The 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.
(b) Magnitude of the force p cos \(\alpha\) = \(\frac{(3)^2 + (2.4)^2 - (1.8(^2}{2(3)(2.5)}\) cos \(\alpha\) = = \(\frac{9 + 5.76 - 3.24}{14.4}\) cos \(\alpha\) = \(\frac{11.52}{14.4}\) cos \(\alpha\) = 0.8 \(\alpha\) = cos \(^{-1}\) 0.8 = 36.86\(^o\) \(\frac{p}{2.4} = cos 36.86^o\) p = 2.4 x 0.8 p = 1.92N
(c) The tension T in the string T\(^2\) = 8\(^2\) + 1.92\(^2\) T\(^2\) = 64 + 3.59 Tv = 67.69 T = \(\sqrt{67.69}\) T = 8.23N
