Moderate -0.3 This is a standard statics problem involving three strings meeting at a point in equilibrium. Students must resolve forces horizontally and vertically at point A, using the given angles (30° and 60°) and the known weight (50N). The solution requires routine application of equilibrium conditions (ΣFₓ = 0, ΣFᵧ = 0) with straightforward trigonometry. While it involves multiple steps and careful angle work, it's a textbook exercise with no novel insight required, making it slightly easier than average for A-level mechanics.
3
\includegraphics[max width=\textwidth, alt={}, center]{2026cad4-8494-4139-ad21-d8a17ac2b955-2_479_771_1356_687}
Each of three light strings has a particle attached to one of its ends. The other ends of the strings are tied together at a point \(A\). The strings are in equilibrium with two of them passing over fixed smooth horizontal pegs, and with the particles hanging freely. The weights of the particles, and the angles between the sloping parts of the strings and the vertical, are as shown in the diagram. Find the values of \(W _ { 1 }\) and \(W _ { 2 }\). [0pt]
[6]
Either: triangle of forces diagram with angles 40°, 80°, 60° and sides \(W_1\)N, \(W_2\)N, 5N
M1
For correct triangle of forces or for resolving forces at the knot either horizontally or vertically
Or: \(W_1\sin40° = W_2\sin60°\) and \(W_1\cos40° + W_2\cos60° = 5\)
A1
For correct angles and sides marked on triangle, or for two correct equations in \(W_1\) and \(W_2\)
\(\frac{W_1}{\sin60°} = \frac{5}{\sin80°}\)
M1
For using the sine rule to obtain an equation in \(W_1\) or \(W_2\) only, or for eliminating \(W_2\) (or \(W_1\)) from simultaneous equations
\(W_1 = 4.40\)
A1
\(\frac{W_2}{\sin40°} = \frac{5}{\sin80°}\)
M1
For using the sine rule again to obtain an equation in \(W_2\) or \(W_1\) only, or for back substitution
\(W_2 = 3.26\)
A1
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either: triangle of forces diagram with angles 40°, 80°, 60° and sides $W_1$N, $W_2$N, 5N | M1 | For correct triangle of forces or for resolving forces at the knot either horizontally or vertically |
| Or: $W_1\sin40° = W_2\sin60°$ and $W_1\cos40° + W_2\cos60° = 5$ | A1 | For correct angles and sides marked on triangle, or for two correct equations in $W_1$ and $W_2$ |
| $\frac{W_1}{\sin60°} = \frac{5}{\sin80°}$ | M1 | For using the sine rule to obtain an equation in $W_1$ or $W_2$ only, or for eliminating $W_2$ (or $W_1$) from simultaneous equations |
| $W_1 = 4.40$ | A1 | |
| $\frac{W_2}{\sin40°} = \frac{5}{\sin80°}$ | M1 | For using the sine rule again to obtain an equation in $W_2$ or $W_1$ only, or for back substitution |
| $W_2 = 3.26$ | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{2026cad4-8494-4139-ad21-d8a17ac2b955-2_479_771_1356_687}
Each of three light strings has a particle attached to one of its ends. The other ends of the strings are tied together at a point $A$. The strings are in equilibrium with two of them passing over fixed smooth horizontal pegs, and with the particles hanging freely. The weights of the particles, and the angles between the sloping parts of the strings and the vertical, are as shown in the diagram. Find the values of $W _ { 1 }$ and $W _ { 2 }$.\\[0pt]
[6]
\hfill \mbox{\textit{CAIE M1 2005 Q3 [6]}}