  \coordinate (B) at (\Wx,\By);
  \coordinate (C) at ({0.35*\Wx},{0.35*\By});
  \coordinate (midAB) at ({0.5*\Wx},{0.5*\By});
  \coordinate (midAC) at ({0.175*\Wx},{0.175*\By});

  % Ground
  \draw (-.5,0) -- ({\Wx+1},0);
  \foreach \x in {-0.3,0.0,...,6.3} {
    \draw (\x,0) -- ++(-.2,-.2);
  }

  % Wall
  \fill[gray!60] (\Wx,0) rectangle ++(0.15,{\By+0.8});
  \draw (\Wx,0) -- (\Wx,{\By+0.8});

  % Ladder
  \draw (A) -- (B);

  % Angle arc at A
  \draw (0.9,0) arc[start angle=0, end angle=30, radius=0.9];
  \node at (1.15,0.22) {$30^\circ$};

  % Height h label
  \draw[<->] ({\Wx+0.6},0) -- ({\Wx+0.6},\By);
  \node[right] at ({\Wx+0.6},{0.5*\By}) {$h$};

  % Label 2a along ladder
  \node[above left] at (midAB) {$2a$};

  % Label d from A to C
  \node[above left] at (midAC) {$d$};

  % Point C dot and label
  \fill (C) circle (2pt);
  \node[below right] at (C) {$C$};

  % Labels for A and B
  \node[below left] at (A) {$A$};
  \node[right] at (B) {$B$};
\end{tikzpicture}

The diagram shows a ladder $AB$, of length $2a$ and mass $m$, resting in equilibrium on a vertical wall of height $h$. The ladder is inclined at an angle of $30°$ to the horizontal. The end $A$ is in contact with horizontal ground. An object of mass $2m$ is placed on the ladder at a point $C$ where $AC = d$.

The ladder is modelled as uniform, the ground is modelled as being rough, and the vertical wall is modelled as being smooth.

\begin{enumerate}[label=(\alph*)]
\item Show that the normal contact force between the ladder and the wall is $\frac{mg(a + 2d)\sqrt{3}}{4h}$. [4]
\end{enumerate}

It is given that the equilibrium is limiting and the coefficient of friction between the ladder and the ground is $\frac{1}{3}\sqrt{3}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumii}{1}
\item Show that $h = k(a + 2d)$, where $k$ is a constant to be determined. [7]

\item Hence find, in terms of $a$, the greatest possible value of $d$. [2]

\item State one improvement that could be made to the model. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/03 2019 Q11 [14]}}