2017 Paper 2 Q9
Year: 2017
Paper: 2
Question Number: 9
Course: LFM Pure and Mechanics
Section: Moments
Problem
- Let \(F\) be the frictional force between one cylinder and the floor, and let \(R\) be the normal reaction between the plank and one cylinder. Show that \[ R\sin\theta = F(1+\cos\theta)\,, \] where \(\theta\) is the acute angle between the plank and the tangent to the cylinder at the point of contact. Deduce that \(2\sin\theta \le 1+\cos\theta\,\).
- Show that \[ N= \left( 1+\frac2 k\right)\left(\frac{1+\cos\theta}{\sin\theta} \right) F \,, \] where \(N\) is the normal reaction between the floor and one cylinder. Write down the condition that the cylinder does not slip on the floor and show that it is satisfied with no extra restrictions on \(\theta\).
- Show that \(\sin\theta\le\frac45\,\) and hence that \(r\le5a\,\).
Solution
- \(\,\) \begin{align*} \text{N2}(\rightarrow, \text{ one cylinder}): && F\cos \theta + F - R \sin \theta &= 0 \\ \Rightarrow && F(1+\cos \theta) &= R \sin \theta \\ && F \leq \tfrac12 R \\ \Rightarrow && R \sin \theta &\leq \frac12 R(1+\cos \theta) \\ \Rightarrow && 2 \sin \theta &\leq 1 + \cos \theta \end{align*}
- \(\,\) \begin{align*} \text{N2}(\uparrow, \text{system}): && 2N-(k+2)W &= 0 \\ \Rightarrow && W &= \left ( \frac{2}{k+2} \right)N \\ \text{N2}(\uparrow, \text{one cylinder}): && N - W - R\cos \theta -F\sin \theta &= 0 \\ \Rightarrow && N - \left ( \frac{2}{k+2} \right)N - F \left ( \frac{1+\cos \theta}{\sin \theta} \right) \cos \theta - F \sin \theta &= 0 \\ \Rightarrow && \left ( \frac{k}{k+2} \right)N &= \left ( \frac{\cos \theta + \cos^2 \theta + \sin^2 \theta}{\sin \theta} \right) F\\ \Rightarrow && N &= \left ( 1 + \frac2{k} \right) \left ( \frac{\cos \theta + 1}{\sin \theta} \right) F \end{align*} The cylinder does not slip if \(F \leq \tfrac12 N\), ie \begin{align*} && N &\leq \left ( 1 + \frac2{k} \right) \left ( \frac{\cos \theta + 1}{\sin \theta} \right) \frac12 N \\ \Rightarrow && 2\sin \theta &\leq \left ( 1 + \frac2{k} \right) \left ( \cos \theta + 1 \right) \end{align*} but since \(2 \sin \theta \leq (1 + \cos \theta)\) and \((1+\frac2k) > 1\) this inequality is obviously satisfied.
- We can notice that \(2\sin \theta = 1 + \cos \theta\) is satisfied by a \(3-4-5\) triangle, where \(\sin \theta = 4/5, \cos \theta = 3/5\) and hence if \(\sin \theta \leq \frac45\) the condition must hold.
So \(\sin \theta = \frac{r-a}{r} \leq \frac45 \Rightarrow 5r-5a \leq 4r \Rightarrow r \leq 5a\)
Examiner's report
— 2017 STEP 2, Question 9This year's paper was, perhaps, slightly more straightforward than usual, with more helpful guidance offered in some of the questions. Thus the mark required for a "1", a Distinction, was 80 (out of 120), around ten marks higher than that which would customarily be required to be awarded this grade. Nonetheless, a three‐figure mark is still a considerable achievement and, of the 1330 candidates sitting the paper, there were 89 who achieved this. At the other end of the scale, there were over 350 who scored 40 or below, including almost 150 who failed to exceed a total score of 25. As a general strategy for success in a STEP examination, candidates should be looking to find four "good" questions to work at (which may be chosen freely by the candidates from a total of 13 questions overall). It is unfortunately the case that so many low‐scoring candidates flit from one question to another, barely starting each one before moving on. There needs to be a willingness to persevere with a question until a measure of understanding as to the nature of the question's purpose and direction begins to emerge. Many low‐scoring candidates fail to deal with those parts of questions which cover routine mathematical processes ‐ processes that should be standard for an A‐level candidate. The significance of the "rule of four" is that four high‐scoring questions (15‐20 marks apiece) obtains you up to around the total of 70 that is usually required for a "1"; and with a couple of supporting starts to questions, such a total should not be beyond a good candidate who has prepared adequately. This year, significantly more than 10% of candidates failed to score at least half marks on any one question; and, given that Q1 (and often Q2 also) is (are) specifically set to give all candidates the opportunity to secure some marks, this indicates that these candidates are giving up too easily. Mathematics is about more than just getting to correct answers. It is about communicating clearly and precisely. Particularly with "show that" questions, candidates need to distinguish themselves from those who are just tracking back from given results. They should also be aware that convincing themselves is not sufficient, and if they are using a result from 3 pages earlier, they should make this clear in their working. A few specifics: In answers to mechanics questions, clarity of diagrams would have helped many students. If new variables or functions are introduced, it is important that students clearly define them. One area which is very important in STEP but which was very poorly done is dealing with inequalities. Although a wide range of approaches such as perturbation theory were attempted, at STEP level having a good understanding of the basics – such as changing the inequality if multiplying by a negative number – is more than enough. In fact, candidates who used more advanced methods rarely succeeded.
