2019 Paper 1 Q9

Year: 2019
Paper: 1
Question Number: 9

Course: UFM Mechanics
Section: Moments

Difficulty: 1500.0 Banger: 1500.0

Problem

A box has the shape of a uniform solid cuboid of height \(h\) and with a square base of side \(b\), where \(h > b\). It rests on rough horizontal ground. A light ladder has its foot on the ground and rests against one of the upper horizontal edges of the box, making an acute angle of \(\alpha\) with the ground, where \(h = b \tan \alpha\). The weight of the box is \(W\). There is no friction at the contact between ladder and box. A painter of weight \(kW\) climbs the ladder slowly. Neither the base of the ladder nor the box slips, but the box starts to topple when the painter reaches height \(\lambda h\) above the ground, where \(\lambda < 1\). Show that:
  1. \(R = k\lambda W \cos \alpha\), where \(R\) is the magnitude of the force exerted by the box on the ladder;
  2. \(2k\lambda \cos 2\alpha + 1 = 0\);
  3. \(\mu \geq \frac{\sin 2\alpha}{1 - 3 \cos 2\alpha}\), where \(\mu\) is the coefficient of friction between the box and the ground.

Solution

TikZ diagram
At the point we are about to topple, reaction and friction forces will be acting at \(C\)
  1. \(\,\) \begin{align*} \overset{\curvearrowright}{X}:&& kW \cdot \lambda h\cos \alpha - R h &= 0 \\ \Rightarrow && R &= k\lambda W \cos \alpha \\ \end{align*}
  2. \(\,\) \begin{align*} \overset{\curvearrowright}{C}:&& R \sin \alpha \cdot h-R\cos \alpha \cdot b-W\frac{b}{2} &= 0 \\ && k\lambda W \cos \alpha \sin \alpha \cdot b \tan \alpha- k\lambda W \cos \alpha\cos \alpha \cdot h-W\frac{b}{2} &= 0 \\ && k \lambda (\cos^2 \alpha - \sin^2 \alpha) +\frac12 &= 0 \\ \Rightarrow && 2k \lambda \cos 2\alpha + 1 &= 0 \end{align*}
  3. \(\,\) \begin{align*} \text{N2}(\uparrow): && R_b -W-R\cos \alpha &= 0 \\ \Rightarrow && R_b &= W + k\lambda W \cos^2 \alpha\\ \text{N2}(\rightarrow): && R\sin \alpha - F_b &= 0 \\ \Rightarrow && F_b &= R \sin \alpha \\ \\ && F_b &\leq \mu R \\ \Rightarrow && k\lambda W \cos \alpha \sin \alpha &= \mu (W + k\lambda W \cos^2 \alpha) \\ \Rightarrow && \mu &\geq \frac{k\lambda \cos \alpha \sin \alpha}{1 + k\lambda \cos^2 \alpha} \\ &&&= \frac{k\lambda \sin 2\alpha}{2 + 2k\lambda cos^2 \alpha} \\ &&&= \frac{k\lambda \sin 2\alpha}{2 + k\lambda (\cos 2 \alpha+1)} \\ &&&= \frac{k\lambda \sin 2\alpha}{-4k\lambda \cos 2 \alpha + k\lambda (\cos 2 \alpha+1)} \\ &&&= \frac{\sin 2 \alpha}{1 -3 \cos 2\alpha} \end{align*}
Examiner's report
— 2019 STEP 1, Question 9
Mean: 3.5 / 20 ~16% attempted (inferred) Inferred ~16% from 'fewer than 500 starts' out of ~3000 entry. Mean of 'almost 3½' stated in intro; two-thirds scored ≤2.

There were fewer than 500 starts at Q9, with almost two-thirds of these attempts earning no more than 2/20 overall. Very few candidates were able to make a substantial attempt at this question; most were stymied by not having any physical intuition regarding the direction and position of the reaction forces at the moment of toppling. Once this was done then the question became mainly angle chasing and taking moments about appropriate points but most candidates lacked the confidence to make much progress. Even the opening result in (i) – which required nothing more than taking moments (once) about a suitably chosen point – offered four marks for obtaining a given answer; this, however, proved too much for the majority of candidates. The real obstacle lay in the widespread reluctance among candidates to draw a good diagram, of a suitable size for labelling all necessary points, angles and forces (including their directions), and clearly labelled; a good diagram is the most important thing that a candidate can do … everything else follows so much more easily from a good diagram. In principle, all that is required to complete this question is to resolve twice, take moments (twice) and then apply the 'Law of Friction' in its F ≤ μR form and then sort out the resulting mix of algebra and trigonometry.

