| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Standard +0.2 This is a straightforward M1 moments question requiring basic equilibrium conditions (sum of forces = 0, sum of moments = 0) and simple algebraic manipulation. The multi-part structure and need to find distances using moment equations is standard textbook fare, slightly easier than average due to the simple arithmetic and clear setup. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(M(A): 80 \times \frac{x}{2} + 20 \times x = 90 \times 2\) | M1 A1 | |
| Solve for \(x\): \(x = 3\) | M1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| By having weight act at \(B\) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(R(\uparrow): R + 3R = 100 \Rightarrow R = 25\) | B1 | |
| \(M(A): 25y + 75 \times 2 = 80 \times 1.5 + 20 \times 3\) | M1 A1 ft | |
| Solve: \(y = 1.2\) m | A1 | (4) |
## Question 4:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $M(A): 80 \times \frac{x}{2} + 20 \times x = 90 \times 2$ | M1 A1 | |
| Solve for $x$: $x = 3$ | M1 A1 | **(4)** |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| By having weight act at $B$ | B1 | **(1)** |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| $R(\uparrow): R + 3R = 100 \Rightarrow R = 25$ | B1 | |
| $M(A): 25y + 75 \times 2 = 80 \times 1.5 + 20 \times 3$ | M1 A1 ft | |
| Solve: $y = 1.2$ m | A1 | **(4)** |
---4.
\begin{tikzpicture}
% Plank
\draw[thick] (0,0) -- (6,0);
% Labels A, C, B on the plank
\node[above] at (0,0.05) {$A$};
\node[above] at (3,0.05) {$C$};
\node[above] at (6,0.05) {$B$};
% Triangular support at A
\draw[fill=white] (0,0) -- (-0.3,-0.6) -- (0.3,-0.6) -- cycle;
% Triangular support at C
\draw[fill=white] (3,0) -- (2.7,-0.6) -- (3.3,-0.6) -- cycle;
% Rock at B (small filled semicircle/bump)
\draw[fill=black] (6,0) .. controls (5.85,0.25) and (6.15,0.25) .. (6,0);
% Dimension: x m (full length A to B)
\draw[<->] (0,0.8) -- (6,0.8);
\node[above] at (3,0.8) {$x$ m};
% Dimension: 2 m (A to C)
\draw[<->] (0,-1.0) -- (3,-1.0);
\node[below] at (1.5,-1.0) {2\,m};
\end{tikzpicture}
A uniform plank $A B$ has weight 80 N and length $x$ metres. The plank rests in equilibrium horizontally on two smooth supports at $A$ and $C$, where $A C = 2 \mathrm {~m}$, as shown in Fig. 2. A rock of weight 20 N is placed at $B$ and the plank remains in equilibrium. The reaction on the plank at $C$ has magnitude 90 N . The plank is modelled as a rod and the rock as a particle.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $x$.
\item State how you have used the model of the rock as a particle.
The support at $A$ is now moved to a point $D$ on the plank and the plank remains in equilibrium with the rock at $B$. The reaction on the plank at $C$ is now three times the reaction at $D$.
\item Find the distance $A D$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2002 Q4 [9]}}