| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform beam on supports |
| Difficulty | Standard +0.3 This is a standard M1 moments question requiring systematic application of equilibrium conditions (sum of moments = 0, reaction forces = 0 at tilting point) across two scenarios to find mass and centre of mass, then using equal reactions for part (b). While it involves multiple steps and careful bookkeeping, it follows a well-established template with no novel insight required—slightly easier than average due to its routine nature. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| M(P): \(50g \times 2 = Mg \times (x - 2)\) | M1 A1 |
| M(Q): \(50g \times 3 = Mg \times (12 - x)\) | M1 A1 |
| (i) \(M = 25\) (kg) | DM1 A1 |
| (ii) \(x = 6\) (m) | DM1 A1 |
| (8) |
| Answer | Marks |
|---|---|
| \((\uparrow) R + R = 25g + 50g\) | M1 A1 ft |
| M(A): \(2R + 12R = 25g \times 6 + 50g \times AX\) | M1 A1 ft |
| \(AX = 7.5\) (m) | DM1 A1 |
| (6) | |
| [14] |
## Part (a)
M(P): $50g \times 2 = Mg \times (x - 2)$ | M1 A1
M(Q): $50g \times 3 = Mg \times (12 - x)$ | M1 A1
(i) $M = 25$ (kg) | DM1 A1
(ii) $x = 6$ (m) | DM1 A1
| (8)
## Part (b)
$(\uparrow) R + R = 25g + 50g$ | M1 A1 ft
M(A): $2R + 12R = 25g \times 6 + 50g \times AX$ | M1 A1 ft
$AX = 7.5$ (m) | DM1 A1
| (6)
| [14]
**Notes for Question 6:**
**Q6(a)**
- First M1 for moments about $P$ equation with usual rules (or moments about a different point AND vertical resolution and $R$ then eliminated) (M0 if non-zero reaction at $Q$)
- Second M1 for moments about $Q$ equation with usual rules (or moments about a different point AND vertical resolution) (M0 if non-zero reaction at $P$)
- Second A1 for a correct equation in $M$ and same unknown.
- Third M1, dependent on first and second M marks, for solving for $M$
- Third A1 for 25 (kg)
- Fourth M1, dependent on first and second M marks, for solving for $x$
- Fourth A1 for 6 (m)
- N.B. No marks available if rod is assumed to be uniform but can score max 5/6 in part (b), provided they have found values for $M$ and $x$ to f.t. on.
- If they have just invented values for $M$ and $x$ in part (a), they can score the M marks in part (b) but not the A marks.
**Q6(b)**
- First M1 for vertical resolution or a moments equation, with usual rules.
- First A1 ft on their $M$ and $x$ from part (a), for a correct equation. (must have equal reactions in vertical resolution to earn this mark)
- Second M1 for a moments equation with usual rules.
- Second A1 ft on their $M$ and $x$ from part (a), for a correct equation in $R$ and same unknown length.
- Third M1, dependent on first and second M marks, for solving for $AX$ (not their unknown length) with $AX \leq 15$
- Third A1 for $AX = 7.5$ (m)
- N.B. If a single equation is used (see below), equating the sum of the moments of the child and the weight about $P$ to the sum of the moments of the child and the weight about $Q$, this can score M2 A2 ft on their $M$ and $x$ from part (a), provided the equation is in one unknown. Any method error, loses both M marks. e.g. $25g.4 + 50g(x - 2) = 25g.6 + 50g(12 - x)$ oe.
---A beam $AB$ has length 15 m. The beam rests horizontally in equilibrium on two smooth supports at the points $P$ and $Q$, where $AP = 2$ m and $QB = 3$ m. When a child of mass 50 kg stands on the beam at $A$, the beam remains in equilibrium and is on the point of tilting about $P$. When the same child of mass 50 kg stands on the beam at $B$, the beam remains in equilibrium and is on the point of tilting about $Q$. The child is modelled as a particle and the beam is modelled as a non-uniform rod.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the mass of the beam.
\item Find the distance of the centre of mass of the beam from $A$. [8]
\end{enumerate}
\end{enumerate}
When the child stands at the point $X$ on the beam, it remains horizontal and in equilibrium. Given that the reactions at the two supports are equal in magnitude,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find $AX$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2013 Q6 [14]}}