| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Topic | Friction |
| Type | Limiting equilibrium with variable angle |
| Difficulty | Standard +0.3 This is a standard mechanics problem involving limiting equilibrium with friction and optimization. Part (i) is routine force diagram drawing, part (ii) requires resolving forces and applying F=μR (standard A-level technique), and part (iii) involves differentiating T with respect to θ and setting to zero—a common calculus application. While it combines mechanics and calculus across multiple parts, each step follows well-established procedures without requiring novel insight or particularly complex reasoning. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates3.03a Force: vector nature and diagrams3.03e Resolve forces: two dimensions3.03i Normal reaction force3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
**Question 9**
**(i)** Diagram with weight, normal contact and friction forces added. — B1 **[1]**
Do not accept both $T$ and the components of $T$ shown.
**(ii)**
$F = T\cos\theta$ — B1 (Resolve horizontally)
$mg = R + T\sin\theta$ — B1 (Resolve vertically)
$F = \mu R$ — M1 (Limiting friction)
$T\cos\theta = \mu(mg - T\sin\theta)$ — M1
$\therefore T = \frac{\mu mg}{\cos\theta + \mu\sin\theta}$ **[4]**
Eliminate $F$ and $R$ and rearrange to given answer. Must be convincing – require at least one intermediate line.
**(iii)** With $\mu = 0.75$, min $T$ occurs at max $(\cos\theta + 0.75\sin\theta)$.
EITHER $-\sin\theta + 0.75\cos\theta = 0$ — M1 (Allow substitution for $\mu$ at any stage), M1 (Differentiate and set $= 0$), A1
$\tan\theta = 0.75\ \therefore \theta = \text{invtan}(0.75) = 36.9°$ — A1 **[4]**
OR Use of $R\cos(\theta - \alpha)$ or $R\sin(\theta + \alpha)$. — M1 (And set $\cos(...)$ or $\sin(...) = 1$)
$\alpha = 36.9°$ or $53.1°$ — A1 (As appropriate)
$\theta = 36.9°$ — A1
**Total: [9]**9 A particle of mass $m \mathrm {~kg}$ rests in equilibrium on a rough horizontal table. There is a string attached to the particle. The tension in the string is $T \mathrm {~N}$ at an angle of $\theta$ to the horizontal, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{2e3f056c-58a2-4466-94ea-3fb873e54752-4_205_547_1027_799}\\
(i) Copy and complete the diagram to show all the forces acting on the particle.\\
(ii) The coefficient of friction between the particle and the table is $\mu$ and the particle is on the point of slipping. Show that $T = \frac { \mu m g } { \cos \theta + \mu \sin \theta }$.\\
(iii) Given that $\mu = 0.75$, find the value of $\theta$ for which $T$ is a minimum.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2013 Q9 [9]}}