| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam suspended by vertical ropes |
| Difficulty | Moderate -0.3 This is a standard M1 moments question requiring taking moments about a point and resolving vertically. Part (a) involves straightforward simultaneous equations, part (b) requires finding the length using moment equilibrium, and part (c) adds a load requiring similar analysis. The techniques are routine for M1 students with no novel problem-solving 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) \(R + 2R = 210 \Rightarrow R = 70 \text{ N}\) | M1 A1 | (2) |
| (b) e.g. M(A): \(140 \times 90 = 210 \times d \Rightarrow d = 60 \Rightarrow AB = 120 \text{ cm}\) | M1 A1∇ ↓ M1 A1 | (4) |
| (c) \(4S = 210 + W\) | M1 A1 | |
| e.g. M(B): \(S \times 120 + 3S \times 30 = 210 \times 60\) | M1 A2,1,0 | ↓ M1 A1 (7) |
| Solve → \((S = 60\) and\() W = 30\) |
| (a) $R + 2R = 210 \Rightarrow R = 70 \text{ N}$ | M1 A1 | (2) |
| (b) e.g. M(A): $140 \times 90 = 210 \times d \Rightarrow d = 60 \Rightarrow AB = 120 \text{ cm}$ | M1 A1∇ ↓ M1 A1 | (4) |
| (c) $4S = 210 + W$ | M1 A1 | |
| e.g. M(B): $S \times 120 + 3S \times 30 = 210 \times 60$ | M1 A2,1,0 | ↓ M1 A1 (7) |
| Solve → $(S = 60$ and$) W = 30$ | | |
**Guidance:** Note that they can take moments legitimately about many points. (a) M1 for a valid method to get $R$ (almost always resolving!). (b) 1st M1 for a valid moments equation. 2nd M1 for complete solution to find $AB$ (or verification). Allow 'verification', e.g. showing $140 \times 90 = 210 \times 60$ M1 A1; $1260 = 1260$ QED M1 A1. (c) In both equations, allow whatever they think $S$ is in their equations for full marks (e.g. if using $S = 70$). 2nd M1 A2 is for a moments equation (which may be about any one of 4+ points!). 1st M1 A1 is for a second equation (resolving or moments). If they have two moments equations, given M1 A2 if possible for the best one. 2 M marks only available *without* using $S = 70$. If take mass as 210 (hence use 210g) consistently: treat as MR, i.e. deduct up to two A marks and treat rest as f.t. (Answers all as given = 9.8). But allow full marks in (b) ($g$'s should all cancel and give correct result).
---5.
\begin{tikzpicture}
% Girder
\draw[thick] (0,0) -- (6,0);
% Cable at A (vertical, upward)
\draw[thick] (0,0) -- (0,2);
% Cable at C (vertical, upward)
\draw[thick] (4.5,0) -- (4.5,2);
% Labels
\node[below] at (0,0) {$A$};
\node[below] at (6,0) {$B$};
\node[below] at (4.5,0) {$C$};
% 90 cm dimension
\draw[<->] (0,0.5) -- (4.5,0.5);
\node[above] at (2.25,0.5) {90\,cm};
\end{tikzpicture}
A steel girder $A B$ has weight 210 N . It is held in equilibrium in a horizontal position by two vertical cables. One cable is attached to the end $A$. The other cable is attached to the point $C$ on the girder, where $A C = 90 \mathrm {~cm}$, as shown in Figure 3. The girder is modelled as a uniform rod, and the cables as light inextensible strings.
Given that the tension in the cable at $C$ is twice the tension in the cable at $A$, find
\begin{enumerate}[label=(\alph*)]
\item the tension in the cable at $A$,
\item show that $A B = 120 \mathrm {~cm}$.
A small load of weight $W$ newtons is attached to the girder at $B$. The load is modelled as a particle. The girder remains in equilibrium in a horizontal position. The tension in the cable at $C$ is now three times the tension in the cable at $A$.
\item Find the value of $W$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2006 Q5 [13]}}