| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions with given angle information (tan α = 3/4 implies sin α = 3/5, cos α = 4/5). Part (a) involves routine application of equilibrium conditions and Pythagoras to find tensions. Part (b) tests conceptual understanding of force systems. Slightly easier than average due to perpendicular cables simplifying the geometry and straightforward trigonometry. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\tan\alpha = \frac{3}{4} (3.4.5 \text{ Pythag. triple}) \text{ so } \sin\alpha = \frac{3}{5}, \cos\alpha = \frac{4}{5}\) | B1 | |
| \(\text{resolve} \rightarrow: T_B\sin\alpha - T_A\cos\alpha = 0\) | M1 | |
| \(\frac{3}{5}T_B = \frac{4}{5}T_A \therefore T_B = \frac{4}{3}T_A\) | A1 | |
| \(\text{resolve } \uparrow: T_A\sin\alpha + T_B\cos\alpha - 1000g = 0\) | M1 | |
| \(\frac{3}{5}T_A + \frac{4}{5}T_A(\frac{4}{3}) = 1000g\) | M1 | |
| \(\frac{5}{3}T_A = 1000g \therefore T_A = 600g = 5880 \text{ N}\) | A1 | |
| \(\text{hence } T_B = \frac{4}{3}T_A = 7840 \text{ N}\) | A1 | |
| (b) tension in both cables will increase | B2 | (9 marks) |
(a) $\tan\alpha = \frac{3}{4} (3.4.5 \text{ Pythag. triple}) \text{ so } \sin\alpha = \frac{3}{5}, \cos\alpha = \frac{4}{5}$ | B1 |
$\text{resolve} \rightarrow: T_B\sin\alpha - T_A\cos\alpha = 0$ | M1 |
$\frac{3}{5}T_B = \frac{4}{5}T_A \therefore T_B = \frac{4}{3}T_A$ | A1 |
$\text{resolve } \uparrow: T_A\sin\alpha + T_B\cos\alpha - 1000g = 0$ | M1 |
$\frac{3}{5}T_A + \frac{4}{5}T_A(\frac{4}{3}) = 1000g$ | M1 |
$\frac{5}{3}T_A = 1000g \therefore T_A = 600g = 5880 \text{ N}$ | A1 |
$\text{hence } T_B = \frac{4}{3}T_A = 7840 \text{ N}$ | A1 |
(b) tension in both cables will increase | B2 | (9 marks)\includegraphics{figure_2}
Figure 2 shows a cable car $C$ of mass 1 tonne which has broken down. The cable car is suspended in equilibrium by two perpendicular cables $AC$ and $BC$ which are attached to fixed points $A$ and $B$, at the same horizontal level on either side of a valley. The cable $AC$ is inclined at an angle $\alpha$ to the horizontal where $\tan \alpha = \frac{3}{4}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the cable $AC$ is 5880 N and find the tension in the cable $BC$. [7 marks]
A gust of wind then blows along the valley.
\item Explain the effect that this will have on the tension in the two cables. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q3 [9]}}