| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Moderate -0.8 This is a standard two-string equilibrium problem requiring resolution of forces in two directions and basic trigonometry. The setup is straightforward with given angles and one tension, making it easier than average—typical textbook exercise with no novel problem-solving required, just methodical application of equilibrium conditions. |
| Spec | 3.03e Resolve forces: two dimensions3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving horizontally: \(T\cos 30° = 6\cos 50°\) | M1A1 | Complete method to find tension; both \(T_{AC}\) and '6' terms resolved |
| \(T = 4.45\) (N), \(4.5\) (N), or better | A1 | Second A1 for 4.5(N), 4.45(N) or better (4.453363194) |
| Alternatives: | ||
| Triangle of Forces: \(\frac{T_{AC}}{\sin 40°} = \frac{6}{\sin 60°}\) | M1A1 | Same equation as resolution |
| Lami's Theorem: \(\frac{T_{AC}}{\sin 140°} = \frac{6}{\sin 120°}\) | M1A1 | Same equation as resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically: \(W = 6\cos 40° + T\cos 60°\) | M1A1 | Correct no. of terms; both \(T_{AC}\) and '6' terms resolved; \(T_{AC}\) does not need to be substituted |
| \(= 6.82\) (N), \(6.8\) (N), or better | A1 | Second A1 for 6.8(N), 6.82(N) or better (6.822948256) |
| Alternatives: | ||
| Triangle of Forces: \(\frac{6}{\sin 60°} = \frac{W}{\sin 80°}\) | M1A1 | |
| Lami's Theorem: \(\frac{6}{\sin 120°} = \frac{W}{\sin 100°}\) | M1A1 | |
| Resolution in another direction e.g. along one of the strings | M1 | A1 for a correct equation |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving horizontally: $T\cos 30° = 6\cos 50°$ | M1A1 | Complete method to find tension; both $T_{AC}$ and '6' terms resolved |
| $T = 4.45$ (N), $4.5$ (N), or better | A1 | Second A1 for 4.5(N), 4.45(N) or better (4.453363194) |
| **Alternatives:** | | |
| Triangle of Forces: $\frac{T_{AC}}{\sin 40°} = \frac{6}{\sin 60°}$ | M1A1 | Same equation as resolution |
| Lami's Theorem: $\frac{T_{AC}}{\sin 140°} = \frac{6}{\sin 120°}$ | M1A1 | Same equation as resolution |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically: $W = 6\cos 40° + T\cos 60°$ | M1A1 | Correct no. of terms; both $T_{AC}$ and '6' terms resolved; $T_{AC}$ does not need to be substituted |
| $= 6.82$ (N), $6.8$ (N), or better | A1 | Second A1 for 6.8(N), 6.82(N) or better (6.822948256) |
| **Alternatives:** | | |
| Triangle of Forces: $\frac{6}{\sin 60°} = \frac{W}{\sin 80°}$ | M1A1 | |
| Lami's Theorem: $\frac{6}{\sin 120°} = \frac{W}{\sin 100°}$ | M1A1 | |
| Resolution in another direction e.g. along one of the strings | M1 | A1 for a correct equation |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ed659098-c1cf-4ee1-a12a-bf8b6c42db95-02_332_921_260_516}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle of weight $W$ newtons is attached at $C$ to two light inextensible strings $A C$ and $B C$. The other ends of the strings are attached to fixed points $A$ and $B$ on a horizontal ceiling. The particle hangs in equilibrium with $A C$ and $B C$ inclined to the horizontal at $30 ^ { \circ }$ and $50 ^ { \circ }$ respectively, as shown in Figure 1.
Given that the tension in $B C$ is 6 N , find
\begin{enumerate}[label=(\alph*)]
\item the tension in $A C$,
\item the value of $W$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q1 [6]}}