| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Moderate -0.3 This is a standard two-string equilibrium problem requiring resolution of forces in two directions. While it involves trigonometry and simultaneous consideration of horizontal and vertical equilibrium, it follows a completely routine method taught in M1 with no problem-solving insight required. The angles (30° and 60°) give nice exact trigonometric values, making calculations straightforward. Slightly easier than average due to its textbook nature. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = 20\sqrt{3}\), 34.6, 34.64,... | M1 A2 (1,0) A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(m = \frac{40}{g}\) (\(\approx\) 4.1), 4.08 | M1 A2 (1,0) A1 | (4) [8] |
**Part (a):**
R($\rightarrow$): $20\cos 30° = T \cos 60°$
$T = 20\sqrt{3}$, 34.6, 34.64,... | M1 A2 (1,0) A1 | (4)
**Part (b):**
R($\uparrow$): $mg = 20\sin 30° + T \sin 60°$
$m = \frac{40}{g}$ ($\approx$ 4.1), 4.08 | M1 A2 (1,0) A1 | (4) [8]\includegraphics{figure_1}
A particle of mass $m$ kg is attached at $C$ to two light inextensible strings $AC$ and $BC$. The other ends of the strings are attached to fixed points $A$ and $B$ on a horizontal ceiling. The particle hangs in equilibrium with $AC$ and $BC$ inclined to the horizontal at 30° and 60° respectively, as shown in Figure 1.
Given that the tension in $AC$ is 20 N, find
\begin{enumerate}[label=(\alph*)]
\item the tension in $BC$, [4]
\item the value of $m$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2010 Q3 [8]}}