| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2019 |
| Session | Specimen |
| Marks | 6 |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard ladder equilibrium problem requiring three equations (horizontal/vertical forces and moments about one point) with straightforward resolution. The setup is clearly defined, the friction coefficient is given, and students simply need to compare the required friction force with the maximum available friction (μR). While it requires multiple steps, it follows a well-practiced textbook method with no novel insight needed. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
**Question 9**
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 $\operatorname { rod } A B$ 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]{adf5bd3c-5408-421d-b7d5-dea2d0f0185b-5_661_655_390_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 2019 Q9 [6]}}