2010 Paper 1 Q9
Year: 2010
Paper: 1
Question Number: 9
Course: UFM Mechanics
Section: Centre of Mass 1
Problem
Solution
Examiner's report
— 2010 STEP 1, Question 9There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Show LaTeX source
Problem source
\begin{center}
\begin{tikzpicture}
\begin{scope}[rotate=60]
\coordinate (O) at (0,0);
\coordinate (A) at (2,0);
\coordinate (B) at (2,6);
\coordinate (C) at (0,6);
\end{scope}
\coordinate (X) at (-1,0);
\draw (O) -- (A) -- (B) -- (C) -- cycle;
\draw (-7, 0) -- (0.5,0);
\draw ({6*cos(150)},0) -- ({6*cos(150)},5);
\node[right] at ($(O)!0.5!(A)$) {$2b$};
\node[below] at ($(O)!0.75!(C)$) {$2a$};
\pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\alpha$"] {angle = C--O--X};
\end{tikzpicture}
\end{center}
The diagram shows a uniform rectangular lamina with sides of lengths $2a$ and $2b$ leaning against a rough vertical wall, with one corner resting on a rough horizontal plane. The plane of the lamina is vertical and perpendicular to the wall, and one edge makes an angle of $\alpha$ with the horizontal plane. Show that the centre of mass of the lamina is a distance $a\cos\alpha + b\sin\alpha$ from the wall.
The coefficients of friction at the two points of contact are each $\mu$ and the friction is limiting at both contacts. Show that
\[
a\cos(2\lambda +\alpha) = b\sin\alpha \,,
\]
where $\tan\lambda = \mu$.
Show also that if the lamina is square, then $\lambda = \frac{1}{4}\pi -\alpha$.Solution source
\begin{center}
\begin{tikzpicture}
\begin{scope}[rotate=60]
\coordinate (O) at (0,0);
\coordinate (A) at (2,0);
\coordinate (B) at (2,6);
\coordinate (C) at (0,6);
\coordinate (G) at (1,3);
\coordinate (Gx) at (0,3);
\end{scope}
\coordinate (X) at (-1,0);
\draw (O) -- (A) -- (B) -- (C) -- cycle;
\draw (-7, 0) -- (0.5,0);
\draw ({6*cos(150)},0) -- ({6*cos(150)},5);
\node[right] at ($(O)!0.5!(A)$) {$2b$};
\node[below] at ($(O)!0.75!(C)$) {$2a$};
\filldraw (G) circle (1pt) node [right] {$G$};
\filldraw (Gx) circle (1pt) node[below] {$X$};
\draw[dashed] (Gx) -- (G);
\draw[dashed] ({6*cos(150)},{3*sin(150)}) -- (Gx);
\pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\alpha$"] {angle = C--O--X};
\end{tikzpicture}
\end{center}
The horizontal distance to $X$ is $a\cos \alpha$.
The horizontal distance to $G$ from $X$ is $b \sin \alpha$, therefore the centre of mass is a distance $a \cos \alpha + b \sin \alpha$ from the wall.
\begin{center}
\begin{tikzpicture}
\begin{scope}[rotate=60]
\coordinate (O) at (0,0);
\coordinate (A) at (2,0);
\coordinate (B) at (2,6);
\coordinate (C) at (0,6);
\coordinate (G) at (1,3);
\coordinate (Gx) at (0,3);
\end{scope}
\coordinate (X) at (-1,0);
\draw (O) -- (A) -- (B) -- (C) -- cycle;
\draw (-7, 0) -- (0.5,0);
\draw ({6*cos(150)},0) -- ({6*cos(150)},5);
\node[right] at ($(O)!0.5!(A)$) {$2b$};
\node[below] at ($(O)!0.75!(C)$) {$2a$};
\filldraw (G) circle (1pt) node [right] {$G$};
\filldraw (Gx) circle (1pt) node[below] {$X$};
\draw[dashed] (Gx) -- (G);
\draw[dashed] ({6*cos(150)},{3*sin(150)}) -- (Gx);
\node[left] at (C) {$Y$};
\pic [draw, angle radius=1cm, angle eccentricity=1.25, "$\alpha$"] {angle = C--O--X};
\draw[-latex, ultra thick, blue] (C) -- ++(1,0) node[right] {$R_W$};
\draw[-latex, ultra thick, blue] (C) -- ++(0,1) node[above] {$F_W$};
\draw[-latex, ultra thick, blue] (O) -- ++(-1,0) node[left] {$F_G$};
\draw[-latex, ultra thick, blue] (O) -- ++(0,1) node[above] {$R_G$};
\draw[-latex, ultra thick, blue] (G) -- ++(0,-1) node[below] {$W$};
\end{tikzpicture}
\end{center}
\begin{align*}
\text{lim eq}: && F_W &= \mu R_W \\
&& F_G &= \mu R_G\\
\text{N2}(\rightarrow): && \mu R _G &= R_W \\
\text{N2}(\uparrow): && \mu R_W + R_G &= W \\
\Rightarrow && (1+\mu^2)R_G &= W \\
