A small smooth ring R of weight 8.5 N is threaded on a light inextensible string. The ends of the
string are attached to ﬁxed points A and B, with A vertically above B. A horizontal force of magnitude
15.5 N acts on R so that the ring is in equilibrium with angle ARB = 90◦
. The part AR of the string
makes an angle θ with the horizontal and the part BR makes an angle θ with the vertical (see diagram).
The tension in the string is T N. Show that T sin θ = 12 and T cos θ = 3.5 and hence ﬁnd θ.

Question

A small smooth ring R of weight 8.5 N is threaded on a light inextensible string. The ends of the
string are attached to ﬁxed points A and B, with A vertically above B. A horizontal force of magnitude
15.5 N acts on R so that the ring is in equilibrium with angle ARB = 90◦
. The part AR of the string
makes an angle θ with the horizontal and the part BR makes an angle θ with the vertical (see diagram).
The tension in the string is T N. Show that T sin θ = 12 and T cos θ = 3.5 and hence ﬁnd θ.