| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 6 |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard statics problem requiring three equilibrium equations (horizontal/vertical forces and moments) with straightforward resolution. The setup is typical textbook material, though it requires careful application of friction inequality and moment calculation about a point. Slightly easier than average due to clear structure and standard method. |
| Spec | 3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces6.04e Rigid body equilibrium: coplanar forces |
Resolve vertically $N_B = 2 \times 10$ [M1]
Take moments about e.g. intersection of normals:
$20 \times 0.2\cos 60 = F \times 0.4\sin 60$ [M1]
Moments equation correct (if in equilibrium) [A1]
$F = 5.77,\ N_B = 20$ [A1]
$F > \mu N_B$ [M1]
Correctly deduce not in equilibrium therefore rod does slip [A1]
**Total: 6 marks**9 The diagram shows a uniform rod $A B$ of length 40 cm and mass 2 kg placed with the end $A$ resting against a smooth vertical wall and the end $B$ on rough horizontal ground. The angle between $A B$ and the horizontal is $60 ^ { \circ }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f4acd242-eb78-4124-bfa2-fdecaa188690-5_657_659_392_705}
Given that the value of the coefficient of friction between the rod and the ground is 0.2 , determine whether the rod slips.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2020 Q9 [6]}}