| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform beam on supports |
| Difficulty | Standard +0.3 This is a straightforward M1 moments problem requiring only two equilibrium equations (vertical forces and moments about one point) to find the center of mass position. The setup is standard with clear given information and the 2:1 force ratio simplifies the algebra, making it slightly easier than average for M1. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| VIAV SIHI NI III IM IONOO | VIAV SIHI NI III IM I ON OO | VIAV SIHI NI III HM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| \((\uparrow),\ R + 2R = Mg\) | M1 A1 | Correct number of terms, dimensionally correct, condone sign errors and missing \(g\); correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{M}(B),\ 6.5R + (2 \times 2R) = Mg(9 - d)\) | M1 A1 | Correct number of terms, dimensionally correct, condone sign errors and missing \(g\); correct equation |
| Answer | Marks |
|---|---|
| \(\text{M}(G),\ R(d - 2.5) = 2R(7 - d)\) | M2 A2,1,0 |
| Answer | Marks | Guidance |
|---|---|---|
| Solve for \(d\), must be numerical: | DM1 | Dependent on previous two M marks |
| \(d = 5.5\) | A1 | \(d = 5.5\) oe, ignore an extra m (but not M) (6 marks) |
## Question 1:
**Setup/Diagram:**
A uniform beam AB of mass $M$ with supports at $C$ and $D$, where $AC = 2.5$ m, $CD = 4.5$ m, $DB = 2$ m. Reactions $R$ at $C$ and $2R$ at $D$. Centre of mass $G$ is at distance $d$ m from $A$.
---
**Possible equations (1st equation in $d$, $R$ and $M$):**
$(\uparrow),\ R + 2R = Mg$ | M1 A1 | Correct number of terms, dimensionally correct, condone sign errors and missing $g$; correct equation
---
**Possible equations (2nd equation in $d$, $R$ and $M$):**
$\text{M}(C),\ 2R \times 4.5 = Mg(d - 2.5)$
$\text{M}(D),\ R \times 4.5 = Mg(7 - d)$
$\text{M}(A),\ 2.5R + (7 \times 2R) = Mgd$
$\text{M}(B),\ 6.5R + (2 \times 2R) = Mg(9 - d)$ | M1 A1 | Correct number of terms, dimensionally correct, condone sign errors and missing $g$; correct equation
---
**Special case:**
$\text{M}(G),\ R(d - 2.5) = 2R(7 - d)$ | M2 A2,1,0 |
---
**Solve for $d$, must be numerical:** | DM1 | Dependent on previous two M marks
$d = 5.5$ | A1 | $d = 5.5$ oe, ignore an extra m (but not M) **(6 marks)**
---
**Notes:**
- Allow M marks if $R_C$ and $R_D$ used
- If $g$ omitted consistently in both equations, all three A marks available
- If $Rg$ used consistently in both equations, all three A marks available
- If three equations present, mark those used to obtain $d$
- If $R$ and $2R$ are consistently swapped, apply scheme unless an MR gives a better total1.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{151d9232-5a78-4bc1-a57e-6c9cae80e473-02_298_1288_264_328}
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\caption{Figure 1}
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A non-uniform rod $A B$ has length 9 m and mass $M \mathrm {~kg}$.\\
The rod rests in equilibrium in a horizontal position on two supports, one at $C$ where $A C = 2.5 \mathrm {~m}$ and the other at $D$ where $D B = 2 \mathrm {~m}$, as shown in Figure 1 .
The magnitude of the force acting on the rod at $D$ is twice the magnitude of the force acting on the $\operatorname { rod }$ at $C$.
The centre of mass of the rod is $d$ metres from $A$.\\
Find the value of $d$.
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VIAV SIHI NI III IM IONOO & VIAV SIHI NI III IM I ON OO & VIAV SIHI NI III HM ION OC \\
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\hfill \mbox{\textit{Edexcel M1 2021 Q1 [6]}}