| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | January |
| Marks | 7 |
| 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 perpendicular directions. Students apply routine mechanics techniques (resolving horizontally and vertically, using given angles) with no conceptual challenges or novel problem-solving required. Easier than the average A-level question due to its straightforward setup and direct application of Newton's first law. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks |
|---|---|
| Resolve parallel to AB: \(T_A \cos 30° = T_B \cos 45°\) | M1A1 |
| Resolve perpendicular to AB: \(W = T_A \sin 30° + T_B \sin 45°\) | M1A1 |
| Solve for \(T_A\) or \(T_B\) | DM1 |
| \(T_A = \frac{W}{1-\sqrt{3}}\) or \(0.73W\) (or better) | A1 |
| \(T_B = \frac{W}{1+\sqrt{3}}\) or \(0.90W\) (or better) | A1 |
| Answer | Marks |
|---|---|
| Sine rule for \(T_A\): \(\frac{T_A}{W} = \frac{\sin 45°}{\sin 75°}\) | M1A1 |
| Sine rule for \(T_B\): \(\frac{T_B}{W} = \frac{\sin 60°}{\sin 75°}\) | M1A1 |
| Solve for \(T_A\) or \(T_B\): \(T_A = 0.73W\) (or better) | DM1A1 |
| \(T_B = 0.90W\) (or better) | A1 |
**Method 1: Resolving forces**
Resolve parallel to AB: $T_A \cos 30° = T_B \cos 45°$ | M1A1
Resolve perpendicular to AB: $W = T_A \sin 30° + T_B \sin 45°$ | M1A1
Solve for $T_A$ or $T_B$ | DM1
$T_A = \frac{W}{1-\sqrt{3}}$ or $0.73W$ (or better) | A1
$T_B = \frac{W}{1+\sqrt{3}}$ or $0.90W$ (or better) | A1
(7 marks)
**Alternative: Triangle of forces**
Sine rule for $T_A$: $\frac{T_A}{W} = \frac{\sin 45°}{\sin 75°}$ | M1A1
Sine rule for $T_B$: $\frac{T_B}{W} = \frac{\sin 60°}{\sin 75°}$ | M1A1
Solve for $T_A$ or $T_B$: $T_A = 0.73W$ (or better) | DM1A1
$T_B = 0.90W$ (or better) | A1
(7 marks)
**Notes for Question 1**
First M1 for resolving horizontally with usual rules.
First A1 for a correct equation.
Second M1 for resolving vertically with usual rules.
Second A1 for a correct equation.
Third DM1, dependent on both previous M marks, for solving for either $T_A$ or $T_B$.
Third A1 for $T_A = 0.73W$ or better or any correct surd answer but A0 for $\frac{W}{k}$ where $k$ is a decimal. Allow invisible brackets.
Fourth A1 for $T_B = 0.90W$ or better (0.9W is A0) or any correct surd answer but A0 for $\frac{W}{k}$ where $k$ is a decimal.
Alternative using sine rule or Lami's Theorem: First M1A1 for $\frac{T_A}{W} = \frac{\sin 45°}{\sin 75°}$ or equivalent (e.g. allow $\sin 105°$ or reciprocals).
Second M1 for $\frac{T_B}{W} = \frac{\sin 60°}{\sin 75°}$ (allow $\sin 30°$ and/or $\sin 105°$).
Second A1 for $\frac{T_B}{W} = \frac{\sin 60°}{\sin 75°}$.
Third DM1, dependent on either previous M mark, for solving for either $T_A$ or $T_B$.
Third A1 for $T_A = 0.73W$ or better or any correct surd answer but A0 for $\frac{W}{k}$ where $k$ is a decimal.
Fourth A1 for $T_B = 0.90W$ or better or any correct surd answer but A0 for $\frac{W}{k}$ where $k$ is a decimal.
**N.B.** If they assume that the tensions are the same, can score max: M0A0M1A0DM0A0A0.
If they use the same angles, can score max: M1A0M1A0DM0A0A0.
---1.
\begin{figure}[h]
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\caption{Figure 1}
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\end{figure}
A particle of weight $W$ 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 the strings in a vertical plane and with $A C$ and $B C$ inclined to the horizontal at $30 ^ { \circ }$ and $45 ^ { \circ }$ respectively, as shown in Figure 1.
Find, in terms of $W$,\\
(i) the tension in $A C$,\\
(ii) the tension in $B C$.\\
VILLI SIHI NITIIIUM ION OC\\
VILV SIHI NI JAHM ION OC\\
VI4V SIHI NI JIIIM ION OC
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\hfill \mbox{\textit{Edexcel M1 2018 Q1 [7]}}