| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Standard +0.3 This is a standard M1 statics problem requiring equilibrium of forces and moments. Part (a) is trivial vertical force balance (3+2+6=11N). Part (b) requires taking moments about one point to find position C, which is routine but involves careful arithmetic with the given ratios. Part (c) extends this with weights added, requiring similar moment calculations. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\mathbf{F} = 3 + 2 + 6 = 11\) N | B1 | |
| (b) Let \(AB = 11x\); \(M(4): 11AC = 3x + 10x + 42x = 55x\) | M1 A1 A1 | |
| \(AC = 5x\), so \(AC : CB = 5 : 6\) | M1 A1 | |
| (c) \(M(C): 3(5x) + 6(2x) = k(6x) + 3(4x)\) | M1 A1 A1 | |
| \(15 + 12 = 6k + 12\) → \(6k = 15\) → \(k = 2.5\) | A1 M1 A1 | 12 marks |
**(a)** $\mathbf{F} = 3 + 2 + 6 = 11$ N | B1 |
**(b)** Let $AB = 11x$; $M(4): 11AC = 3x + 10x + 42x = 55x$ | M1 A1 A1 |
$AC = 5x$, so $AC : CB = 5 : 6$ | M1 A1 |
**(c)** $M(C): 3(5x) + 6(2x) = k(6x) + 3(4x)$ | M1 A1 A1 |
$15 + 12 = 6k + 12$ → $6k = 15$ → $k = 2.5$ | A1 M1 A1 | 12 marks |$AB$ is a light rod. Forces $\mathbf{F}$, $\mathbf{G}$ and $\mathbf{H}$, of magnitudes $3$ N, $2$ N and $6$ N respectively, act upwards at right angles to the rod in a vertical plane at points dividing $AB$ in the ratio $1:4:2:4$, as shown.
\includegraphics{figure_4}
A single force $\mathbf{P}$ is applied downwards at the point $C$ to keep the rod horizontal in equilibrium.
\begin{enumerate}[label=(\alph*)]
\item State the magnitude of $\mathbf{P}$. [1 mark]
\item Show that $AC:CB = 5:6$. [5 marks]
\end{enumerate}
Two particles, of weights $3$ N and $k$ N, are now placed on the rod at $A$ and $B$ respectively, while the same upward forces $\mathbf{F}$, $\mathbf{G}$ and $\mathbf{H}$ act as before. It is found that a single downward force at the same point $C$ as before keeps $AB$ horizontal under gravity.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $k$. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q4 [12]}}