June 2025 Paper 3 Q10
10
The diagram shows a non-uniform rod \(AB\) of mass 4 kg and length 2 m. The end \(A\) of the rod rests against a rough vertical wall. The rod is held in a horizontal position, perpendicular to the wall, by a light inextensible string attached to the rod at \(B\). The other end of the string is attached to the wall at a point vertically above \(A\). The string is inclined at an angle of \(\theta^\circ\) to the horizontal, where \(\sin\theta = \frac{4}{5}\).
The rod rests in equilibrium in a vertical plane perpendicular to the wall. The rod’s weight acts at the point \(G\) on \(AB\).
You are given that the magnitude of the moment of the rod’s weight about \(A\) is 47.04 N m.
| Scheme | Marks | AO |
|---|---|---|
| \(AG \times 4g = 47.04\) or \((AG =)\ \dfrac{47.04}{4g}\) or \(4g \times 1.2 = 47.04\) | B1 | 3.1b |
| [1] |
Notes
B1: Taking moments about \(A\) for the rod’s weight – allow \(\dfrac{47.04}{4 \times 9.8}\) oe (so using 9.8 for \(g\)) but B0 for only \(\dfrac{47.04}{39.2}\) oe
B0 for \(AB \times 4g = 47.04\) or similar
AG – www but need not see 1.2 explicitly stated
| Scheme | Marks | AO |
|---|---|---|
| \(2 \times \frac{4}{5}T = 47.04\) or \(2 \times \frac{4}{5}T = 1.2 \times 4g\) | M1 | 3.3 |
| \(T = 29.4\) (N) | A1 | 1.1 |
| [2] |
Notes
M1: Taking moments about \(A\) for the rod – M1 for \(2 \times \frac{4}{5}T = 47.04\) or \(2 \times T\sin 53 = 47.04\) or \(2 \times \frac{3}{5}T = 47.04\) or \(2 \times T\cos 53 = 47.04\) only (or \(1.2 \times 4g\) for 47.04)
Allow use of 53 or better (53.1301…) for \(\theta\) - note that e.g. \(2 \times T\sin 37 = 47.04\) or \(2 \times T\cos 37 = 47.04\) (therefore using the complementary angle) is M1
A1: cao exact e.g. \(\frac{147}{5}\) or awrt 29.4
| Scheme | Marks | AO |
|---|---|---|
| \(X = \pm\text{‘}29.4\text{’} \times \frac{3}{5}\) or \(\pm\text{‘}29.4\text{’} \times \cos 53 \quad (= \pm 17.64)\) | B1FT | 3.3 |
| Resolve vertically or take moments about \(B\) or \(G\) for the forces acting on the rod to form an equation or an expression for \(Y\) | M1* | 3.3 |
| \(\mathrm{R}(\uparrow)\): \(\text{‘}29.4\text{’} \times \frac{4}{5} + Y = 4g\) or \(\text{‘}29.4\text{’} \times \frac{4}{5} = 4g + Y\) or \(\mathfrak{M}(B)\): \(0.8 \times 4g = 2Y\) or \(\mathfrak{M}(G)\): \(0.8 \times \text{‘}29.4\text{’} \times \frac{4}{5} = Y \times 1.2\) | A1FT | 1.1 |
| \(\sqrt{17.64^2 + 15.68^2}\) | M1dep* | 3.4 |
| Magnitude is 23.6 (N) | A1 | 1.1 |
| [5] |
Notes
B1FT: Follow through their value of \(T\) only. Allow use of 53 or better (53.1301…) for \(\theta\)
\(X\) is the horizontal component of the contact force at \(A\)
M1*: Must be the correct number of dimensionally correct terms but allow sign errors and sin/cos confusion only – all values must be substituted. If taking moments about \(B\) or \(G\) allow sin/cos confusion only
\(Y\) is the vertical contact force at \(A\)
A1FT: Correct equation for \(Y\) follow through their value of \(T\) only – if correct then \(Y = \pm 15.68\)
Allow \(\sin 53\) or \(\cos 37\) or better for \(\frac{4}{5}\)
M1dep*: Correct use of Pythagoras with their values of \(X\) and \(Y\) which must come from equations with the correct number of dimensionally correct relevant terms
Relevant for \(X\) requires \(\text{‘}29.4\text{’} \times \frac{3}{5}\) or \(\text{‘}29.4\text{’} \times \frac{4}{5}\) oe only
A1: awrt 23.6 or exact e.g. \(\dfrac{49\sqrt{145}}{25}\) oe
23.60152…