| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions and basic trigonometry. Part (a) involves resolving with one unknown, part (b) requires solving simultaneous equations with two unknowns. While it requires careful setup and algebraic manipulation, it follows a well-practiced procedure with no novel insight needed, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4\sin 30 + F\sin 60 - 6 = 0\) | M1 | Resolve forces vertically and equate to zero |
| Correct equation | A1 | |
| \(F = 4.62\) | A1 | Allow \(F = \dfrac{8}{\sqrt{3}}\) or \(F = \dfrac{8}{3}\sqrt{3}\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resolve forces either vertically or horizontally | M1 | |
| \(F\sin\alpha + 4\sin 30 - 6 = 0\) and \(F\cos\alpha + 3 - 4\cos 30 = 0\) | A1 | Both equations correct: \([F\sin\alpha = 4]\), \([F\cos\alpha = 0.464102...]\) |
| \(F^2 = 4^2 + 0.464^2\) or \(F = \dfrac{4}{\sin 83.4} = \dfrac{0.464}{\cos 83.4}\) | M1 | Attempt to solve for \(F\) using Pythagoras or from a value found for \(\alpha\) |
| \(\alpha = \tan^{-1}\!\left(\dfrac{4}{0.464}\right)\) or \(\alpha = \sin^{-1}\!\left(\dfrac{4}{4.03}\right) = \cos^{-1}\!\left(\dfrac{0.464}{4.03}\right)\) | M1 | Attempt to solve for \(\alpha\) using trigonometry or from a value found for \(F\) |
| \(F = 4.03\) and \(\alpha = 83.4\) | A1 | Both correct as shown \([F = 4.0268\ldots,\ \alpha = 83.382\ldots]\) |
| Total: 5 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4\sin 30 + F\sin 60 - 6 = 0$ | M1 | Resolve forces vertically and equate to zero |
| Correct equation | A1 | |
| $F = 4.62$ | A1 | Allow $F = \dfrac{8}{\sqrt{3}}$ or $F = \dfrac{8}{3}\sqrt{3}$ |
| **Total: 3** | | |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolve forces either vertically or horizontally | M1 | |
| $F\sin\alpha + 4\sin 30 - 6 = 0$ and $F\cos\alpha + 3 - 4\cos 30 = 0$ | A1 | Both equations correct: $[F\sin\alpha = 4]$, $[F\cos\alpha = 0.464102...]$ |
| $F^2 = 4^2 + 0.464^2$ **or** $F = \dfrac{4}{\sin 83.4} = \dfrac{0.464}{\cos 83.4}$ | M1 | Attempt to solve for $F$ using Pythagoras or from a value found for $\alpha$ |
| $\alpha = \tan^{-1}\!\left(\dfrac{4}{0.464}\right)$ **or** $\alpha = \sin^{-1}\!\left(\dfrac{4}{4.03}\right) = \cos^{-1}\!\left(\dfrac{0.464}{4.03}\right)$ | M1 | Attempt to solve for $\alpha$ using trigonometry or from a value found for $F$ |
| $F = 4.03$ and $\alpha = 83.4$ | A1 | Both correct as shown $[F = 4.0268\ldots,\ \alpha = 83.382\ldots]$ |
| **Total: 5** | | |5\\
\includegraphics[max width=\textwidth, alt={}, center]{9ac08732-e825-473a-943c-8ad8e9e0bc17-08_572_572_262_790}
Coplanar forces, of magnitudes $F \mathrm {~N} , 3 \mathrm {~N} , 6 \mathrm {~N}$ and 4 N , act at a point $P$, as shown in the diagram.
\begin{enumerate}[label=(\alph*)]
\item Given that $\alpha = 60$, and that the resultant of the four forces is in the direction of the 3 N force, find $F$.
\item Given instead that the four forces are in equilibrium, find the values of $F$ and $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q5 [8]}}