A particle of mass 8 kg is suspended in equilibrium by two light inextensible strings which make angles of 60° and 45° above the horizontal.
(a) Draw a diagram showing the forces acting on the particle.
(b) Find the tensions in the strings.
Solution
To solve for the tensions in the strings, we resolve the forces horizontally and vertically.
Let the tensions in the strings be \(T_1\) and \(T_2\).
Resolving horizontally:
\(T_1 \cos 60^{\circ} = T_2 \cos 45^{\circ}\)
Resolving vertically:
\(T_1 \sin 60^{\circ} + T_2 \sin 45^{\circ} = 8g\)
Substituting \(g = 9.8 \text{ m/s}^2\), we have:
\(T_1 \cos 60^{\circ} = T_2 \cos 45^{\circ}\)
\(T_1 \sin 60^{\circ} + T_2 \sin 45^{\circ} = 8 \times 9.8\)
Solving these equations simultaneously gives:
\(T_1 = 58.6 \text{ N}\)
\(T_2 = 41.4 \text{ N}\)
An alternative method using Lami's Theorem:
\(\frac{T_1}{\sin 135^{\circ}} = \frac{T_2}{\sin 150^{\circ}} = \frac{80}{\sin 75^{\circ}}\)
Solving for \(T_1\) and \(T_2\) gives the same results:
\(T_1 = 58.6 \text{ N}\)
\(T_2 = 41.4 \text{ N}\)
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