| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Moderate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions to find two unknowns. Part (i) involves standard simultaneous equations from ΣFₓ=0 and ΣFᵧ=0, while part (ii) is immediate from Newton's third law (the resultant equals the removed force). The techniques are routine for M1 with no geometric insight or extended reasoning required. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| \([7 = F\cos\theta \text{ and } 4 = F\sin\theta \Rightarrow F^2 = 7^2 + 4^2 \text{ (or } \tan\theta = 4/7)]\) | M1 | For stating \(F^2 = 7^2 + 4^2\) directly or for resolving in the i and j directions and eliminating \(\theta\) or F |
| \(F = 8.06\) | A1 | |
| \([7 = 8.06\cos\theta \text{ or } 4 = 8.06\sin\theta]\) or (or 7 = Fcos29.7° or 4 = Fsin29.7°) | M1 | For stating \(\tan\theta = 4/7\) directly or for substituting for F or for θ into \(7 = F\cos\theta\) or \(4 = F\sin\theta\) |
| \(\theta = 29.7\) | A1 | 4 |
| SR for candidates who mix sine and cosine (max 3/4) | ||
| Fsin θ = 7, Fcos θ = 4 ⇒ F² = 7² + 4² | M1 | |
| For tan θ = 7/4 | M1 | |
| For F = 7 and θ = 60.3° | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Magnitude 7 N | B1 | 2 |
| Direction opposite to that of the force of magnitude 7 N | B1 |
**(i)**
$[7 = F\cos\theta \text{ and } 4 = F\sin\theta \Rightarrow F^2 = 7^2 + 4^2 \text{ (or } \tan\theta = 4/7)]$ | M1 | For stating $F^2 = 7^2 + 4^2$ directly or for resolving in the i and j directions and eliminating $\theta$ or F
$F = 8.06$ | A1 |
$[7 = 8.06\cos\theta \text{ or } 4 = 8.06\sin\theta]$ or (or 7 = Fcos29.7° or 4 = Fsin29.7°) | M1 | For stating $\tan\theta = 4/7$ directly or for substituting for F or for θ into $7 = F\cos\theta$ or $4 = F\sin\theta$
$\theta = 29.7$ | A1 | 4 | Allow 29.8 from sin⁻¹(4 ÷ 8.06)
| | | SR for candidates who mix sine and cosine (max 3/4)
| | | Fsin θ = 7, Fcos θ = 4 ⇒ F² = 7² + 4² | M1 |
| | | For tan θ = 7/4 | M1 |
| | | For F = 7 and θ = 60.3° | A1 |
**(ii)**
Magnitude 7 N | B1 | 2 | Any equivalent form
Direction opposite to that of the force of magnitude 7 N | B1 |\includegraphics{figure_3}
A particle is in equilibrium on a smooth horizontal table when acted on by the three horizontal forces shown in the diagram.
\begin{enumerate}[label=(\roman*)]
\item Find the values of $F$ and $\theta$. [4]
\item The force of magnitude 7 N is now removed. State the magnitude and direction of the resultant of the remaining two forces. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2007 Q3 [6]}}