| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Lamina hinged at point with string support |
| Difficulty | Standard +0.3 This is a standard moments equilibrium problem requiring students to take moments about the hinge, resolve forces, and find the resultant. The geometry is straightforward (given angles and dimensions), and the method is routine for M2 students: moments about A to find tension, then resolve horizontally and vertically to find reaction components. Slightly easier than average due to clear setup and standard technique. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \((A_1 + A_2)\bar{x} = A_1\bar{x}_1 + A_2\bar{x}_2\) \([0.09\bar{x} = 0.05 \times 0.25 + 0.04 \times 0.45]\) | M1 | |
| \(\bar{x} = 0.0305/0.09\) (or \(\bar{x}^2)\) from B) \(( = 61/180 \text{ or } 0.3388889)\) | A1 | |
| Alternatively for the above 2 marks: Splits the lamina into 2 rectangles of weights 5 N and 4 N or considers it as a square of weight 25 N from which a square of weight 16 N is removed | M1 | |
| Obtains moment distances as 0.25 and 0.45 or 0.2 and 0.45 (2 rectangles cases) or 0.25 and 0.2 (2 squares case) (distances may be implied) | A1 | |
| Takes moments about O to obtain an equation for \(T\) \(9 \times 0.0305/0.09 = 0.57\sin 30°\) or \(5 \times 0.25 + 4 \times 0.45 = 0.57 \sin 30°\) or \(4 \times 0.2 + 5 \times 0.45 = 0.57\sin 30°\) or \(25 \times 0.25 - 16 \times 0.2 = 0.57\sin 30°\) | M1 | |
| Obtains tension as 12.2 N | A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains vertical component of force at A as 2.9 N (ft for 9 - \(\frac{1}{2}T\)) | B1 ft | |
| Obtains horizontal component of force at A as \(6.1\sqrt{3}\) N \(( = 10.5655)\) (ft for \(\frac{1}{2}T\sqrt{3}\)) | B1 ft | |
| Uses \(F^2 = H^2 + V^2\) | M1 | |
| Obtains magnitude as 11.0 N (10.95628) | A1 | 4 marks |
**(i)**
| Uses $(A_1 + A_2)\bar{x} = A_1\bar{x}_1 + A_2\bar{x}_2$ $[0.09\bar{x} = 0.05 \times 0.25 + 0.04 \times 0.45]$ | M1 |
| $\bar{x} = 0.0305/0.09$ (or $\bar{x}^2)$ from B) $( = 61/180 \text{ or } 0.3388889)$ | A1 |
| Alternatively for the above 2 marks: Splits the lamina into 2 rectangles of weights 5 N and 4 N or considers it as a square of weight 25 N from which a square of weight 16 N is removed | M1 |
| Obtains moment distances as 0.25 and 0.45 or 0.2 and 0.45 (2 rectangles cases) or 0.25 and 0.2 (2 squares case) (distances may be implied) | A1 |
| Takes moments about O to obtain an equation for $T$ $9 \times 0.0305/0.09 = 0.57\sin 30°$ or $5 \times 0.25 + 4 \times 0.45 = 0.57 \sin 30°$ or $4 \times 0.2 + 5 \times 0.45 = 0.57\sin 30°$ or $25 \times 0.25 - 16 \times 0.2 = 0.57\sin 30°$ | M1 |
| Obtains tension as 12.2 N | A1 | 5 marks |
**(ii)**
| Obtains vertical component of force at A as 2.9 N (ft for 9 - $\frac{1}{2}T$) | B1 ft |
| Obtains horizontal component of force at A as $6.1\sqrt{3}$ N $( = 10.5655)$ (ft for $\frac{1}{2}T\sqrt{3}$) | B1 ft |
| Uses $F^2 = H^2 + V^2$ | M1 |
| Obtains magnitude as 11.0 N (10.95628) | A1 | 4 marks |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{3e7472a8-df1e-45c4-81fb-e4397bddf5ad-3_590_754_1425_699}
A uniform lamina of weight 9 N has dimensions as shown in the diagram. The lamina is freely hinged to a fixed point at $A$. A light inextensible string has one end attached to $B$, and the other end attached to a fixed point $C$, which is in the same vertical plane as the lamina. The lamina is in equilibrium with $A B$ horizontal and angle $A B C = 150 ^ { \circ }$.\\
(i) Show that the tension in the string is 12.2 N .\\
(ii) Find the magnitude of the force acting on the lamina at $A$.
\hfill \mbox{\textit{CAIE M2 2002 Q5 [9]}}