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Show LaTeX source
Problem source
Two identical rough cylinders of radius
$r$
and weight
$W$
rest,
not
touching each other
but a
negligible distance apart,
on a horizontal floor.
A thin flat rough plank of width $2a$, where
$a < r$,
and weight
$kW$
rests symmetrically and horizontally
on the cylinders, with its length parallel
to the axes of the cylinders and its faces horizontal.
A vertical cross-section is shown in the diagram below.
\begin{center}
\begin{tikzpicture}
% Define the radius of the circles
\def\R{2}
% Draw the horizontal base line
\draw (-1.5*\R, 0) -- (3.5*\R, 0);
% Draw the two mutually tangent circles resting on the line
% Center of the left circle is at (0, R)
\draw (0, \R) circle (\R);
% Center of the right circle is at (2R, R)
\draw (2*\R, \R) circle (\R);
% Draw the ultra-thick horizontal segment connecting the circles.
% Placed at 60 degrees from the centers to match the visual proportions perfectly.
\draw[line width=4pt] ({\R*cos(60)}, {\R + \R*sin(60)}) -- ({2*\R - \R*cos(60)}, {\R + \R*sin(60)});
\end{tikzpicture}
\end{center}
The coefficient of friction at all four
contacts is
$\frac12$.
The system is in equilibrium.
\begin{questionparts}
\item
Let $F$ be the frictional force between one cylinder and the floor,
and let $R$ be the normal reaction between the plank
and one cylinder.
Show that
\[
R\sin\theta = F(1+\cos\theta)\,,
\]
where $\theta$ is the acute angle between the plank and the tangent to
the cylinder at the point of contact.
Deduce that $2\sin\theta \le 1+\cos\theta\,$.
\item
Show that
\[
N=
\left( 1+\frac2 k\right)\left(\frac{1+\cos\theta}{\sin\theta} \right) F
\,,
\]
where $N$ is the normal reaction between the
floor and one cylinder.
Write down the condition
that the cylinder does
not slip on the floor and
show that it is satisfied with no extra restrictions on $\theta$.
\item
Show that $\sin\theta\le\frac45\,$ and hence that $r\le5a\,$.
\end{questionparts}Solution source
\begin{center}
\begin{tikzpicture}
% Define the radius of the circles
\def\R{2}
% Draw the horizontal base line
\draw (-1.5*\R, 0) -- (3.5*\R, 0);
\draw[-latex, ultra thick, blue] (\R, {\R + \R*sin(60)}) -- ++ (0, -1.5) node[below] {$kW$};
% Draw the two mutually tangent circles resting on the line
% Center of the left circle is at (0, R)
\draw (0, \R) circle (\R);
% Center of the right circle is at (2R, R)
\draw (2*\R, \R) circle (\R);
\draw[-latex, ultra thick, blue] (2*\R, 0) -- (2*\R, 0.4*\R) node[above, right] {$N$};
\draw[-latex, ultra thick, blue] (0, 0) -- (0, 0.4*\R) node[above, right] {$N$};
\draw[-latex, ultra thick, blue] (2*\R, \R) -- (2*\R, 0.6*\R) node[below, left] {$W$};
\draw[-latex, ultra thick, blue] (0, \R) -- (0, 0.6*\R) node[below, left] {$W$};
\draw[-latex, ultra thick, blue] ({\R*cos(60)}, {\R + \R*sin(60)}) -- ({\R*0.3*cos(60)}, {\R + \R*0.3*sin(60)}) node[below, left] {$R$};
\draw[-latex, ultra thick, blue] ({\R*cos(60)}, {\R + \R*sin(60)}) -- ({\R*cos(60)+\R*0.4*sin(60)}, {\R + \R*sin(60)-\R*0.4*cos(60)}) node[right, below] {$F$};
\draw[-latex, ultra thick, blue] ({2*\R - \R*cos(60)}, {\R + \R*sin(60)}) -- ({2*\R - \R*0.3*cos(60)}, {\R + \R*0.3*sin(60)}) node[below, left] {$R$};
\draw[-latex, ultra thick, blue] ({2*\R - \R*cos(60)}, {\R + \R*sin(60)}) -- ({2*\R - \R*cos(60)-\R*0.4*sin(60)}, {\R + \R*sin(60)-\R*0.4*cos(60)}) node[right, below] {$F$};
\draw[-latex, ultra thick, blue] (2*\R, 0) -- ({2*\R-0.5\R}, 0) node[left, below] {$F$};
\draw[-latex, ultra thick, blue] (0, 0) -- ({0.5*\R}, 0) node[right, below] {$F$};
% Draw the ultra-thick horizontal segment connecting the circles.