In order to get the fullest picture, this document should be read in conjunction with the question paper, the marking scheme and (for comments on the underlying purpose and motivation for finding the right solution-approaches to questions) the Hints and Solutions document; all of which are available from the STEP and Cambridge Examinations Board websites. The purpose of the STEPs is to learn what students are able to achieve mathematically when applying the knowledge, skills and techniques that they have learned within their standard A-level (or equivalent) courses … but seldom within the usual range of familiar settings. STEP questions require candidates to work at an extended piece of mathematics, often with the minimum of specific guidance, and to make the necessary connections. This requires a very different mind-set to that which is sufficient for success at A-level, and the requisite skills tend only to develop with prolonged and determined practice at such longer questions for several months beforehand. One of the most crucial features of the STEPs is that the routine technical and manipulative skills are almost taken for granted; it is necessary for candidates to produce them with both speed and accuracy so that the maximum amount of time can be spent in thinking their way through the problem and the various hurdles and obstacles that have been set before them. Most STEP questions begin by asking the solver to do something relatively routine or familiar before letting them loose on the real problem. Almost always, such an opening has not been put there to allow one to pick up a few easy marks, but rather to point the solver in the right direction for what follows. Very often, the opening result or technique will need to be used, adapted or extended in the later parts of the question, with the demands increasing the further on that one goes. So it is that a candidate should never think that they are simply required to 'go through the motions' but must expect, sooner or later, to be required to show either genuine skill or real insight in order to make a reasonably complete effort. The more successful candidates are the ones who manage to figure out how to move on from the given starting-point. Finally, reading through a finished solution is often misleading – even unhelpful – unless you have attempted the problem for yourself. This is because the thinking has been done for you. When you read through the report and look at the solutions (either in the mark-scheme or the Hints & Solutions booklet), try to figure out how you could have arrived at the solution, learn from your mistakes and pick up as many tips as you can whilst working through past paper questions. This year's paper produced the usual sorts of outcomes, with far too many candidates wasting valuable time by attempting more than six questions, and with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. Around one candidate in eight failed to hit the 30 mark overall, though this is an improvement on last year. Most candidates were able to produce good attempts at two or more questions. At the top end of the scale, around a hundred candidates scored 100 or more out of 120, with four hitting the maximum of 120 and many others not far behind. The paper is constructed so that question 1 is very approachable indeed, the intention being to get everyone started with some measure of success; unsurprisingly, Q1 was the most popular question of all, although under two-thirds of the entry attempted it this year, and it also turned out to be the most successful question on the paper with a mean score of about 12 out of 20. In order of popularity, Q1 was followed by Qs.3, 4 and 2. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the applied questions combined scoring fewer 'hits' than any one of the first four questions on its own. Though slightly more popular than the applied questions, the least successful question of all was Q5, on vectors. This question was attempted by almost 750 candidates, but 70% of these scored no more than 2 marks, leaving it with a mean score of just over 3 out of 20. Q9 (a statics question) was found only marginally more appetising, with a mean score of almost 3½ out of 20. In general, it was found that explanations were poorly supplied, with many candidates happy to overlook completely any requests for such details.

Source: Cambridge STEP 2019 Examiner's Report · 2019-p1.pdf
Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

Show LaTeX source
Problem source
A box has the shape of a uniform solid cuboid of height $h$ and with a square base of side $b$, where $h > b$. It rests on rough horizontal ground. A light ladder has its foot on the ground and rests against one of the upper horizontal edges of the box, making an acute angle of $\alpha$ with the ground, where $h = b \tan \alpha$. The weight of the box is $W$. There is no friction at the contact between ladder and box.
A painter of weight $kW$ climbs the ladder slowly. Neither the base of the ladder nor the box slips, but the box starts to topple when the painter reaches height $\lambda h$ above the ground, where $\lambda < 1$.
Show that:
\begin{questionparts}
\item $R = k\lambda W \cos \alpha$, where $R$ is the magnitude of the force exerted by the box on the ladder;
\item $2k\lambda \cos 2\alpha + 1 = 0$;
\item $\mu \geq \frac{\sin 2\alpha}{1 - 3 \cos 2\alpha}$, where $\mu$ is the coefficient of friction between the box and the ground.
\end{questionparts}
Solution source



\begin{center}
    \begin{tikzpicture}
        \def\a{3};
        \def\c{2};
        \def\b{-2};
        \coordinate (A) at (0,0);
        \coordinate (B) at (0,\a);
        \coordinate (C) at (\c,\a);
        \coordinate (D) at (\c,0);

