| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform beam on supports |
| Difficulty | Standard +0.3 This is a standard M1 moments question requiring equilibrium equations and taking moments about a point. Part (a) involves routine algebraic manipulation with given constraint 2R_D = 3R_C, while part (b) requires recognizing that 'point of tilting' means one reaction becomes zero. Both parts use familiar techniques with clear setup and straightforward calculation, making it slightly easier than average for A-level. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve vertically or take moments | M1 | First equation in \(R_C\) and/or \(R_D\). Dimensionally correct, correct no. of terms. Condone sign errors. N.B. \(3R_C + 2R_D = 75g\) or \(5R_C = 75g\) are both M0A0 unless they recover |
| \(\uparrow R_C + R_D = 50g + 25g\ (= 75g)\) | A1 | Correct unsimplified equation but A0 if they assume \(R_C = R_D\). N.B. This mark can be awarded even if they clearly have \(R_C = 3X\) and \(R_D = 2X\) |
| Form a moments equation or resolve vertically | M1 | Second equation in \(R_C\) and/or \(R_D\). Dimensionally correct, correct no. of terms. Condone sign errors |
| \(M(D): 50gx + 25g \times 1.2 = 3.3R_C\) \(M(A): 0.9R_C + 4.2R_D = 3 \times 25g + (4.2-x)50g\) \(M(B): 5.1R_C + 1.8R_D = 3 \times 25g + (1.8+x)50g\) \(M(C): 3.3R_D = 2.1 \times 25g + (3.3-x)50g\) \(M(E): 2.1R_C = (x-1.2)50g + 1.2R_D\) \(M(G): R_D x = 25g(x-1.2) + R_C(3.3-x)\) | A1 | Correct unsimplified equation in \(R_D\) or \(R_C\) seen but give A0 if equation is incorrect. A0 if they assume \(R_C = R_D\) |
| \((2R_D = 3R_C \Rightarrow R_C = 30g)\) \(50x + 30 = 99\) | M1 | Complete method, using either \(2R_D = 3R_C\) or \(2R_C = 3R_D\) to find equation in \(x\) only |
| \(x = \frac{69}{50} = 1.38\) | A1* | Obtain given answer from correct working, with no incorrect equations seen |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to form an equation in \(M\) only | M1 | e.g. moments about \(D\) or vertical resolution and moments about another point or two moments equations. Dimensionally correct equation. Condone sign errors |
| \(M(D)\ 25g \times 1.2 + 50g \times 1.38 = 1.8Mg\) OR any two of: \((50+25+M)g = R_D\) \(M(A)\ 4.2R_D = 3\times25g + (4.2-1.38)50g + 6Mg\) \(M(B)\ 1.8R_D = 3\times25g + (1.8+1.38)50g\) \(M(C)\ 3.3R_D = 2.1\times25g + (3.3-1.38)50g + 5.1Mg\) \(M(E)\ 3Mg = (1.38-1.2)50g + 1.2R_D\) \(M(G)\ R_D\times1.38 = 25g(1.38-1.2) + Mg(1.38+1.8)\) AND \(R_D\) eliminated | A1 | Correct unsimplified equation in \(M\) |
| \((M) = 55\) | A1 | Correct only |
| [3] |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically or take moments | M1 | First equation in $R_C$ and/or $R_D$. Dimensionally correct, correct no. of terms. Condone sign errors. N.B. $3R_C + 2R_D = 75g$ or $5R_C = 75g$ are both M0A0 unless they recover |
| $\uparrow R_C + R_D = 50g + 25g\ (= 75g)$ | A1 | Correct unsimplified equation but A0 if they assume $R_C = R_D$. N.B. This mark can be awarded even if they clearly have $R_C = 3X$ and $R_D = 2X$ |
| Form a moments equation or resolve vertically | M1 | Second equation in $R_C$ and/or $R_D$. Dimensionally correct, correct no. of terms. Condone sign errors |
| $M(D): 50gx + 25g \times 1.2 = 3.3R_C$ $M(A): 0.9R_C + 4.2R_D = 3 \times 25g + (4.2-x)50g$ $M(B): 5.1R_C + 1.8R_D = 3 \times 25g + (1.8+x)50g$ $M(C): 3.3R_D = 2.1 \times 25g + (3.3-x)50g$ $M(E): 2.1R_C = (x-1.2)50g + 1.2R_D$ $M(G): R_D x = 25g(x-1.2) + R_C(3.3-x)$ | A1 | Correct unsimplified equation in $R_D$ or $R_C$ seen but give A0 if equation is incorrect. A0 if they assume $R_C = R_D$ |
| $(2R_D = 3R_C \Rightarrow R_C = 30g)$ $50x + 30 = 99$ | M1 | Complete method, using either $2R_D = 3R_C$ or $2R_C = 3R_D$ to find equation in $x$ only |
| $x = \frac{69}{50} = 1.38$ | A1* | Obtain given answer from correct working, with no incorrect equations seen |
| **[6]** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to form an equation in $M$ only | M1 | e.g. moments about $D$ or vertical resolution and moments about another point or two moments equations. Dimensionally correct equation. Condone sign errors |
| $M(D)\ 25g \times 1.2 + 50g \times 1.38 = 1.8Mg$ **OR any two of:** $(50+25+M)g = R_D$ $M(A)\ 4.2R_D = 3\times25g + (4.2-1.38)50g + 6Mg$ $M(B)\ 1.8R_D = 3\times25g + (1.8+1.38)50g$ $M(C)\ 3.3R_D = 2.1\times25g + (3.3-1.38)50g + 5.1Mg$ $M(E)\ 3Mg = (1.38-1.2)50g + 1.2R_D$ $M(G)\ R_D\times1.38 = 25g(1.38-1.2) + Mg(1.38+1.8)$ AND $R_D$ eliminated | A1 | Correct unsimplified equation in $M$ |
| $(M) = 55$ | A1 | Correct only |
| **[3]** | | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2f2f89a6-cec4-444d-95d9-0112887d87eb-04_282_1075_296_495}
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\caption{Figure 1}
\end{center}
\end{figure}
A non-uniform beam $A B$ has length 6 m and mass 50 kg . The beam rests horizontally on two supports at $C$ and $D$, where $A C = 0.9 \mathrm {~m}$ and $D B = 1.8 \mathrm {~m}$.
A child of mass 25 kg stands on the beam at $E$, where $A E = E B = 3 \mathrm {~m}$, as shown in Figure 1.
The beam is in equilibrium.\\
The magnitude of the normal reaction between the beam and the support at $C$ is $R _ { C }$ newtons.
The magnitude of the normal reaction between the beam and the support at $D$ is $R _ { D }$ newtons.
The beam is modelled as a rod and the child is modelled as a particle.\\
The centre of mass of the beam is between $C$ and $D$ and is a distance $x$ metres from $D$.\\
Given that $2 R _ { D } = 3 R _ { C }$
\begin{enumerate}[label=(\alph*)]
\item show that $x = 1.38$
The child remains at $E$ and a block of mass $M \mathrm {~kg}$ is placed on the beam at $B$.\\
The block is modelled as a particle.\\
Given that the beam is on the point of tilting,
\item find the value of $M$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2024 Q2}}