Answer:
The value is [tex]F = 562.7 \ N[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 10 \ mm = 0.01 \ m[/tex]
The distance between the poles is L = 50 m
The velocity of the wind is [tex]v = 34 \ m/s[/tex]
Generally the horizontal force the cable puts on each other is mathematically represented as
[tex]F = \frac{1}{2} * C_D * \rho * v^2 * A[/tex]
Here A is the area of the rope(i.e assuming the rope to be a cylinder ) which is mathematically represented as
[tex]A = L * d[/tex]
=> [tex]A = 0.01 * 50[/tex]
=> [tex]A = 0.5[/tex]
Here [tex]C_D[/tex] is the coefficient of discharge of the rope and the value is [tex]C_D = 1.5[/tex]
[tex]\rho[/tex] is the density of air with value [tex]\rho = 1.225 \ kg/m^3[/tex]
So
[tex]F = \frac{1}{2} * C_D * \rho * v^2 * A[/tex]
=> [tex]F = \frac{1}{2} * 1.5 * 1.225 * 34^2 * 0.5[/tex]
=> [tex]F = 562.7 \ N[/tex]
