A uniform chain of mass M = 4.8 kg hangs in vertical plane as shown in the fig.
(a) Show that horizontal component of tension is same throughout the chain.
(b) Find tension in the chain at point P where the chain makes an angle `theta = 15^(@)` with horizontal.
(c) Find mass of segment AP of the chain.
[Take `g = 10 m//s^(2), "cos" 15^(@) = 0.96 , "sin"15^(@) = 0.25`]

Answer

(b) `T_(P) = 21. 65 N`
(c) 3.05 kg