Question

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

Answer 1

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

Step-by-step 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_(B)` - 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_(A)`

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