| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Standard +0.3 This is a standard M1 moments problem requiring taking moments about a point and using equilibrium conditions. Part (a) involves straightforward moment equation setup with one unknown, while part (b) requires recognizing that equal reactions mean the resultant force acts at the midpoint. Both parts use routine mechanics techniques with clear problem structure and no novel insight required. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(C)\ 140(a-2) + 30(2a-2) = 120\times4\) | M1 | Moments or alternative complete method to form equation in \(a\) only. Dimensionally correct. Condone sign error(s). No missing/additional terms. Condone common factor of \(g\) |
| \(M(G)\ 50(a-2) + 30a = 120(6-a)\) | ||
| \(M(D)\ 4\times50 + 30(2a-6) = 140(6-a)\) | ||
| \(M(B)\ 140a = 120(a-6) + 50(2a-2)\) | ||
| \(M(A)\ 50\times2 + 120\times6 = 140a + 30\times2a\) | ||
| At most one error | A1 | |
| \((200a = 820)\) | A1 | Correct unsimplified equation in \(a\) |
| \(a = 4.1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\uparrow),\ (2R = 170 \Rightarrow) R = 85\) | B1 | Or a correct second moments equation in their \(a\) to achieve 2 equations in 2 unknowns |
| \(M(A)\ 85\times2 + 85\times x = 140\times a + 30\times2a\) | M1 | Moments equation with equal reactions in \(a\) or their \(a\). Dimensionally correct. No missing/additional terms. Condone sign error(s). Accept alternative complete method to form equation in different horizontal distance to \(E\). Condone incorrect \(R\), \(R\neq120, R\neq50\). Condone common factor of \(g\) |
| \(M(C)\ 85(x-2) = 140(a-2) + (2a-2)\times30\) | ||
| \(M(G)\ 85(a-2) + 30\times a = 85(x-a)\) | ||
| \(M(E)\ 30(2a-x) + 85(x-2) = 140(x-a)\) | ||
| \(M(B)\ 85\times(2a-2) + 85(2a-x) = 140\times a\) | ||
| At most one error. Follow their \(a\) and their \(R\neq120, R\neq50\) | A1ft | |
| Correct unsimplified equation in \(AE\). Follow their \(a\) and their \(R\neq120, R\neq50\) | A1ft | |
| \(AE = \dfrac{130}{17}\text{ m}\) (7.6 m or better) | A1 | If they find different \(x\), e.g. \(CE=5.6\) and go no further, score 4/5 |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(C)\ 140(a-2) + 30(2a-2) = 120\times4$ | M1 | Moments or alternative complete method to form equation in $a$ only. Dimensionally correct. Condone sign error(s). No missing/additional terms. Condone common factor of $g$ |
| $M(G)\ 50(a-2) + 30a = 120(6-a)$ | | |
| $M(D)\ 4\times50 + 30(2a-6) = 140(6-a)$ | | |
| $M(B)\ 140a = 120(a-6) + 50(2a-2)$ | | |
| $M(A)\ 50\times2 + 120\times6 = 140a + 30\times2a$ | | |
| At most one error | A1 | |
| $(200a = 820)$ | A1 | Correct unsimplified equation in $a$ |
| $a = 4.1$ | A1 | |
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\uparrow),\ (2R = 170 \Rightarrow) R = 85$ | B1 | Or a correct second moments equation in their $a$ to achieve 2 equations in 2 unknowns |
| $M(A)\ 85\times2 + 85\times x = 140\times a + 30\times2a$ | M1 | Moments equation with equal reactions in $a$ or their $a$. Dimensionally correct. No missing/additional terms. Condone sign error(s). Accept alternative complete method to form equation in different horizontal distance to $E$. Condone incorrect $R$, $R\neq120, R\neq50$. Condone common factor of $g$ |
| $M(C)\ 85(x-2) = 140(a-2) + (2a-2)\times30$ | | |
| $M(G)\ 85(a-2) + 30\times a = 85(x-a)$ | | |
| $M(E)\ 30(2a-x) + 85(x-2) = 140(x-a)$ | | |
| $M(B)\ 85\times(2a-2) + 85(2a-x) = 140\times a$ | | |
| At most one error. Follow their $a$ and their $R\neq120, R\neq50$ | A1ft | |
| Correct unsimplified equation in $AE$. Follow their $a$ and their $R\neq120, R\neq50$ | A1ft | |
| $AE = \dfrac{130}{17}\text{ m}$ (7.6 m or better) | A1 | If they find different $x$, e.g. $CE=5.6$ and go no further, score 4/5 |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5c3869c7-008f-4131-b68d-8ecdd4da3377-04_429_1298_255_324}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A wooden beam $A B$ has weight 140 N and length $2 a$ metres. The beam rests horizontally in equilibrium on two supports at $C$ and $D$, where $A C = 2 \mathrm {~m}$ and $A D = 6 \mathrm {~m}$. A block of weight 30 N is placed on the beam at $B$ and the beam remains horizontal and in equilibrium, as shown in Figure 2. The reaction on the beam at $D$ has magnitude 120 N . The block is modelled as a particle and the beam is modelled as a uniform rod.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$.
The support at $D$ is now moved to a point $E$ on the beam and the beam remains horizontal and in equilibrium with the block at $B$. The magnitude of the reaction on the beam at $C$ is now equal to the magnitude of the reaction on the beam at $E$.
\item Find the distance $A E$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2017 Q2 [9]}}