Theory question 115
StaticsThe diagram is that of a light inextensible string of length 4.2m, whose ends are attached to two fixed points X and Y, 3m apart, and on the same horizontal level. A body of mass 800g is hung on the string at a point O, 2.4m from Y. If the system is kept in equilibrium by a horizontal force P acting on the body and the tensions are equal, calculate: (a) ∠XOY; (b) the magnitude of the force P; (c) the tension T in the string.
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OX = 1.8 m, OY = 2.4 m, XY = 3 m.
(a) 1.8² + 2.4² = 9 = 3², so ∠XOY = 90°.
OX makes 53.1° with the horizontal (cos = 0.6) and OY makes 36.9° (cos = 0.8). Weight = 0.8 × 10 = 8 N.
(c) Vertically: T sin 53.1° + T sin 36.9° = 8, so 1.4T = 8 and T ≈ 5.71 N.
(b) Horizontally: P = T cos 36.9° − T cos 53.1° = 0.2T ≈ 1.14 N.
Resolve the forces at O horizontally and vertically; take g = 10 ms⁻².