N objeto de 15 kg esta suspendido de una cuerda A, de la que tira horizontalmente meidante de la cuerda B de manera que a cuerda forma un angulo de 30, encontrar a tension de la cuerda A y B

T_{A} = 294 N ,    T_{B}  = 254.6 N    

Explanation:

Let's apply the condition of static equilibrium, in attached we can see a diagram of the problem. Let's apply each axis

X axis

          - T_{Ax} = 0

Y axis y

        T_{Ay} - W = 0

Let's use trigonometry to find the components of the stress Ta

The angle is measured with respect to the horizontal

       cos 30 = T_{Ax} / T_{A}

       sin 30 = T_{Ay} /

       T_{Ax} = T_{A} cos 30

       T_{Ay} = T_{A} sin 30

let's substitute

       T_{B} = T_{A} cos 30

       T_{A}  sin 30 = m g

let's calculate

       T_{A} = m g / sin 30

       T_{A} = 15 9.8 / sin 30

       T_{A} = 294 N

       T_{B} = 294 cos 30

       T_{B}  = 254.6 N

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~