% Placed at 60 degrees from the centers to match the visual proportions perfectly.
\draw[line width=4pt] ({\R*cos(60)}, {\R + \R*sin(60)}) -- ({2*\R - \R*cos(60)}, {\R + \R*sin(60)});
\end{tikzpicture}
\end{center}
First, notice that by taking moments about the centre of one of the cylinders the two frictional forces must be equal to each other, say $F$.
\begin{questionparts}
\item $\,$
\begin{align*}
\text{N2}(\rightarrow, \text{ one cylinder}): && F\cos \theta + F - R \sin \theta &= 0 \\
\Rightarrow && F(1+\cos \theta) &= R \sin \theta \\
&& F \leq \tfrac12 R \\
\Rightarrow && R \sin \theta &\leq \frac12 R(1+\cos \theta) \\
\Rightarrow && 2 \sin \theta &\leq 1 + \cos \theta
\end{align*}
\item $\,$
\begin{align*}
\text{N2}(\uparrow, \text{system}): && 2N-(k+2)W &= 0 \\
\Rightarrow && W &= \left ( \frac{2}{k+2} \right)N \\
\text{N2}(\uparrow, \text{one cylinder}): && N - W - R\cos \theta -F\sin \theta &= 0 \\
\Rightarrow && N - \left ( \frac{2}{k+2} \right)N - F \left ( \frac{1+\cos \theta}{\sin \theta} \right) \cos \theta - F \sin \theta &= 0 \\
\Rightarrow && \left ( \frac{k}{k+2} \right)N &= \left ( \frac{\cos \theta + \cos^2 \theta + \sin^2 \theta}{\sin \theta} \right) F\\
\Rightarrow && N &= \left ( 1 + \frac2{k} \right) \left ( \frac{\cos \theta + 1}{\sin \theta} \right) F
\end{align*}
The cylinder does not slip if $F \leq \tfrac12 N$, ie
\begin{align*}
&& N &\leq \left ( 1 + \frac2{k} \right) \left ( \frac{\cos \theta + 1}{\sin \theta} \right) \frac12 N \\
\Rightarrow && 2\sin \theta &\leq \left ( 1 + \frac2{k} \right) \left ( \cos \theta + 1 \right)
\end{align*}
but since $2 \sin \theta \leq (1 + \cos \theta)$ and $(1+\frac2k) > 1$ this inequality is obviously satisfied.
\item We can notice that $2\sin \theta = 1 + \cos \theta$ is satisfied by a $3-4-5$ triangle, where $\sin \theta = 4/5, \cos \theta = 3/5$ and hence if $\sin \theta \leq \frac45$ the condition must hold.
\begin{center}
\begin{tikzpicture}
% \draw (0,0) -- (4, 0) -- ({4-2*sin(50)}, {2*sin(50)}) -- ({2*cos(50)}, {2*sin(50)}) -- cycle;
\draw (0,0) -- (4,0) node[pos=0.5, below] {$2r$};
\draw (0,0) -- ({2*cos(50)}, {2*sin(50)}) node[pos=0.5, left] {$r$};
\draw ({4-2*cos(50)}, {2*sin(50)}) -- ({2*cos(50)}, {2*sin(50)}) node[pos=0.5, above] {$2a$};
\draw ({4-2*cos(50)}, {2*sin(50)}) -- (4,0);
\draw[dashed] ({2*cos(50)}, {2*sin(50)}) -- ({2*cos(50)}, {0});
\draw (0,0) -- ({2*cos(50)}, {0}) node[pos=0.5, below] {$r-a$};
\end{tikzpicture}
\end{center}
So $\sin \theta = \frac{r-a}{r} \leq \frac45 \Rightarrow 5r-5a \leq 4r \Rightarrow r \leq 5a$
\end{questionparts}
This question was the least popular of the mechanics questions. Even amongst those who attempted this question only about a quarter made any meaningful attempt. As in so many mechanics questions a good, clearly labelled diagram often meant the difference between candidates making no progress and good attempts. It does seem that many students are reluctant to attempt these types of problems. It is hoped that looking at the hints and solutions should help candidates see that some resolving and taking of moments often leads to efficient solutions of problems like this.