| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| 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 problem requiring vertical force equilibrium and taking moments about one point. The setup is straightforward with clear given information, and the solution follows a routine two-equation approach (ΣF=0 and Στ=0). Slightly easier than average due to the simple relationship between reactions (one is twice the other) and no complications like angles or multiple objects. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\uparrow \quad 2X + X = 4.5g\) | M1 A1 | M1 for complete method finding \(R_Q\); A1 for correct equation in \(R_Q\) or \(R_P\) only |
| \(X = \frac{3g}{2}\) or 14.7 or 15 (N) | A1 (3) | A0 for negative answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(M(A): \quad 4.5g \times AG = (2X) \times 0.8 + X \times 2.4\) | M1 A2 ft (1,0) | M1 for moments about any point; A2 ft for correct equation (A1A0 one error, A0A0 two or more errors) |
| \(AG = \frac{4}{3}\) (m), 1.3, 1.33,... | A1 (4) | Need '\(AG=\)' or \(x\) marked on diagram |
## Question 2:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\uparrow \quad 2X + X = 4.5g$ | M1 A1 | M1 for complete method finding $R_Q$; A1 for correct equation in $R_Q$ or $R_P$ only |
| $X = \frac{3g}{2}$ or 14.7 or 15 (N) | A1 **(3)** | A0 for negative answer |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $M(A): \quad 4.5g \times AG = (2X) \times 0.8 + X \times 2.4$ | M1 A2 ft (1,0) | M1 for moments about any point; A2 ft for correct equation (A1A0 one error, A0A0 two or more errors) |
| $AG = \frac{4}{3}$ (m), 1.3, 1.33,... | A1 **(4)** | Need '$AG=$' or $x$ marked on diagram |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5c908e75-73df-46be-93bb-09dba2cb3b7e-03_215_716_233_614}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A non-uniform rod $A B$ has length 3 m and mass 4.5 kg . The rod rests in equilibrium, in a horizontal position, on two smooth supports at $P$ and at $Q$, where $A P = 0.8 \mathrm {~m}$ and $Q B = 0.6 \mathrm {~m}$, as shown in Figure 1. The centre of mass of the rod is at $G$. Given that the magnitude of the reaction of the support at $P$ on the rod is twice the magnitude of the reaction of the support at $Q$ on the rod, find
\begin{enumerate}[label=(\alph*)]
\item the magnitude of the reaction of the support at $Q$ on the rod,
\item the distance $A G$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2012 Q2 [7]}}