Moderate -0.3 This is a standard two-string equilibrium problem requiring basic trigonometry to find angles, then resolving forces horizontally and vertically. The geometry is straightforward (Pythagoras to find horizontal distances, then simple trig), and the force resolution is routine. Slightly easier than average due to the clear setup and standard method, though it requires careful arithmetic with the given numerical values.
4
\includegraphics[max width=\textwidth, alt={}, center]{631ddcd9-17c0-4a15-8671-40788c3a84d3-2_396_880_1996_630}
A particle \(P\) of weight 21 N is attached to one end of each of two light inextensible strings, \(S _ { 1 }\) and \(S _ { 2 }\), of lengths 0.52 m and 0.25 m respectively. The other end of \(S _ { 1 }\) is attached to a fixed point \(A\), and the other end of \(S _ { 2 }\) is attached to a fixed point \(B\) at the same horizontal level as \(A\). The particle \(P\) hangs in equilibrium at a point 0.2 m below the level of \(A B\) with both strings taut (see diagram). Find the tension in \(S _ { 1 }\) and the tension in \(S _ { 2 }\).
Tension in \(S_1\) is 13 N, tension in \(S_2\) is 20 N
A1
6 marks total
$[T_1 \sin APN = T_2 \sin BPN]$ | M1 | For resolving forces horizontally
$(12 \div 13)T_1 = (15 \div 25)T_2$ or $T_1 \sin 67.4° = T_2 \sin 36.9°$ | A1 | AEF
$[T_1 \cos APN + T_2 \cos BPN = 21]$ | M1 | For resolving forces vertically
$(5 \div 13)T_1 + (20 \div 25)T_2 = 21$ or $T_1 \cos 67.4° + T_2 \cos 36.9° = 21$ | A1 | AEF
$\quad$ | M1 | For solving for $T_1$ and $T_2$
Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total
### Alternative solution using Lami's Theorem
$[T_1/\sin(180 - BPN) = 21/\sin(APN + BPN)]$ | M1 | For using Lami's Theorem to form an equation in $T_1$
$T_1/\sin(180 - \cos^{-1}(20/25)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_1/\sin(180 - 36.9) = 21/\sin(36.9 + 67.4)$ | A1 | AEF
$[T_2/\sin(180 - APN) = 21/\sin(APN + BPN)]$ | M1 | For using Lami's Theorem to form an equation in $T_2$
$T_2/\sin(180 - \cos^{-1}(20/25)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_2/\sin 67.4° = 21/\sin(180 - (36.9 + 67.4))$ | A1 | AEF
$\quad$ | M1 | For solving for $T_1$ and $T_2$
Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total
### Alternative solution using Sine Rule
$[T_1/\sin BPN = 21/\sin(180 - (APN + BPN))]$ | M1 | For using the Sine Rule on a triangle of forces to form an equation in $T_1$
$T_1/(15/25) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_1/\sin 36.9° = 21/\sin(180 - (36.9 + 67.4))$ | A1 | AEF
$[T_2/\sin APN = 21/\sin(180 - (APN + BPN))]$ | M1 | For using the Sine Rule to form an equation in $T_2$
$T_2/(12/13) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_2/\sin 67.4° = 21/\sin(180 - (36.9 + 67.4))$ | A1 | AEF
$\quad$ | M1 | For solving for $T_1$ and $T_2$
Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total
---
4\\
\includegraphics[max width=\textwidth, alt={}, center]{631ddcd9-17c0-4a15-8671-40788c3a84d3-2_396_880_1996_630}
A particle $P$ of weight 21 N is attached to one end of each of two light inextensible strings, $S _ { 1 }$ and $S _ { 2 }$, of lengths 0.52 m and 0.25 m respectively. The other end of $S _ { 1 }$ is attached to a fixed point $A$, and the other end of $S _ { 2 }$ is attached to a fixed point $B$ at the same horizontal level as $A$. The particle $P$ hangs in equilibrium at a point 0.2 m below the level of $A B$ with both strings taut (see diagram). Find the tension in $S _ { 1 }$ and the tension in $S _ { 2 }$.
\hfill \mbox{\textit{CAIE M1 2012 Q4 [6]}}