Year: 2024
Paper: 3
Question Number: 10
Course: UFM Mechanics
Section: Moments
No solution available for this problem.
The total entry was an increase on that of 2023 by more than 10%. One question was attempted by more than 98% of candidates, another two by about 80%, and another five by between 50% and 70%. The remaining four questions were attempted by between 5% and 30% of candidates, these being from Section B: Mechanics, and Section C: Probability and Statistics, though the Statistics questions were in general attempted more often and more successfully. All questions were perfectly solved by some candidates. About 84% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A cube of uniform density $\rho$ is placed on a horizontal plane and a second cube, also of uniform density $\rho$, is placed on top of it. The lower cube has side length $1$ and the upper cube has side length $a$, with $a \leqslant 1$. The centre of mass of the upper cube is vertically above the centre of mass of the lower cube and all the edges of the upper cube are parallel to the corresponding edges of the lower cube. The contacts between the two cubes, and between the lower cube and the plane, are rough, with the same coefficient of friction $\mu < 1$ in each case. The midpoint of the base of the upper cube is $X$ and the midpoint of the base of the lower cube is $Y$.
A horizontal force $P$ is exerted, perpendicular to one of the vertical faces of the upper cube, at a point halfway between the two vertical edges of this face, and a distance $h$, with $h < a$, above the lower edge of this face.
\begin{questionparts}
\item Show that, if the two cubes remain in equilibrium, the normal reaction of the plane on the lower cube acts at a point which is a distance
\[\frac{P(1+h)}{(1+a^3)\rho g}\]
from $Y$, and find a similar expression for the distance from $X$ of the point at which the normal reaction of the lower cube on the upper cube acts.
\end{questionparts}
The force $P$ is now gradually increased from zero.
\begin{questionparts}
\setcounter{enumi}{1}
\item Show that, if neither cube topples, equilibrium will be broken by the slipping of the upper cube on the lower cube, and not by the slipping of the lower cube on the ground.
\item Show that, if $a = 1$, then equilibrium will be broken by the slipping of the upper cube on the lower cube if $\mu(1+h) < 1$ and by the toppling of the lower and upper cube together if $\mu(1+h) > 1$.
\item Show that, in a situation where $a < 1$ and $h\bigl(1 + a^3(1-a)\bigr) > a^4$, and no slipping occurs, equilibrium will be broken by the toppling of the upper cube.
\item Show, by considering $a = \frac{1}{2}$ and choosing suitable values of $h$ and $\mu$, that the situation described in \textbf{(iv)} can in fact occur.
\end{questionparts}
This was the least popular question on the paper by some way, being attempted by fewer than 6% of the candidates. It was attempted only a little more successfully than question 9 scoring a mean of about 5.5/20. Some of the few attempts were little more than a poor diagram and nothing further. If it was setup correctly, the candidates did fairly well, despite losing marks for not drawing everything required on the diagram, though there was some leniency about drawing equal and opposite forces (e.g. the reaction force from the top cube down onto the bottom cube). It should be stressed that very few did this so it could be a point of focus when preparing candidates for STEP mechanics. The only common error found once the first part was complete was mostly to do with reading carefully. In part (iii), most did not check that the upper cube could not topple without the lower toppling first, they just compared toppling of bottom cube and slipping. The main challenge in this sort of question is in the initial setup, after which the techniques required are not particularly difficult. Candidates who were able to interpret the context and setup the situation usually did very well.