Year: 2020
Paper: 3
Question Number: 9
Course: UFM Mechanics
Section: Moments
No solution available for this problem.
In spite of the change to criteria for entering the paper, there was still a very healthy number of candidates, and the vast majority handled the protocols for the online testing very well. Just over half the candidates attempted exactly six questions, and whilst about 10% attempted a seventh, hardly any did more than seven. With 20% attempting five questions, and 10% attempting only four, overall, there were very few candidates not attempting the target number. There was a spread of popularity across the questions, with no question attracting more than 90% of candidates and only one less than 10%, but every question received a good number of attempts. Likewise, there was a spread of success on the questions, though every question attracted at least one perfect solution.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Two inclined planes $\Pi_1$ and $\Pi_2$ meet in a horizontal line at the lowest points of both planes and lie on either side of this line. $\Pi_1$ and $\Pi_2$ make angles of $\alpha$ and $\beta$, respectively, to the horizontal, where $0 < \beta < \alpha < \frac{1}{2}\pi$.
A uniform rigid rod $PQ$ of mass $m$ rests with $P$ lying on $\Pi_1$ and $Q$ lying on $\Pi_2$ so that the rod lies in a vertical plane perpendicular to $\Pi_1$ and $\Pi_2$ with $P$ higher than $Q$.
\begin{questionparts}
\item It is given that both planes are smooth and that the rod makes an angle $\theta$ with the horizontal. Show that $2\tan\theta = \cot\beta - \cot\alpha$.
\item It is given instead that $\Pi_1$ is smooth, that $\Pi_2$ is rough with coefficient of friction $\mu$ and that the rod makes an angle $\phi$ with the horizontal. Given that the rod is in limiting equilibrium, with $P$ about to slip down the plane $\Pi_1$, show that
\[ \tan\theta - \tan\phi = \frac{\mu}{(\mu + \tan\beta)\sin 2\beta} \]
where $\theta$ is the angle satisfying $2\tan\theta = \cot\beta - \cot\alpha$.
\end{questionparts}
Just over 10% of the candidates attempted this question, with most understanding the setup and writing down some resolve and moment equations. A few candidates misunderstood the setup, specifically the "(planes) meet in a horizontal line at the lowest points of both planes and lie on either side of this line" and sketched a 'tilted wedge'. There was a moderate degree of success with the mean score being just short of half marks. Part (i) was generally done well. Candidates who were unable to progress usually forgot about the moment equation. The resolve equations were done in various ways, with the most popular being horizontal-vertical and perp-parallel to the rod. All kinds of combinations of resolve and moment equations were used. With the horizontal resolve equation and moments about the centre of mass one only needs two equations to do this part. Most candidates were able to do the required algebra, and the failure to reach the answer usually stemmed from incorrect trigonometry in the equilibrium equations. Most candidates who succeeded in part (i) then proceeded to do part (ii). Most candidates were successful at incorporating the friction and writing down the new equations. At this point trig errors were common, and people who were resolving perp-parallel to the rod made more errors. Many candidates were put off by the difficult algebra that was about to follow. Of those who persisted, a good number arrived at the final answer, with some submitting many pages of attempts to do the algebra. The most common mistake was failing to eliminate α systematically.