| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | Specimen |
| Marks | 12 |
| Topic | Moments |
| Type | Rod or block on rough surface in limiting equilibrium (no wall) |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem involving equilibrium on an inclined plane with a pulley system. While it requires resolving forces in two directions and careful geometric reasoning about the string angle, the techniques are standard A-level mechanics. Parts (i)-(ii) are structured 'show that' questions guiding students through the solution. Part (iv) requires optimization using calculus, which adds some challenge, but the overall problem follows familiar patterns for Further Maths mechanics questions. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces6.04e Rigid body equilibrium: coplanar forces |
**(i)** As system is in equilibrium, tension in string is $T = mg$ **B1**
Resolving at right angles to the plane: $R + T\sin\alpha = 2mg\cos\alpha$ **M1**
giving $R = mg(2\cos\alpha - \sin\alpha)$ AG **A1**
**(ii)** By implication $\alpha \leqslant 45°$ **M1**
$\cos\alpha \geqslant \frac{1}{\sqrt{2}}$; $\sin\alpha \leqslant \frac{1}{\sqrt{2}}$ **A1**
$R \geqslant mg\left(\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)$ AG **A1**
**(iii)** Resolving up the slope $F = 2mg\sin\alpha - T\cos\alpha = mg(2\sin\alpha - \cos\alpha)$ **M1**
For this to be non-negative **A1**
and combined with first line of solution to **(ii)** $0.5 \leqslant \tan\alpha \leqslant 1$ AG **A1**
**(iv)** Using $F = \mu R$ **M1**
$\mu = \frac{2\sin\alpha - \cos\alpha}{2\cos\alpha - \sin\varepsilon} = \frac{2t-1}{2-t}$ **A1**
Max value of $\mu$ is $1$ when $t = 1$ **A1**10\\
\includegraphics[max width=\textwidth, alt={}, center]{b18b1bc5-bf26-4161-b5a5-764b00e97bea-6_490_661_267_703}
Particles $A$ and $B$ of masses $2 m$ and $m$, respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley $P$. The particle $A$ rests in equilibrium on a rough plane inclined at an angle $\alpha$ to the horizontal, where $\alpha \leqslant 45 ^ { \circ }$ and $B$ is above the plane. The vertical plane defined by $A P B$ contains a line of greatest slope of the plane, and $P A$ is inclined at angle $2 \alpha$ to the horizontal (see diagram).\\
(i) Show that the normal reaction $R$ between $A$ and the plane is $m g ( 2 \cos \alpha - \sin \alpha )$.\\
(ii) Show that $R \geqslant \frac { 1 } { 2 } m g \sqrt { 2 }$.
The coefficient of friction between $A$ and the plane is $\mu$. The particle is about to slip down the plane.\\
(iii) Show that $0.5 < \tan \alpha \leqslant 1$.\\
(iv) Express $\mu$ as a function of $\tan \alpha$ and deduce its maximum value as $\alpha$ varies.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q10 [12]}}