| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Range of equilibrium positions |
| Difficulty | Standard +0.3 This is a standard M1 moments problem requiring taking moments about two points and applying equilibrium conditions. While it involves two parts and careful consideration of limiting cases (reactions becoming zero or maximum), the techniques are routine and well-practiced. The setup is clear and the mathematical steps are straightforward applications of ΣM=0 and ΣF=0, making it slightly easier than average. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| \(x\) is greatest when rod is about to tip about \(B\) i.e. \(R_A = 0\) (can be implied) | B1 |
| \(M(B), 2W(x - 2l) = W \frac{1}{2}l\) | M1 A1 |
| \(x = 2.25l\) | DM1 A1 (5) |
| Answer | Marks |
|---|---|
| Use of \(R_A = 2W\) in an equation | M1 |
| \(M(B), 2W(2l - x) + W \frac{1}{2}l = 2W.2l\) | M1 A1 A1 |
| \(x = 0.25l\) | A1 (5) |
## Part (a)
$x$ is greatest when rod is about to tip about $B$ i.e. $R_A = 0$ (can be implied) | B1 |
$M(B), 2W(x - 2l) = W \frac{1}{2}l$ | M1 A1 |
$x = 2.25l$ | DM1 A1 (5) |
## Part (b)
Use of $R_A = 2W$ in an equation | M1 |
$M(B), 2W(2l - x) + W \frac{1}{2}l = 2W.2l$ | M1 A1 A1 |
$x = 0.25l$ | A1 (5) |
**Notes for Question 6(a):**
B1 for $x$ greatest when $R_A = 0$ (usually implied in moments equation) or correct use of $R_A \geq 0$.
First M1 for an equation in $x$ and / ONLY (usually moments about $B$ but could come from two equations). Allow if there is $W$ (uncancelled) in each term. (M0 if $R_A$ term included unless it subsequently becomes zero)
First A1 for a correct equation –again allow even if $W$ has not been cancelled.
Second M1, dependent on previous M, for solving for $x$ in terms of $l$.
Second A1 for $x = 2.25l$.
N.B. If '$l$' omitted consistently and then inserted at end award full marks. If not inserted then can score max B1M1A0M1A0
**Notes for Question 6(b) Scheme change:**
First M1 for use of $R_A = 2W$ in any equation (vertical resolution or moments) or for correct use of $R_A \leq 2W$.
Second M1 for an equation in $x$ and / ONLY (usually moments about $B$ but could come from two equations). Allow if there is $W$ (uncancelled) in each term.
A2 for the equation, again allow even if $W$ has not been cancelled, -1 each error.
Third A1 for $x = 0.25l$.
N.B. If '$l$' omitted consistently and then inserted at end award full marks. If not inserted then can score max M1M1A0A0A0.
---A uniform rod $AC$, of weight $W$ and length $3l$, rests horizontally on two supports, one at $A$ and one at $B$, where $AB = 2l$. A particle of weight $2W$ is placed on the rod at a distance $x$ from $A$. The rod remains horizontal and in equilibrium.
\begin{enumerate}[label=(\alph*)]
\item Find the greatest possible value of $x$. [5]
\end{enumerate}
The magnitude of the reaction of the support at $A$ is $R$. Due to a weakness in the support at $A$, the greatest possible value of $R$ is $4W$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the least possible value of $x$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2015 Q6 [10]}}