A weight of mass m is suspended from a cord so that one section of it makes an angle `alph_(1)` with the horizontal and the other - an angle `alpha_(2)` (Fig.). Find the tension of these sections of the cord.

Answer

`T_(1)=(P cos alpha_(2))/(sin (alpha_(1)+alpha_(2))), T_(2)=(P cos alpha_(1))/(sin alpha_(1)+alpha_(2))`

Explanation

Using the sine rule, we obtain from Fig.
`(T_(1))/(sin beta_(2)) = (T_(2))/( sin beta_(1)) = (P)/(sin[pi" "(beta_(1)"|"beta_(2))])`
Noting that `beta_(1)=(pi)/(2) - alpha_(1), beta_(2)=(pi)/(2) -alpha_(2)`, we obtain the given answer.