| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle with string at angle to wall |
| Difficulty | Moderate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions with a simple geometry (60° angle). The setup is standard M1 fare with only two unknowns and basic trigonometry, making it easier than average but not trivial since students must correctly identify force directions and apply equilibrium conditions. |
| Spec | 3.03b Newton's first law: equilibrium3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = 5\cos 30°\) or Lami's: \(\frac{F}{\sin 120°} = \frac{5}{\sin 90°}\) | M1 A1 | Complete method for equation in \(F\) only; condone sign errors and sin/cos confusion |
| OR \(\frac{F\sin 30°}{\sin 60°}\cos 60° + F\cos 30° = 5\) | If resolving horizontally and vertically, must eliminate \(T\) | |
| \(F = \frac{5\sqrt{3}}{2}\), 4.3 or better | A1 | cao (4.3301...) |
| N.B. \(F\sin 30° = T\sin 60°\) | M0 if using wrong angles e.g. 45° |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = 5\sin 30°\) or Lami's: \(\frac{T}{\sin 150°} = \frac{5}{\sin 90°}\) | M1 A1 | Complete method for equation in \(T\) only; condone sign errors and sin/cos confusion |
| OR \(T\cos 60° + \frac{T\sin 60°}{\sin 30°}\cos 30° = 5\) | If resolving horizontally and vertically, must eliminate \(F\) | |
| \(T = \frac{5}{2}\) (N) | A1 | cao |
| N.B. \(F\sin 30° = T\sin 60°\) | M0 if using wrong angles e.g. 45° |
## Question 1:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = 5\cos 30°$ or Lami's: $\frac{F}{\sin 120°} = \frac{5}{\sin 90°}$ | M1 A1 | Complete method for equation in $F$ only; condone sign errors and sin/cos confusion |
| **OR** $\frac{F\sin 30°}{\sin 60°}\cos 60° + F\cos 30° = 5$ | | If resolving horizontally and vertically, must eliminate $T$ |
| $F = \frac{5\sqrt{3}}{2}$, 4.3 or better | A1 | cao (4.3301...) |
| **N.B.** $F\sin 30° = T\sin 60°$ | | M0 if using wrong angles e.g. 45° |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 5\sin 30°$ or Lami's: $\frac{T}{\sin 150°} = \frac{5}{\sin 90°}$ | M1 A1 | Complete method for equation in $T$ only; condone sign errors and sin/cos confusion |
| **OR** $T\cos 60° + \frac{T\sin 60°}{\sin 30°}\cos 30° = 5$ | | If resolving horizontally and vertically, must eliminate $F$ |
| $T = \frac{5}{2}$ (N) | A1 | cao |
| **N.B.** $F\sin 30° = T\sin 60°$ | | M0 if using wrong angles e.g. 45° |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1bdc84b-c8a1-4e7c-a2ba-48b40c6a6d36-02_486_638_248_653}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle $P$ of weight 5 N is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point $O$. The particle $P$ is held in equilibrium by a force of magnitude $F$ newtons. The direction of this force is perpendicular to the string and $O P$ makes an angle of $60 ^ { \circ }$ with the vertical, as shown in Figure 1.
Find
\begin{enumerate}[label=(\alph*)]
\item the value of $F$
\item the tension in the string.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q1 [6]}}