| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 8 |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard ladder equilibrium problem requiring resolution of forces, taking moments about a point, and comparing friction with limiting friction. Part (i) is routine mechanics bookwork; part (ii) tests conceptual understanding of friction direction but is straightforward once the setup is visualized. Slightly above average due to the two-part nature and the conceptual element in (ii). |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
**(i)** Resolve vertically $N_B = 2 \times 10$ **M1**
Take moments about e.g. intersection of normals: $N_B \times 0.2\cos60° = F \times 0.4\sin60°$ **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**
**(ii)** Consider forces horizontally $N_A$ non-zero **M1**
Correctly deduce impossibility hence leftwards force required for equilibrium **A1**8 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 $60 ^ { \circ }$.\\
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(i) Given that the value of the coefficient of friction between the rod and the ground is 0.2 , determine whether the rod slips.\\
(ii) Explain why it is impossible for the rod to be in equilibrium with one end on smooth horizontal ground and the other against a rough vertical wall.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q8 [8]}}