| Exam Board | OCR MEI |
|---|---|
| Module | M2 (Mechanics 2) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with straight rod/wire components only |
| Difficulty | Standard +0.3 This is a systematic centre of mass problem requiring careful bookkeeping of multiple components (lamina + 6 rods) in both 2D and 3D configurations. While it involves many parts and 3D coordinates, the methodology is entirely standard: identify component masses and positions, apply the centre of mass formula, then use equilibrium conditions. The calculations are lengthy but straightforward with no novel insights required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \((30+10)\cos 40° - 10\sin 40°\) ... leading to moment | M1 | |
| \(60(40\cos 40° - 10\sin 40°) = 60(30.64... - 6.43) = 9.93\) N m | A1 A1 | shown |
| Answer | Marks |
|---|---|
| Moment of \(P\) about A: \(P \times (10+10)\sin 40° + P \times 80\cos 40°\)... | M1 |
| Taking moments: \(P \times d = 9.93\), \(P = \frac{9.93}{...}\) | M1 A1 |
| Answer | Marks |
|---|---|
| Resolve vertically at A: \(V_A = 60 -\) (vertical component of peg reaction) | M1 |
| Taking moments about B for \(P=0\), find peg reaction, resolve vertically | M1 A1 A1 |
| Answer | Marks |
|---|---|
| Resolve perpendicular to plane: \(N = 60\cos 40° + 200\sin 40°\) | M1 A1 |
| Resolve along plane: \(200\cos 40° - 60\sin 40° - F = ma\) | M1 A1 |
| \(F = \mu N\), substitute and solve for \(\mu\) | M1 |
| \(\mu \approx 0.508\) | A1 A1 A1 |
# Question 3:
## Part (i)
Perpendicular distance from A to line of action of weight:
$(30+10)\cos 40° - 10\sin 40°$ ... leading to moment | M1 |
$60(40\cos 40° - 10\sin 40°) = 60(30.64... - 6.43) = 9.93$ N m | A1 A1 | **shown**
## Part (ii)
Moment of $P$ about A: $P \times (10+10)\sin 40° + P \times 80\cos 40°$... | M1 |
Taking moments: $P \times d = 9.93$, $P = \frac{9.93}{...}$ | M1 A1 |
## Part (iii)
Resolve vertically at A: $V_A = 60 -$ (vertical component of peg reaction) | M1 |
Taking moments about B for $P=0$, find peg reaction, resolve vertically | M1 A1 A1 |
## Part (iv)
Normal reaction $N$ and friction $F$ on panel from plane:
Resolve perpendicular to plane: $N = 60\cos 40° + 200\sin 40°$ | M1 A1 |
Resolve along plane: $200\cos 40° - 60\sin 40° - F = ma$ | M1 A1 |
$F = \mu N$, substitute and solve for $\mu$ | M1 |
$\mu \approx 0.508$ | A1 A1 A1 |
---3 Fig. 3.1 shows an object made up as follows. ABCD is a uniform lamina of mass $16 \mathrm {~kg} . \mathrm { BE } , \mathrm { EF }$, FG, HI, IJ and JD are each uniform rods of mass 2 kg . ABCD, BEFG and HIJD are squares lying in the same plane. The dimensions in metres are shown in the figure.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5dd6ba0d-e516-4b9e-ba19-6e90520b171b-004_627_648_429_735}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}
(i) Find the coordinates of the centre of mass of the object, referred to the axes shown in Fig.3.1.
The rods are now re-positioned so that BEFG and HIJD are perpendicular to the lamina, as shown in Fig. 3.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5dd6ba0d-e516-4b9e-ba19-6e90520b171b-004_442_666_1510_722}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}
(ii) Find the $x$-, $y$-and $z$-coordinates of the centre of mass of the object, referred to the axes shown in Fig. 3.2. Calculate the distance of the centre of mass from A .
The object is now freely suspended from A and hangs in equilibrium with AC at $\alpha ^ { \circ }$ to the vertical.\\
(iii) Calculate $\alpha$.
\hfill \mbox{\textit{OCR MEI M2 Q3}}