| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Standard +0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions to find two unknowns (P and θ). Part (ii) tests basic understanding that the resultant force determines acceleration direction. Straightforward application of trigonometry and simultaneous equations, slightly above average due to the algebraic manipulation required. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
**Question 11(i)**
Attempt to resolve in $x$ or $y$ direction **M1**
$P\cos\theta + 4\cos30 = 8$ **A1** $P\cos\theta = 8 - 2\sqrt{3} = 4.535898385$
$P\sin\theta = 4\sin30$ **A1** $P\sin\theta = 2$
Attempt to solve for $P$ or $\theta$ **M1**
$P = 4.96$ **A1**
$\theta = 23.8°$ **A1**
OR
Use cosine rule **M1**
Get $P^2 = 8^2 + 4^2 - 2 \times 8 \times 4 \times \cos30$ oe **A1**
$P = 4.96$ **A1**
Use sine or cosine rule **M1**
Get $\frac{\sin\theta}{4} = \frac{\sin30}{P}$ or $4^2 = 8^2 + P^2 - 2 \times 8 \times P \times \cos\theta$ oe **A1**
$\theta = 23.8°$ **A1**
**Question 11(ii)**
Right **B1** Allow east, positive x-direction, 090°11\\
\includegraphics[max width=\textwidth, alt={}, center]{35d24778-1203-4d5d-be4b-bb375344fe09-4_285_700_1043_721}
Three forces are acting on a particle $A$ as shown in the diagram. The forces act in the same plane and the particle is in equilibrium.\\
(i) Evaluate $P$ and $\theta$.
The 8 N force is removed.\\
(ii) State the direction of the instantaneous acceleration of $A$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2018 Q11 [13]}}