| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 14 |
| 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 to find tension expressions, then applying constraints. Part (a) requires understanding the linear relationship, part (b) involves algebraic manipulation of inequalities, and part (c) applies the same method. While multi-part, each step follows routine mechanics procedures with no novel insight required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| As rock moves further from A, tension at A decreases linearly and is a minimum when rock reaches B. | B3 |
| [Diagram showing T decreasing linearly from positive value at O to zero at x = 8] |
| Answer | Marks |
|---|---|
| Max. tension when rock at one end (A, say) | B1 |
| Moments about B: \(50g(4) + Mg(8) - T_A(8) = 0\) | M1 |
| \(8Mg = 8T_A - 200g \therefore Mg = T_A - 25g\) | M1 |
| Given \(T_A \leq 40g\); \(Mg \leq 40g - 25g\) (= 15g) | M1 |
| i.e. \(M \leq 15\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Assume rock placed as close to A as poss. so that \(T_A = 40g\) | M1 A1 | |
| Resolve (↑): \(T_A + T_B = 50g + 20g = 70g \therefore T_B = 30g\) | M1 | |
| Moments about centre of plank: \(T_A(4) - T_B(4) - 20g(d) = 0\) | M1 A1 | |
| \(160g - 120g - 20gd = 0 \therefore d = 2\) | M1 A1 | |
| Rock can be 2 m either side of centre i.e. \(\frac{4}{8}\) plank | A1 | (14) |
**Part (a)**
As rock moves further from A, tension at A decreases linearly and is a minimum when rock reaches B. | B3 |
[Diagram showing T decreasing linearly from positive value at O to zero at x = 8] | |
**Part (b)**
Max. tension when rock at one end (A, say) | B1 |
Moments about B: $50g(4) + Mg(8) - T_A(8) = 0$ | M1 |
$8Mg = 8T_A - 200g \therefore Mg = T_A - 25g$ | M1 |
Given $T_A \leq 40g$; $Mg \leq 40g - 25g$ (= 15g) | M1 |
i.e. $M \leq 15$ | A1 |
**Part (c)**
Assume rock placed as close to A as poss. so that $T_A = 40g$ | M1 A1 |
Resolve (↑): $T_A + T_B = 50g + 20g = 70g \therefore T_B = 30g$ | M1 |
Moments about centre of plank: $T_A(4) - T_B(4) - 20g(d) = 0$ | M1 A1 |
$160g - 120g - 20gd = 0 \therefore d = 2$ | M1 A1 |
Rock can be 2 m either side of centre i.e. $\frac{4}{8}$ plank | A1 | (14)6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4fe54579-ac77-46f9-85e1-2e95963d6b3e-4_288_1275_201_410}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Figure 2 shows a uniform plank $A B$ of length 8 m and mass 50 kg suspended horizontally by two light vertical inextensible strings attached at either end of the plank. The maximum tension that either string can support is 40 gN .
A rock of mass $M \mathrm {~kg}$ is placed on the plank at $A$ and rolled along the plank to $B$ without either string breaking.
\begin{enumerate}[label=(\alph*)]
\item Explain, with the aid of a sketch-graph, how the tension in the string at $A$ varies with $x$, the distance of the rock from $A$.
\item Show that $M \leq 15$.
The first rock is removed and a second rock of mass 20 kg is placed on the plank.
\item Find the fraction of the plank on which the rock can be placed without one of the strings breaking.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [14]}}