Moderate -0.3 This is a standard equilibrium problem requiring resolution of forces in two directions and Pythagoras to find angles. The setup is straightforward with clear geometry, requiring only basic trigonometry and simultaneous equations. Slightly easier than average due to the simple geometry and routine application of equilibrium conditions, though the numerical work with non-standard values prevents it from being trivial.
3
\includegraphics[max width=\textwidth, alt={}, center]{fd534430-2619-4078-ad0a-2355e656e121-2_307_857_1695_644}
A particle \(P\) of mass 1.05 kg is attached to one end of each of two light inextensible strings, of lengths 2.6 m and 1.25 m . The other ends of the strings are attached to fixed points \(A\) and \(B\), which are at the same horizontal level. \(P\) hangs in equilibrium at a point 1 m below the level of \(A\) and \(B\) (see diagram). Find the tensions in the strings.
For using Lami's rule to find equations for \(T_A\) and \(T_B\)
M1
For solving for \(T_A\) and \(T_B\)
Tension in AP is 6.5 N and tension in BP is 10 N
A1
6 marks
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| | M1 | For resolving forces on P vertically |
| $T_A(1/2.6) + T_B(1/1.25) = 10.5$ | A1 | |
| | M1 | For resolving forces on P horizontally |
| $T_A(2.4/2.6) = T_B(0.75/1.25)$ | A1 | |
| | M1 | For solving for $T_A$ and $T_B$ |
| Tension in AP is 6.5 N and tension in BP is 10 N | A1 | 6 marks |
**First Alternative:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $75.7(5)°$ opposite to 10.5 N; $36.8(7)°$ opposite to $T_A$; $67.3(8)°$ opposite to $T_B$ | M1, A1 | For finding two angles in the triangle of forces |
| $T_A \div \sin 36.8(7) = 10.5 \div \sin 75.7(5)$ and $T_B \div \sin 67.3(8) = 10.5 \div \sin 75.7(5)$ | M1, A1 | For using the sine rule to find equations for $T_A$ and $T_B$ |
| | M1 | For solving for $T_A$ and $T_B$ |
| Tension in AP is 6.5 N and tension in BP is 10 N | A1 | 6 marks |
**Second Alternative:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $104.2(5)°$ opposite to 10.5 N; $143.1(3)°$ opposite to $T_A$; $112.6(2)°$ opposite to $T_B$ | M1, A1 | For finding angles at P in the space diagram |
| $T_A \div \sin 143.1(3) = 10.5 \div \sin 104.2(5)$ and $T_B \div \sin 112.6(2) = 10.5 \div \sin 104.2(5)$ | M1, A1 | For using Lami's rule to find equations for $T_A$ and $T_B$ |
| | M1 | For solving for $T_A$ and $T_B$ |
| Tension in AP is 6.5 N and tension in BP is 10 N | A1 | 6 marks |
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3\\
\includegraphics[max width=\textwidth, alt={}, center]{fd534430-2619-4078-ad0a-2355e656e121-2_307_857_1695_644}
A particle $P$ of mass 1.05 kg is attached to one end of each of two light inextensible strings, of lengths 2.6 m and 1.25 m . The other ends of the strings are attached to fixed points $A$ and $B$, which are at the same horizontal level. $P$ hangs in equilibrium at a point 1 m below the level of $A$ and $B$ (see diagram). Find the tensions in the strings.
\hfill \mbox{\textit{CAIE M1 2013 Q3 [6]}}