2019 Paper 2 Q10
Year: 2019
Paper: 2
Question Number: 10
Course: LFM Pure and Mechanics
Section: Moments
Difficulty: 1500.0
Banger: 1500.0
Problem
A small light ring is attached to the end \(A\) of a uniform rod \(AB\) of weight \(W\) and length \(2a\). The ring can slide on a rough horizontal rail.
One end of a light inextensible string of length \(2a\) is attached to the rod at \(B\) and the other end is attached to a point \(C\) on the rail so that the rod makes an angle of \(\theta\) with the rail, where \(0 < \theta < 90^{\circ}\). The rod hangs in the same vertical plane as the rail.
A force of \(kW\) acts vertically downwards on the rod at \(B\) and the rod is in equilibrium.
- You are given that the string will break if the tension \(T\) is greater than \(W\). Show that (assuming that the ring does not slip) the string will break if $$2k + 1 > 4 \sin \theta.$$
- Show that (assuming that the string does not break) the ring will slip if $$2k + 1 > (2k + 3)\mu \tan \theta,$$ where \(\mu\) is the coefficient of friction between the rail and the ring.
- You are now given that \(\mu \tan \theta < 1\). Show that, when \(k\) is increased gradually from zero, the ring will slip before the string breaks if $$\mu < \frac{2 \cos \theta}{1 + 2 \sin \theta}.$$
Solution
- \(\,\) \begin{align*} \overset{\curvearrowright}{A}:&& W \cos \theta \cdot a + kW \cos \theta \cdot 2a - T \cos \theta \sin \theta \cdot 2a - T \sin \theta \cos \theta \cdot 2a &= 0 \\ && (2k+1) \cos \theta W &= T \cos \theta \cdot 4 \sin \theta \\ \Rightarrow && T &= \frac{2k+1}{4 \sin \theta} W \\ \Rightarrow && \text{breaks if }\quad \quad 2k+1 &> 4 \sin \theta \end{align*}
- \(\,\) \begin{align*} \text{N2}(\uparrow): && R - W - kW - T \sin \theta &= 0 \\ \Rightarrow && R &= (k+1)W - T \sin \theta \\ &&&= (k+1)W - \frac{2k+1}{4} W \\ &&&= \frac{2k+3}{4}W \\ \text{N2}(\leftarrow): && F_A - T \cos \theta &= 0 \\ \Rightarrow && F_A &= \frac{2k+1}{4 }\cot \theta \\ \Rightarrow && \text{slips if }\quad \quad\quad \quad\quad \quad F_A &> \mu R \\ \Rightarrow && \text{slips if }\quad \quad \frac{2k+1}{4 }\cot \theta &> \mu \frac{2k+3}{4}W \\ \Rightarrow && 2k+1 &> (2k+3) \mu \tan \theta \end{align*}
- The condition for breaking is \(k > 2\sin \theta -\frac12\). The condition for slipping is equivalent to: \begin{align*} && 2k+1 &> (2k+3) \mu \tan \theta \\ \Leftrightarrow && 2k(1- \mu \tan \theta) &> 3 \mu \tan \theta-1 \\ \Leftrightarrow && k &> \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} \end{align*} Therefore we will slip first if: \begin{align*} && \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} &< 2 \sin \theta - \frac12 \\ \Leftrightarrow && 3 \mu \tan \theta-1 &< 4 \sin \theta (1- \mu \tan \theta) - (1- \mu \tan \theta) \\ &&&=4 \sin \theta - 1 + \mu \tan \theta (1-4 \sin \theta) \\ \Leftrightarrow && 3 \mu \tan \theta &< 4 \sin \theta + \mu \tan \theta (1- 4 \sin \theta) \\ \Leftrightarrow && \mu \tan \theta(3-1+4\sin \theta) &< 4 \sin \theta \\ \Leftrightarrow && \mu &< \frac{2 \cos \theta}{1+2 \sin \theta} \end{align*}
Examiner's report
— 2019 STEP 2, Question 10
~15% attempted (inferred)
Inferred ~15%: non-Pure, not Q9 → attempted by fewer than a quarter of candidates.
From the paper introduction
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Source: Cambridge STEP 2019 Examiner's Report
· 2019-p2.pdf
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
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Problem source
A small light ring is attached to the end $A$ of a uniform rod $AB$ of weight $W$ and length $2a$. The ring can slide on a rough horizontal rail.
One end of a light inextensible string of length $2a$ is attached to the rod at $B$ and the other end is attached to a point $C$ on the rail so that the rod makes an angle of $\theta$ with the rail, where $0 < \theta < 90^{\circ}$. The rod hangs in the same vertical plane as the rail.
A force of $kW$ acts vertically downwards on the rod at $B$ and the rod is in equilibrium.
\begin{questionparts}
\item You are given that the string will break if the tension $T$ is greater than $W$. Show that (assuming that the ring does not slip) the string will break if
$$2k + 1 > 4 \sin \theta.$$
\item Show that (assuming that the string does not break) the ring will slip if
$$2k + 1 > (2k + 3)\mu \tan \theta,$$
where $\mu$ is the coefficient of friction between the rail and the ring.
\item You are now given that $\mu \tan \theta < 1$.