        \coordinate (X) at (\b, 0);
        \coordinate (Y) at ($(X)!1.4!(B)$);
        \coordinate (Z) at ($(X)!.6!(B)$);

        \draw (A) -- (B) -- (C) -- (D);

        \draw[ultra thick] (X) -- (Y);

        \draw (A) -- (B) node[pos=0.5, left] {$h$};
        \draw (B) -- (C) node[pos=0.5, above] {$b$};

        \node[below right] at (D) {$C$};
        \node[below] at (X) {$X$};

        \draw[-latex, ultra thick, blue] ({\c/2},{\a/2}) -- ++(0, -1) node[below] {$W$};
        \draw[-latex, ultra thick, red] (Z) -- ++(0, -1) node[below] {$kW$};
        \draw[-latex, ultra thick, red] (X) -- ++(0, 1) node[left] {$R_l$};
        \draw[-latex, ultra thick, red] (X) -- ++(1, 0) node[below] {$F_l$};
        \draw[-latex, ultra thick, red] (B) -- ++({-1}, {1/abs(\a/\b)}) node[right] {$R$};
        \draw[-latex, ultra thick, blue] (B) -- ++({1}, {-1/abs(\a/\b)}) node[right] {$R$};
        \draw[-latex, ultra thick, blue] (D) -- ++(0, 1) node[right] {$R_b$};
        \draw[-latex, ultra thick, blue] (D) -- ++(-1, 0) node[below] {$F_b$};

        \draw ($(A)!-1!(D)$) -- ($(A)!2!(D)$);

        \filldraw (Z) circle (1.5pt);


        \pic [draw, angle radius=.8cm, angle eccentricity=.7, "$\alpha$"] {angle = A--X--B};
        
    \end{tikzpicture}
\end{center}

At the point we are about to topple, reaction and friction forces will be acting at $C$ 

\begin{questionparts}
\item $\,$
\begin{align*}
    \overset{\curvearrowright}{X}:&& kW \cdot \lambda h\cos \alpha - R h &= 0 \\
    \Rightarrow && R &= k\lambda W \cos \alpha \\
\end{align*}
\item $\,$
\begin{align*}
    \overset{\curvearrowright}{C}:&& R \sin \alpha \cdot h-R\cos \alpha \cdot  b-W\frac{b}{2} &= 0 \\
    &&  k\lambda W \cos \alpha \sin \alpha \cdot b \tan \alpha- k\lambda W \cos \alpha\cos \alpha \cdot  h-W\frac{b}{2} &= 0 \\
    && k \lambda (\cos^2 \alpha - \sin^2 \alpha) +\frac12 &= 0 \\
    \Rightarrow && 2k \lambda \cos 2\alpha + 1 &= 0
\end{align*}
\item $\,$
\begin{align*}
    \text{N2}(\uparrow): && R_b -W-R\cos \alpha &= 0 \\
    \Rightarrow && R_b &= W + k\lambda W \cos^2 \alpha\\
    \text{N2}(\rightarrow): && R\sin \alpha - F_b &= 0 \\
    \Rightarrow && F_b &= R \sin \alpha \\
    \\
    && F_b &\leq \mu R \\
    \Rightarrow && k\lambda W \cos \alpha \sin \alpha &= \mu (W + k\lambda W \cos^2 \alpha) \\
    \Rightarrow && \mu &\geq \frac{k\lambda \cos \alpha \sin \alpha}{1 + k\lambda \cos^2 \alpha} \\
    &&&= \frac{k\lambda \sin 2\alpha}{2 + 2k\lambda cos^2 \alpha} \\
    &&&= \frac{k\lambda \sin 2\alpha}{2 + k\lambda (\cos 2 \alpha+1)} \\
    &&&= \frac{k\lambda \sin 2\alpha}{-4k\lambda \cos 2 \alpha + k\lambda (\cos 2 \alpha+1)} \\
    &&&= \frac{\sin 2 \alpha}{1 -3 \cos 2\alpha}
\end{align*}

\end{questionparts}