| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Range of equilibrium positions |
| Difficulty | Standard +0.3 This is a standard M1 moments question requiring taking moments about two points and using equilibrium conditions. Part (a)-(b) involve straightforward simultaneous equations with the 2:1 reaction ratio given, and part (c) is a routine tipping condition (reaction at B = 0). 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 |
| M1 | Resolve vertically to form equation in \(R_C\) or \(R_D\). All terms required. Condone sign errors | |
| \(2T + T = 6g + 15g\) | A1 | Correct unsimplified equation \(\left(R+\frac{1}{2}R = 6g+15g\right)\) |
| \(2T = 14g = 137\) N or \(140\) N | A1 | |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| M1 | Take moments — all terms must be present and of correct structure. Form equation with one unknown length | |
| \(M(A)\): \(15g\cdot AX + 6g\times 2 = (2T\times 1.5)+4T=7T\) | A2 | \(-1\) each error. Follow their \(T\). NB: use of correct reactions the wrong way round is one error. \((15g\approx 147,\ 6g\approx 58.8,\ 12g\approx 117.6)\) |
| \(M(B)\): \(15gd + 6g\times 2 = 2T\times 2.5\) | ||
| \(M(\text{c of m})\): \(2T\times 0.5 + 15gd = 2\times T\) | ||
| \(M(C)\): \(6g\times 0.5 + 15g(x-1.5) = T\times 2.5\) | ||
| M1 | Substitute for \(T\) and solve for \(AX\) | |
| \(AX = \frac{37}{15}\) m \(= 2.5\) m (or better) | A1 | \(2.4\dot{6}\) |
| Total: (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(M(C)\): \(15g\cdot YC = 6g\times 0.5\) | M1 | Requires both terms present and of correct structure. No additional terms (using \(R_C = 21g\), \(R_B = 0\)) |
| A1 | Correct unsimplified equation | |
| \(YC = 0.2\) m | A1 | |
| \(AY = 1.3\) m | A1 | |
| Total: (4) |
# Question 6(a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| | M1 | Resolve vertically to form equation in $R_C$ or $R_D$. All terms required. Condone sign errors |
| $2T + T = 6g + 15g$ | A1 | Correct unsimplified equation $\left(R+\frac{1}{2}R = 6g+15g\right)$ |
| $2T = 14g = 137$ N or $140$ N | A1 | |
| **Total: (3)** | | |
---
# Question 6(b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| | M1 | Take moments — all terms must be present and of correct structure. Form equation with one unknown length |
| $M(A)$: $15g\cdot AX + 6g\times 2 = (2T\times 1.5)+4T=7T$ | A2 | $-1$ each error. Follow their $T$. NB: use of correct reactions the wrong way round is one error. $(15g\approx 147,\ 6g\approx 58.8,\ 12g\approx 117.6)$ |
| $M(B)$: $15gd + 6g\times 2 = 2T\times 2.5$ | | |
| $M(\text{c of m})$: $2T\times 0.5 + 15gd = 2\times T$ | | |
| $M(C)$: $6g\times 0.5 + 15g(x-1.5) = T\times 2.5$ | | |
| | M1 | Substitute for $T$ and solve for $AX$ |
| $AX = \frac{37}{15}$ m $= 2.5$ m (or better) | A1 | $2.4\dot{6}$ |
| **Total: (5)** | | |
---
# Question 6(c):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $M(C)$: $15g\cdot YC = 6g\times 0.5$ | M1 | Requires both terms present and of correct structure. No additional terms (using $R_C = 21g$, $R_B = 0$) |
| | A1 | Correct unsimplified equation |
| $YC = 0.2$ m | A1 | |
| $AY = 1.3$ m | A1 | |
| **Total: (4)** | | |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3280fdf1-d81a-4729-b065-e84dece6a220-10_238_1258_267_342}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A plank $A B$ has length 4 m and mass 6 kg . The plank rests in a horizontal position on two supports, one at $B$ and one at $C$, where $A C = 1.5 \mathrm {~m}$. A load of mass 15 kg is placed on the plank at the point $X$, as shown in Figure 2, and the plank remains horizontal and in equilibrium. The plank is modelled as a uniform rod and the load is modelled as a particle. The magnitude of the reaction on the plank at $C$ is twice the magnitude of the reaction on the plank at $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the reaction on the plank at $C$.
\item Find the distance $A X$.
The load is now moved along the plank to a point $Y$, between $A$ and $C$. Given that the plank is on the point of tipping about $C$,
\item find the distance $A Y$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2015 Q6 [12]}}