7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del extremo izquierdo de la viga, como se muestra en la figura. La cuerda que sostiene la viga forma un ángulo θ = 60° con la horizontal. Todo el sistema se encuentra en equilibrio mecánico e. Si el alambre puede soportar una tensión máxima de 900 N, determinar ¿cuál es la distancia máxima “x” a la que se puede colocar el bloque antes de que se rompa el alambre? Sugerencia: aplicar nuevamente la condición de equilibrio rotacional tomando como referencia el extremo izquierdo de la viga, donde esta vez la incógnita es x.

 x =  0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

      sin 60 = / T

      T_{y} = T sin 60

       cos 60 = Tₓ / T

      Tₓ = T cos 60

we apply the equation

       ∑ τ = 0

       -W L / 2 - w x + T_{y} L = 0

the length of the bar is L = 6m

           -Mg 6/2 - m g x + T sin 60 6 = 0

             x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

             x = (4676 - 5880) / 6860

             x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

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