| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Moderate -0.3 This is a standard M1 moments question requiring taking moments about two points and resolving vertically. Part (a) involves routine application of equilibrium conditions with given masses and distances. Part (b) adds a simple algebraic twist (equal reactions) but remains straightforward. Slightly easier than average due to clear setup and standard method. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| (i) \(M(D)\) \(3R_C + 1 \times 3g = 2 \times 4g + 5 \times 2g\) | M1 | e.g. Take moments about \(D\) – requires all 4 terms of the correct form, but condone sign errors. \(1\times\) need not be seen. |
| (correct unsimplified equation) | A1 | Correct unsimplified equation |
| \(R_C = 5g\) or \(49\) N | A1 | |
| (ii) \(R(\uparrow)\) \(R_C + R_D = 4g + 2g + 3g\) | M1 | e.g. Resolve vertically to form an equation in \(R_C\) and \(R_D\), requires all 5 terms |
| (correct unsimplified equation) | A1 | Correct unsimplified equation |
| \(R_D = 4g\) or \(39\) or \(39.2\) N | A1 (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(M(A)\) \(3 \times 4g + 6 \times 3g = 2R_C + 5R_D\ (= 30g)\) | M1A1 | Two equations – M1A1 for each |
| \(M(B)\) \(3 \times 4g + 6 \times 2g = R_D + 4R_C\ (= 24g)\) | M1A1 | |
| \(R_C = 5g\) or \(49\) N, \(R_D = 4g\) or \(39\) or \(39.2\) N | A1, A1 | Solve simultaneously for \(R_C\) and \(R_D\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(M(D)\) \(3R_C + xg = 8g + 10g \quad (3R_C = (18-x)g)\) | M1 | First equation in \(x\) and \(R\) (or \(R_C\) and \(R_D\)) – correct terms required but condone sign slips. |
| \(R(\uparrow)\) \(R_C + R_D = 4g + 2g + xg\) | M1 | A second equation, correct terms required but condone sign slips. |
| Alternatives: \(M(B)\) \(4R_C + R_D = 12g + 12g\) | ||
| \(M(A)\): \(2R_C + 5R_D = 6xg + 3 \times 4g\) | ||
| \(M(C)\): \(2 \times 2g + 3R_D = 4xg + 1 \times 4g\) | ||
| \(2(18-x)g = 3(6+x)g\) | DM1 | Use \(R_C = R_D\) and solve for \(x\). (as far as \(x = \ldots\)). Dependent on the two previous M marks. |
| \(x = 3.6\) | A1 (4) |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| (i) $M(D)$ $3R_C + 1 \times 3g = 2 \times 4g + 5 \times 2g$ | M1 | e.g. Take moments about $D$ – requires all 4 terms of the correct form, but condone sign errors. $1\times$ need not be seen. |
| (correct unsimplified equation) | A1 | Correct unsimplified equation |
| $R_C = 5g$ or $49$ N | A1 | |
| (ii) $R(\uparrow)$ $R_C + R_D = 4g + 2g + 3g$ | M1 | e.g. Resolve vertically to form an equation in $R_C$ and $R_D$, requires all 5 terms |
| (correct unsimplified equation) | A1 | Correct unsimplified equation |
| $R_D = 4g$ or $39$ or $39.2$ N | A1 (6) | |
**Alt:**
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $M(A)$ $3 \times 4g + 6 \times 3g = 2R_C + 5R_D\ (= 30g)$ | M1A1 | Two equations – M1A1 for each |
| $M(B)$ $3 \times 4g + 6 \times 2g = R_D + 4R_C\ (= 24g)$ | M1A1 | |
| $R_C = 5g$ or $49$ N, $R_D = 4g$ or $39$ or $39.2$ N | A1, A1 | Solve simultaneously for $R_C$ and $R_D$ |
## Part (b)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $M(D)$ $3R_C + xg = 8g + 10g \quad (3R_C = (18-x)g)$ | M1 | First equation in $x$ and $R$ (or $R_C$ and $R_D$) – correct terms required but condone sign slips. |
| $R(\uparrow)$ $R_C + R_D = 4g + 2g + xg$ | M1 | A second equation, correct terms required but condone sign slips. |
| Alternatives: $M(B)$ $4R_C + R_D = 12g + 12g$ | | |
| $M(A)$: $2R_C + 5R_D = 6xg + 3 \times 4g$ | | |
| $M(C)$: $2 \times 2g + 3R_D = 4xg + 1 \times 4g$ | | |
| $2(18-x)g = 3(6+x)g$ | DM1 | Use $R_C = R_D$ and solve for $x$. (as far as $x = \ldots$). Dependent on the two previous M marks. |
| $x = 3.6$ | A1 (4) | |
**Total: [10]**4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{edcc4603-f006-4c4f-a4e5-063cab41da98-06_262_1132_223_415}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A plank $A B$, of length 6 m and mass 4 kg , rests in equilibrium horizontally on two supports at $C$ and $D$, where $A C = 2 \mathrm {~m}$ and $D B = 1 \mathrm {~m}$. A brick of mass 2 kg rests on the plank at $A$ and a brick of mass 3 kg rests on the plank at $B$, as shown in Figure 2. The plank is modelled as a uniform rod and all bricks are modelled as particles.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the reaction exerted on the plank
\begin{enumerate}[label=(\roman*)]
\item by the support at $C$,
\item by the support at $D$.
The 3 kg brick is now removed and replaced with a brick of mass $x \mathrm {~kg}$ at $B$. The plank remains horizontal and in equilibrium but the reactions on the plank at $C$ and at $D$ now have equal magnitude.
\end{enumerate}\item Find the value of $x$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q4 [10]}}