(a) A body P of mass q kg is suspended by two light inextensible strings AB and DB attached to a horizontal table. The strings are inclined at 30° and 60° respectively to the horizontal and the tension in AB is 48N. If the system is in equilibrium :
(i) sketch a diagram to represent the information ; (ii) calculate the tension in DB ;
Explanation
(i) 
(ii) Using Lami's theorem,
\(\frac{48}{\sin 150} = \frac{T}{\sin 120}\)
\(T = \frac{48 \sin 120}{\sin 150}\)
\(T = 83.14N\)
The tension in DB is 83.14N.
(iii) Using Lami's theorem,
\(\frac{10q}{\sin 90} = \frac{48}{\sin 150}\)
\(10q = \frac{48 \sin 90}{\sin 150}\)
\(10q = 96 \implies q = 9.6kg\)