| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particle attached to lamina - find mass/position |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass problem requiring systematic calculation of moments about an axis, followed by solving for an unknown mass using equilibrium conditions. While it involves multiple rods and requires careful bookkeeping of the geometry and different mass densities, the techniques are standard for FM2: finding centroids of composite bodies and applying the principle that the vertical line through the suspension point passes through the centre of mass. The geometry is given clearly, and the method is algorithmic rather than requiring novel insight. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Moments about \(AC\) | M1 | Dimensionally correct equation for moments about \(AC\) or parallel axis. All terms needed and horizontal distances. Must use mass ratio. Allow slips but not consistently density and not consistently lengths. Condone without \(a\) |
| \(8a\times 4a + 2\times 3a\times 4a + 2\times 5a\times 2a + 10a\times 4a + 2\times 4a\times 2a = 48a\bar{x}\) | A1 | One side of the equation correct |
| A1 | Both sides of the equation correct | |
| \((132a = 48\bar{x}) \Rightarrow \bar{x} = \frac{11}{4}a\) | A1 | Or equivalent single term. Condone if \(a\) is missing in working and appears at end |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Moments about \(F\): | M1 | Dimensionally correct moments equation. Accept any complete alternative method using \(M\) and \(kM\) to obtain equation in \(k\) only. Condone if \(g\) and/or \(M\) cancelled throughout. Condone incorrect distances. Condone if use \(M = 48\) throughout |
| \(Mg(4a - \bar{x}) = kMg \times 4a\) | A1ft | Correct unsimplified equation (accept without \(g\) and/or \(M\)). Correct mass and distance combination for their \(\bar{x}\) |
| \(\Rightarrow k = \frac{5}{16}\) | A1 | Or \(0.3125\). Condone \(0.31\) or \(0.313\) |
| Alternative — Moments about \(C\): | ||
| \(4a(M + kM) = M\bar{x} + 8akM\) | A1ft | Correct unsimplified equation |
| \(\Rightarrow k = \frac{5}{16}\) | A1 | Or \(0.3125\). Condone \(0.31\) or \(0.313\) |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Moments about $AC$ | M1 | Dimensionally correct equation for moments about $AC$ or parallel axis. All terms needed and horizontal distances. Must use mass ratio. Allow slips but not consistently density and not consistently lengths. Condone without $a$ |
| $8a\times 4a + 2\times 3a\times 4a + 2\times 5a\times 2a + 10a\times 4a + 2\times 4a\times 2a = 48a\bar{x}$ | A1 | One side of the equation correct |
| | A1 | Both sides of the equation correct |
| $(132a = 48\bar{x}) \Rightarrow \bar{x} = \frac{11}{4}a$ | A1 | Or equivalent single term. Condone if $a$ is missing in working and appears at end |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Moments about $F$: | M1 | Dimensionally correct moments equation. Accept any complete alternative method using $M$ and $kM$ to obtain equation in $k$ only. Condone if $g$ and/or $M$ cancelled throughout. Condone incorrect distances. Condone if use $M = 48$ throughout |
| $Mg(4a - \bar{x}) = kMg \times 4a$ | A1ft | Correct unsimplified equation (accept without $g$ and/or $M$). Correct mass and distance combination for their $\bar{x}$ |
| $\Rightarrow k = \frac{5}{16}$ | A1 | Or $0.3125$. Condone $0.31$ or $0.313$ |
| **Alternative — Moments about $C$:** | | |
| $4a(M + kM) = M\bar{x} + 8akM$ | A1ft | Correct unsimplified equation |
| $\Rightarrow k = \frac{5}{16}$ | A1 | Or $0.3125$. Condone $0.31$ or $0.313$ |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1f39620e-c10f-4344-89f1-626fff36d187-08_517_753_258_657}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Nine uniform rods are joined together to form the rigid framework $A B C D E F A$, with $A B = B C = D F = 3 a , B F = C D = D E = 4 a$ and $A F = F E = C F = 5 a$, as shown in Figure 1. All nine rods lie in the same plane.
The mass per unit length of each of the rods $B F , C F$ and $D F$ is twice the mass per unit length of each of the other six rods.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the framework from $A C$
The mass of the framework is $M$. A particle of mass $k M$ is attached to the framework at $E$ to form a loaded framework.
When the loaded framework is freely suspended from $F$, it hangs in equilibrium with $C E$ horizontal.
\item Find the exact value of $k$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2022 Q3 [7]}}