| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina in equilibrium with applied force |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass question requiring (i) standard centroid calculation for a trapezium, (ii) basic trigonometry with the lamina hanging freely, and (iii) resolving forces and moments with an applied horizontal force. Part (iii) involves solving a quadratic equation arising from two equilibrium conditions, which elevates it slightly above routine but remains a standard M2 exercise with well-defined steps. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Rectangle: Area \(= 1.2 \times 1.8 = 2.16\), \(y = \frac{1.8}{2} = 0.9\) | B1 | |
| Triangle(s): Area \(= 1.2 \times \frac{1.8}{2} = 1.08\), \(y = \frac{1.8}{3} = 0.6\) | B1 | |
| \((2.16 + 1.08)Y = 2.16 \times 0.9 + 1.08 \times 0.6\) | M1 | Take moments about \(AD\) |
| \(Y = 0.8\) m | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AG\sin30 = 0.8\) | M1 | Use trigonometry of a right angled triangle |
| \(AG = 1.6\) m | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AD\) makes an angle of \(40°\) or \(20°\) with the vertical | B1 | |
| \(W \times AG\sin10 = 7 \times 2.4\cos40\) | M1 | Take moments about \(A\) |
| \(W = 46.3\) N | A1 | |
| \(W \times AG\sin10 = 7 \times 2.4\cos20\) | M1 | Take moments about \(A\) |
| \(W = 56.8\) N | A1 | |
| Total: 5 |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Rectangle: Area $= 1.2 \times 1.8 = 2.16$, $y = \frac{1.8}{2} = 0.9$ | B1 | |
| Triangle(s): Area $= 1.2 \times \frac{1.8}{2} = 1.08$, $y = \frac{1.8}{3} = 0.6$ | B1 | |
| $(2.16 + 1.08)Y = 2.16 \times 0.9 + 1.08 \times 0.6$ | M1 | Take moments about $AD$ |
| $Y = 0.8$ m | A1 | AG |
| **Total: 4** | | |
---
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $AG\sin30 = 0.8$ | M1 | Use trigonometry of a right angled triangle |
| $AG = 1.6$ m | A1 | |
| **Total: 2** | | |
---
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $AD$ makes an angle of $40°$ or $20°$ with the vertical | B1 | |
| $W \times AG\sin10 = 7 \times 2.4\cos40$ | M1 | Take moments about $A$ |
| $W = 46.3$ N | A1 | |
| $W \times AG\sin10 = 7 \times 2.4\cos20$ | M1 | Take moments about $A$ |
| $W = 56.8$ N | A1 | |
| **Total: 5** | | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{9daebcbe-826e-4eda-afa7-c935c6ea2bfc-10_451_574_258_781}\\
$A B C D$ is a uniform lamina in the shape of a trapezium which has centre of mass $G$. The sides $A D$ and $B C$ are parallel and 1.8 m apart, with $A D = 2.4 \mathrm {~m}$ and $B C = 1.2 \mathrm {~m}$ (see diagram).\\
(i) Show that the distance of $G$ from $A D$ is 0.8 m .\\
The lamina is freely suspended at $A$ and hangs in equilibrium with $A D$ making an angle of $30 ^ { \circ }$ with the vertical.\\
(ii) Calculate the distance $A G$.\\
With the lamina still freely suspended at $A$ a horizontal force of magnitude 7 N acting in the plane of the lamina is applied at $D$. The lamina is in equilibrium with $A G$ making an angle of $10 ^ { \circ }$ with the downward vertical.\\
(iii) Find the two possible values for the weight of the lamina.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M2 2019 Q7 [11]}}