CAIE M2 2014 November — Question 4 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2014
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeResultant force on lamina
DifficultyStandard +0.8 This is a composite body centre of mass problem requiring decomposition into rectangle and triangle, calculation of individual centres of mass, and then finding limiting equilibrium conditions (toppling about two different edges). The multi-step nature, spatial reasoning about toppling conditions, and careful moment calculations make this moderately challenging but still within standard M2 scope.
Spec3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass

\includegraphics{figure_4} \(ABCDEF\) is the cross-section through the centre of mass of a uniform solid prism. \(ABCF\) is a rectangle in which \(AB = CF = 1.6\) m, and \(BC = AF = 0.4\) m. \(CDE\) is a triangle in which \(CD = 1.8\) m, \(CE = 0.4\) m, and angle \(DCE = 90°\). The prism stands on a rough horizontal surface. A horizontal force of magnitude \(T\) N acts at \(B\) in the direction \(CB\) (see diagram). The prism is in equilibrium.
  1. Show that the distance of the centre of mass of the prism from \(AB\) is \(0.488\) m. [4]
  2. Given that the weight of the prism is \(100\) N, find the greatest and least possible values of \(T\). [3]

Question 4:

(ii) ---
4 (i)
AnswerMarks
(ii)ABCF area = 0.64 and CDE = 0.36
0.4 1.8
(0.64 + 0.36)d = 0.64× + 0.36×(0.4 + )
2 3
d = 0.488 m AG
0.488 × 100 = 1.6T
T = 30.5 N
(0.488 – 0.4) × 100 = 1.6T
AnswerMarks
T = 5.5B1
M1
A1
A1
[4]
M1
A1
AnswerMarks
A1 [3]Both areas correct
Table of moments idea
All terms correct
Either limiting case
(no turning about A)
(no turning about F)
AnswerMarks Guidance
Page 5Mark Scheme Syllabus
Cambridge International A Level – October/November 20149709 52
Question 4:
--- 4 (i)
(ii) ---
4 (i)
(ii) | ABCF area = 0.64 and CDE = 0.36
0.4 1.8
(0.64 + 0.36)d = 0.64× + 0.36×(0.4 + )
2 3
d = 0.488 m AG
0.488 × 100 = 1.6T
T = 30.5 N
(0.488 – 0.4) × 100 = 1.6T
T = 5.5 | B1
M1
A1
A1
[4]
M1
A1
A1 [3] | Both areas correct
Table of moments idea
All terms correct
Either limiting case
(no turning about A)
(no turning about F)
Page 5 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9709 | 52
\includegraphics{figure_4}

$ABCDEF$ is the cross-section through the centre of mass of a uniform solid prism. $ABCF$ is a rectangle in which $AB = CF = 1.6$ m, and $BC = AF = 0.4$ m. $CDE$ is a triangle in which $CD = 1.8$ m, $CE = 0.4$ m, and angle $DCE = 90°$. The prism stands on a rough horizontal surface. A horizontal force of magnitude $T$ N acts at $B$ in the direction $CB$ (see diagram). The prism is in equilibrium.

\begin{enumerate}[label=(\roman*)]
\item Show that the distance of the centre of mass of the prism from $AB$ is $0.488$ m. [4]
\item Given that the weight of the prism is $100$ N, find the greatest and least possible values of $T$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2014 Q4 [7]}}