CAIE M2 2009 November — Question 5 8 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2009
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeLamina hinged at point with string support
DifficultyStandard +0.3 This is a standard moments problem requiring knowledge of the center of mass of a circular sector (given in formula booklet), taking moments about the hinge, and resolving forces. It involves multiple steps but uses routine mechanics techniques without requiring novel insight or complex problem-solving.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model6.04d Integration: for centre of mass of laminas/solids

5 \includegraphics[max width=\textwidth, alt={}, center]{68acf474-5da2-4949-b3b2-fc42cd73bd4a-3_405_545_630_799} A uniform lamina \(A O B\) is in the shape of a sector of a circle with centre \(O\) and radius 0.5 m , and has angle \(A O B = \frac { 1 } { 3 } \pi\) radians and weight 3 N . The lamina is freely hinged at \(O\) to a fixed point and is held in equilibrium with \(A O\) vertical by a force of magnitude \(F \mathrm {~N}\) acting at \(B\). The direction of this force is at right angles to \(O B\) (see diagram). Find
  1. the value of \(F\),
  2. the magnitude of the force acting on the lamina at \(O\).

Question 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
(i) \(OG = 2 \times 0.5\sin30° / (3 \times (\pi/6))\) \((= 1/\pi)\)B1
M1For taking moments about \(O\)
\(3 \times (\sin30°/\pi) = F \times 0.5\)A1\(\sqrt{}\)
\(F = 0.955\)A1 [4]
(ii)M1 For resolving forces on the lamina horizontally and vertically
\(X = F\cos60°\) \((= 0.477)\)A1
\(Y = 3 - F\sin60°\) \((= 2.17)\)A1
Magnitude is \(2.22\) NA1\(\sqrt{}\) ft \((F^2 - 3\sqrt{3}\,F + 9)^{\frac{1}{2}}\) [4]
## Question 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $OG = 2 \times 0.5\sin30° / (3 \times (\pi/6))$ $(= 1/\pi)$ | B1 | |
| | M1 | For taking moments about $O$ |
| $3 \times (\sin30°/\pi) = F \times 0.5$ | A1$\sqrt{}$ | |
| $F = 0.955$ | A1 | **[4]** |
| **(ii)** | M1 | For resolving forces on the lamina horizontally and vertically |
| $X = F\cos60°$ $(= 0.477)$ | A1 | |
| $Y = 3 - F\sin60°$ $(= 2.17)$ | A1 | |
| Magnitude is $2.22$ N | A1$\sqrt{}$ | ft $(F^2 - 3\sqrt{3}\,F + 9)^{\frac{1}{2}}$ **[4]** |

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5\\
\includegraphics[max width=\textwidth, alt={}, center]{68acf474-5da2-4949-b3b2-fc42cd73bd4a-3_405_545_630_799}

A uniform lamina $A O B$ is in the shape of a sector of a circle with centre $O$ and radius 0.5 m , and has angle $A O B = \frac { 1 } { 3 } \pi$ radians and weight 3 N . The lamina is freely hinged at $O$ to a fixed point and is held in equilibrium with $A O$ vertical by a force of magnitude $F \mathrm {~N}$ acting at $B$. The direction of this force is at right angles to $O B$ (see diagram). Find\\
(i) the value of $F$,\\
(ii) the magnitude of the force acting on the lamina at $O$.

\hfill \mbox{\textit{CAIE M2 2009 Q5 [8]}}