| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Moderate -0.8 This is a standard M1 moments question requiring equilibrium conditions (sum of forces = 0, sum of moments = 0) with straightforward setup. The equal forces at P and Q simplify the problem significantly, and all calculations involve basic arithmetic with moments about a point. While it's a multi-step problem worth 10 marks, it follows a completely standard template with no conceptual challenges or novel insights required. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| \(M(R)\): \(8X + 2X = 40g \times 6 + 20g \times 4\) | M1 A2 |
| solving for \(X\), \(X = 32g = 314\) or \(310 \text{ N}\) | M1 A1 |
| \((\dagger) X + X = 40g + 20g + Mg\) (or another moments equation) | M1 A2 |
| solving for \(M\), \(M = 4\) | M1 A1 |
| (10) |
| Answer | Marks |
|---|---|
| \(M(P)\): \(6X = 40g \times 2 + 20g \times 4 + Mg \times 8\) | M1 A2 |
| solving for \(X\), \(X = 32g = 314\) or \(310 \text{ N}\) | M1 A1 |
| \((\dagger) X + X = 40g + 20g + Mg\) (or another moments equation) | M1 A2 |
| solving for \(M\), \(M = 4\) | M1 A1 |
| (10) |
| Answer | Marks |
|---|---|
| Masses concentrated at a point or weights act at a point | B1 |
| (1) | |
| 11 |
## (a)
### (i)
**EITHER**
$M(R)$: $8X + 2X = 40g \times 6 + 20g \times 4$ | M1 A2 |
solving for $X$, $X = 32g = 314$ or $310 \text{ N}$ | M1 A1 |
$(\dagger) X + X = 40g + 20g + Mg$ (or another moments equation) | M1 A2 |
solving for $M$, $M = 4$ | M1 A1 |
| (10) |
**OR**
$M(P)$: $6X = 40g \times 2 + 20g \times 4 + Mg \times 8$ | M1 A2 |
solving for $X$, $X = 32g = 314$ or $310 \text{ N}$ | M1 A1 |
$(\dagger) X + X = 40g + 20g + Mg$ (or another moments equation) | M1 A2 |
solving for $M$, $M = 4$ | M1 A1 |
| (10) |
## (b)
Masses concentrated at a point or weights act at a point | B1 |
| (1) |
| **11** |
---A plank $PQR$, of length 8 m and mass 20 kg, is in equilibrium in a horizontal position on two supports at $P$ and $Q$, where $PQ = 6$ m.
A child of mass 40 kg stands on the plank at a distance of 2 m from $P$ and a block of mass $M$ kg is placed on the plank at the end $R$. The plank remains horizontal and in equilibrium. The force exerted on the plank by the support at $P$ is equal to the force exerted on the plank by the support at $Q$.
By modelling the plank as a uniform rod, and the child and the block as particles,
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item find the magnitude of the force exerted on the plank by the support at $P$,
\item find the value of $M$. [10]
\end{enumerate}
\item State how, in your calculations, you have used the fact that the child and the block can be modelled as particles. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2011 Q5 [11]}}