Show that, when $k$ is increased gradually from zero, the ring will slip before the string breaks if
$$\mu < \frac{2 \cos \theta}{1 + 2 \sin \theta}.$$
\end{questionparts}Solution source
\begin{center}
\begin{tikzpicture}
\def\a{-2};
\def\b{-1};
\coordinate (A) at (\a, 0);
\coordinate (B) at (0, \b);
\coordinate (C) at (-\a, 0);
\draw[thin] ($(A)!-0.5!(C)$) -- ($(A)!1.5!(C)$);
\filldraw (A) circle (1.5pt) node[above left] {$A$};
\filldraw (B) circle (1.5pt) node[above] {$B$};
\filldraw (C) circle (1.5pt) node[above] {$C$};
\draw[ultra thick] (A) -- (B);
\draw[thick] (B) -- (C);
\draw[-latex, ultra thick, blue] (A) -- ++(0, 1) node[above] {$R$};
\draw[-latex, ultra thick, blue] (A) -- ++(-1, 0) node[above] {$F_A$};
\draw[-latex, ultra thick, blue] (B) -- ++(0, -1) node[below] {$kW$};
\draw[-latex, ultra thick, blue] (B) -- ($(B)!0.5!(C)$) node[right] {$T$};
\draw[-latex, ultra thick, blue] ($(A)!0.5!(B)$) -- ++(0, -1.5) node[below] {$W$};
\pic [draw, angle radius=1.2cm, angle eccentricity=.6, "$\theta$"] {angle = B--A--C};
\end{tikzpicture}
\end{center}
\begin{questionparts}
\item $\,$
\begin{align*}
\overset{\curvearrowright}{A}:&& W \cos \theta \cdot a + kW \cos \theta \cdot 2a - T \cos \theta \sin \theta \cdot 2a - T \sin \theta \cos \theta \cdot 2a &= 0 \\
&& (2k+1) \cos \theta W &= T \cos \theta \cdot 4 \sin \theta \\
\Rightarrow && T &= \frac{2k+1}{4 \sin \theta} W \\
\Rightarrow && \text{breaks if }\quad \quad 2k+1 &> 4 \sin \theta
\end{align*}
\item $\,$ \begin{align*}
\text{N2}(\uparrow): && R - W - kW - T \sin \theta &= 0 \\
\Rightarrow && R &= (k+1)W - T \sin \theta \\
&&&= (k+1)W - \frac{2k+1}{4} W \\
&&&= \frac{2k+3}{4}W \\
\text{N2}(\leftarrow): && F_A - T \cos \theta &= 0 \\
\Rightarrow && F_A &= \frac{2k+1}{4 }\cot \theta \\
\Rightarrow && \text{slips if }\quad \quad\quad \quad\quad \quad F_A &> \mu R \\
\Rightarrow && \text{slips if }\quad \quad \frac{2k+1}{4 }\cot \theta &> \mu \frac{2k+3}{4}W \\
\Rightarrow && 2k+1 &> (2k+3) \mu \tan \theta
\end{align*}
\item The condition for breaking is $k > 2\sin \theta -\frac12$. The condition for slipping is equivalent to:
\begin{align*}
&& 2k+1 &> (2k+3) \mu \tan \theta \\
\Leftrightarrow && 2k(1- \mu \tan \theta) &> 3 \mu \tan \theta-1 \\
\Leftrightarrow && k &> \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)}
\end{align*}
Therefore we will slip first if:
\begin{align*}
&& \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} &< 2 \sin \theta - \frac12 \\
\Leftrightarrow && 3 \mu \tan \theta-1 &< 4 \sin \theta (1- \mu \tan \theta) - (1- \mu \tan \theta) \\
&&&=4 \sin \theta - 1 + \mu \tan \theta (1-4 \sin \theta) \\
\Leftrightarrow && 3 \mu \tan \theta &< 4 \sin \theta + \mu \tan \theta (1- 4 \sin \theta) \\
\Leftrightarrow && \mu \tan \theta(3-1+4\sin \theta) &< 4 \sin \theta \\
\Leftrightarrow && \mu &< \frac{2 \cos \theta}{1+2 \sin \theta}
\end{align*}
\end{questionparts}
As with so many questions, the big stumbling block for students was drawing a good diagram from the information, including all the relevant forces. With "show that" questions it is beholden on candidates to explain their working. Equations which just appear and lead to the correct answer are not sufficient. In mechanics, it would be very helpful for students to say, for example, "Taking moments about point A for the rod" or "Resolving for the string-rod system vertically" to give some sense of where an equation arises. The flow of logic is a fundamental idea in mathematics, but it was clear in this question that it was not familiar to the vast majority of students. The questions effectively asked "if <given condition> show that <mechanical outcome>". Most students reversed this to show that "if <mechanical outcome> then <given condition>". In this question, most arguments were reversible, but it still demonstrated a fundamental misunderstanding of what was being asked. The other issue which flummoxed students was dealing with inequalities. There are different rules of algebra associated with inequalities and this is something which is frequently tested in STEP. Candidates would benefit from thinking carefully about things like when can one inequality be substituted into another, or when can an inequality be squared. The intuition from equalities was too often applied without thinking.