CAIE M2 2018 November — Question 6 8 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2018
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeResultant force on lamina
DifficultyStandard +0.3 This is a standard composite body centre of mass problem followed by a routine toppling equilibrium calculation. Part (i) requires subtracting areas and moments (standard technique), while part (ii) involves taking moments about point D with the prism on the point of toppling (a common mechanics scenario). The calculations are straightforward with no novel insight required, making it slightly easier than average.
Spec6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_6} Fig. 1 shows the cross-section \(ABCDE\) through the centre of mass \(G\) of a uniform prism. The cross-section consists of a rectangle \(ABCF\) from which a triangle \(DEF\) has been removed; \(AB = 0.6\text{ m}\), \(BC = 0.7\text{ m}\) and \(DF = EF = 0.3\text{ m}\).
  1. Show that the distance of \(G\) from \(BC\) is \(0.276\text{ m}\), and find the distance of \(G\) from \(AB\). [5]
  2. The prism is placed with \(CD\) on a rough horizontal surface. A force of magnitude \(2\text{ N}\) acting in the plane of the cross-section is applied to the prism. The line of action of the force passes through \(G\) and is perpendicular to \(DE\) (see Fig. 2). The prism is on the point of toppling about the edge through \(D\). Calculate the weight of the prism. [3]

Question 6:

AnswerMarks Guidance
6(i)Area of cross-section of prism = 0.5×0.6 – 0.3×0.3/2 = 0.375 m2 B1
0.375y = 0.42×0.6/2 –0.045(0.6 – 0.3/3)M1 Take moments about BC
y = 0.276 m AGA1
0.375x = 0.42×0.7/2 – 0.045(0.7 – 0.3/3)M1 Take moments about AB
x = 0.32 mA1
5

AnswerMarks Guidance
6(ii)M1 Attempt to take moments about D
2cos45× (0.7 – 0.32) = 2cos45× (0.3 – 0.276) + W(0.3 – 0.276)A1
W = 21(.0) NA1
3
AnswerMarks Guidance
QuestionAnswer Marks
Question 6:
--- 6(i) ---
6(i) | Area of cross-section of prism = 0.5×0.6 – 0.3×0.3/2 = 0.375 m2 | B1 | Area of cross-section of prism = area of rectangle – area of triangle
0.375y = 0.42×0.6/2 –0.045(0.6 – 0.3/3) | M1 | Take moments about BC
y = 0.276 m AG | A1
0.375x = 0.42×0.7/2 – 0.045(0.7 – 0.3/3) | M1 | Take moments about AB
x = 0.32 m | A1
5
--- 6(ii) ---
6(ii) | M1 | Attempt to take moments about D
2cos45× (0.7 – 0.32) = 2cos45× (0.3 – 0.276) + W(0.3 – 0.276) | A1
W = 21(.0) N | A1
3
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

Fig. 1 shows the cross-section $ABCDE$ through the centre of mass $G$ of a uniform prism. The cross-section consists of a rectangle $ABCF$ from which a triangle $DEF$ has been removed; $AB = 0.6\text{ m}$, $BC = 0.7\text{ m}$ and $DF = EF = 0.3\text{ m}$.

\begin{enumerate}[label=(\roman*)]
\item Show that the distance of $G$ from $BC$ is $0.276\text{ m}$, and find the distance of $G$ from $AB$. [5]

\item The prism is placed with $CD$ on a rough horizontal surface. A force of magnitude $2\text{ N}$ acting in the plane of the cross-section is applied to the prism. The line of action of the force passes through $G$ and is perpendicular to $DE$ (see Fig. 2). The prism is on the point of toppling about the edge through $D$.
Calculate the weight of the prism. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2018 Q6 [8]}}