The free-body diagram of the forces acting on the flag is in the picture in attachment.
We have: the weight, downward, with magnitude
[tex]W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N[/tex]
the force of the wind F, acting horizontally, with intensity
[tex]F=12 N[/tex]
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
[tex]T \cos \alpha -F=0[/tex]
[tex]T \sin \alpha -W=[/tex]
By dividing the second equation by the first one, we get
[tex]\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04 [/tex]
From which we find
[tex]\alpha = 63.8 ^{\circ}[/tex]
which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope:
[tex]T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N [/tex]