\overset{\curvearrowleft}{Y}: && R_G 2a \cos \alpha - F_G 2a \sin \alpha - W (a \cos \alpha + b \sin \alpha) &= 0 \\
\Leftrightarrow && 2a R_G \cos \alpha -2a \mu R_G \sin \alpha - (1+\mu^2)R_G(a \cos \alpha + b \sin \alpha) &= 0 \\
\Leftrightarrow && a(1-\mu^2)\cos \alpha - (b(1+\mu^2)+2a\mu) \sin \alpha &= 0 \\
\Leftrightarrow && a(1-\tan^2 \lambda )\cos \alpha - (b(1-\tan^2 \lambda)+2a\tan \lambda) \sin \alpha &= 0 \\
\Leftrightarrow&& a(2-\sec^2 \lambda) \cos \alpha - (b\sec^2 \lambda+2a\mu) \sin \alpha &= 0 \\
\Leftrightarrow && a (2\cos \lambda - 1)\cos \alpha - 2a \sin \lambda \cos \lambda \sin \alpha &= b \sin \alpha \\
\Leftrightarrow && a\cos 2 \lambda \cos \alpha - a\sin 2 \lambda \sin \alpha &= b \sin \alpha \\
\Leftrightarrow && a\cos (2 \lambda +\alpha) &= b \sin \alpha
\end{align*}
as required.
If the lamina is a square, $a = b$, so
\begin{align*}
&& \cos(2\lambda + \alpha) &= \sin \alpha \\
\Rightarrow && 0 &= \cos(2\lambda + \alpha) -\sin \alpha \\
&&&= \sin \left (\frac{\pi}{2} - 2 \lambda - \alpha \right )-\sin \alpha \\
&&&= 2 \cos\left ( \frac{\frac{\pi}{2} - 2 \lambda - \alpha +\alpha}{2} \right) \sin\left ( \frac{\frac{\pi}{2} - 2 \lambda - \alpha -\alpha}{2} \right) \\
&&&= 2 \cos\left ( \frac{\pi}4 -\lambda\right) \sin\left ( \frac{\pi}4 -\lambda-\alpha \right) \\
\Rightarrow && \lambda -\frac{\pi}{4} = -\frac{\pi}{2} & \text{ or } \frac{\pi}{4} - \lambda - \alpha = 0 \\
\Rightarrow && \alpha &= \frac{\pi}{4}-\lambda
\end{align*}
This was the most popular of the mechanics questions, being attempted by about one-third of candidates (though this question counted towards the final mark of only two-thirds of these). The mean mark was about 6/20. A number of the attempts struggled to calculate the distance of the centre of mass from the wall, though most were able to do so using a quick sketch. Unfortunately, many candidates gave up at this point, unsure of what to do next. It should be second-nature that for a large-body question, the "right" thing to do is "resolve twice, moments once". Others tried this but failed to do so correctly: first of all, many drew poor or confusing diagrams; it is vital that candidates draw diagrams which are clear enough to understand what is going on at every point. It was sadly common to see friction labelled as Fr, where this could easily be confused with F × r in some contexts: students should always be taught to use single-letter variable names. Further, many students were inconsistent with their force labelling: some labelled the reaction as RA at A but the friction as RB at B (where A and B are the points of contact) or similar gaffes—this, of course, led to confusion and errors later. Others used the same variable for two different reaction forces or two different friction forces. Yet others left out one of the forces. Even with the hurdle of an accurate diagram overcome, many only resolved once rather than twice, and a few tried taking moments around two or three different points but never resolving. (The latter can be made to work, but is usually far more effort than necessary.) Taking moments was also found to be challenging: most attempts failed by forgetting a force or by not understanding the meaning of "perpendicular distance". Some candidates also got their signs wrong. A surprising number switched the a and b at some point in the question. Most of those who correctly reached this point were able to make good progress towards the required conclusions (though a few, sadly, did not attempt the final part of the question, even though it was fairly straightforward). They showed a good command of the trigonometric identities required and were confident in manipulating the equations to eliminate the forces. Unsurprisingly, there were almost no candidates who used the Three Forces Theorem approach.