| 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 with string at angle to wall |
| Difficulty | Moderate -0.8 This is a straightforward two-force equilibrium problem requiring resolution of forces in two perpendicular directions and basic trigonometry. It's a standard M1 exercise with clear setup, routine application of Newton's first law, and simple algebraic manipulation—easier than average A-level questions which typically require more steps or conceptual insight. |
| Spec | 3.03e Resolve forces: two dimensions3.03f Weight: W=mg3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Resolving horizontally: \(5 = T\cos 65°\) | M1A1 | First M1 for resolving horizontally with correct no. of terms and \(T\) term resolved. First A1 for a correct equation in \(T\) only |
| \(T = 12, 11.8,\) or better (N) | A1 | Second A1 for 12 (N) or 11.8 (N) or better |
| Alternative: Lami's Theorem \(\frac{T}{\sin 90°} = \frac{5}{\sin 155°}\) | M1A1 | Same equation as resolution |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Resolving vertically: \(W = T\cos 25°\) | M1A1 | First M1 for resolving vertically with correct no. of terms and \(T\) term resolved. First A1 for correct equation in \(T\) only |
| \(= 11.8\cos 25° = 11, 10.7\) or better (N) | A1 | Second A1 for 11 (N), 10.7 (N) or better |
| Alternative: Triangle of forces: \(W = 5\tan 65°\) | M1A1 | |
| Alternative: Lami's \(\frac{T}{\sin 90°} = \frac{W}{\sin 115°}\) | M1A1 | |
| Or: Resolution along the string | M1 A1 | M1 (usual criteria), A1 for correct equation |
# Question 1:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| Resolving horizontally: $5 = T\cos 65°$ | M1A1 | First M1 for resolving horizontally with correct no. of terms and $T$ term resolved. First A1 for a correct equation in $T$ only |
| $T = 12, 11.8,$ or better (N) | A1 | Second A1 for 12 (N) or 11.8 (N) or better |
| Alternative: Lami's Theorem $\frac{T}{\sin 90°} = \frac{5}{\sin 155°}$ | M1A1 | Same equation as resolution |
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| Resolving vertically: $W = T\cos 25°$ | M1A1 | First M1 for resolving vertically with correct no. of terms and $T$ term resolved. First A1 for correct equation in $T$ only |
| $= 11.8\cos 25° = 11, 10.7$ or better (N) | A1 | Second A1 for 11 (N), 10.7 (N) or better |
| Alternative: Triangle of forces: $W = 5\tan 65°$ | M1A1 | |
| Alternative: Lami's $\frac{T}{\sin 90°} = \frac{W}{\sin 115°}$ | M1A1 | |
| Or: Resolution along the string | M1 A1 | M1 (usual criteria), A1 for correct equation |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b896c631-00a0-46c5-bce9-16d65f6e3095-02_586_506_285_708}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle $P$ of weight $W$ newtons is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point $O$. A horizontal force of magnitude 5 N is applied to $P$. The particle $P$ is in equilibrium with the string taut and with $O P$ making an angle of $25 ^ { \circ }$ to the downward vertical, as shown in Figure 1.
Find
\begin{enumerate}[label=(\alph*)]
\item the tension in the string,
\item the value of $W$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q1 [6]}}