| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | January |
| Marks | 16 |
| 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 M2 moments question involving finding centre of mass of a composite body (lamina plus particles), then applying equilibrium conditions. Part (a) is routine calculation, parts (b-c) are standard hanging equilibrium, parts (d-e) involve resolving forces and taking moments with a horizontal force applied. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (a) AD: \(10m\overline{x} = 3m \times \frac{5a}{2} + 3m \times 5a\) | M1 A1 | |
| \(\overline{x} = 2.25a\) * | A1 | (3) |
| (b) AB: \(10m\overline{y} = 2m \times 2a + 3m \times a\) | M1 | |
| \(\overline{y} = 0.7a\) | A1 | (2) |
| (c) \(\tan \theta = \frac{2.5a - \overline{x}}{\overline{y}}\) | M1 A1 f.t. | |
| \(\theta = 20°\) | A1 | (3) |
| \(M(O), 10mg \times \frac{a}{4} = P \times 2a\) | M1 A1 A1 | |
| (OR: \(4mg \times \frac{5a}{2} - 3mg \times \frac{5a}{2} = P \times 2a\)) | ||
| \(P = \frac{5mg}{4}\) * (exact) | A1 | (4) |
| (e) \(S = \frac{5mg}{4}\); \(R = 10mg\) | B1; B1 | |
| \(F = \sqrt{S^2 + R^2} = \frac{5mg\sqrt{65}}{4}\) (10.1 mg) | M1 A1 | (4) |
| (16 marks) |
**(a)** AD: $10m\overline{x} = 3m \times \frac{5a}{2} + 3m \times 5a$ | M1 A1 |
$\overline{x} = 2.25a$ * | A1 | (3)
**(b)** AB: $10m\overline{y} = 2m \times 2a + 3m \times a$ | M1 |
$\overline{y} = 0.7a$ | A1 | (2)
**(c)** $\tan \theta = \frac{2.5a - \overline{x}}{\overline{y}}$ | M1 A1 f.t. |
$\theta = 20°$ | A1 | (3)
| | $M(O), 10mg \times \frac{a}{4} = P \times 2a$ | M1 A1 A1 |
| | (OR: $4mg \times \frac{5a}{2} - 3mg \times \frac{5a}{2} = P \times 2a$) | |
| | $P = \frac{5mg}{4}$ * (exact) | A1 | (4)
**(e)** $S = \frac{5mg}{4}$; $R = 10mg$ | B1; B1 |
$F = \sqrt{S^2 + R^2} = \frac{5mg\sqrt{65}}{4}$ (10.1 mg) | M1 A1 | (4)
| | | **(16 marks)** |7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{fe64e6f1-e36b-465d-a41c-ac834439623b-6_428_947_404_566}
\end{center}
\end{figure}
A loaded plate $L$ is modelled as a uniform rectangular lamina $A B C D$ and three particles. The sides $C D$ and $A D$ of the lamina have lengths $5 a$ and $2 a$ respectively and the mass of the lamina is $3 m$. The three particles have mass $4 m , m$ and $2 m$ and are attached at the points $A , B$ and $C$ respectively, as shown in Fig. 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of $L$ from $A D$ is $2.25 a$.
\item Find the distance of the centre of mass of $L$ from $A B$.
The point $O$ is the mid-point of $A B$. The loaded plate $L$ is freely suspended from $O$ and hangs at rest under gravity.
\item Find, to the nearest degree, the size of the angle that $A B$ makes with the horizontal.
A horizontal force of magnitude $P$ is applied at $C$ in the direction $C D$. The loaded plate $L$ remains suspended from $O$ and rests in equilibrium with $A B$ horizontal and $C$ vertically below $B$.
\item Show that $P = \frac { 5 } { 4 } \mathrm { mg }$.
\item Find the magnitude of the force on $L$ at $O$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2004 Q7 [16]}}