Question

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the tension in the rope and the angle the rope makes with the horizontal

Answer 1

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude

`W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N`
the force of the wind F, acting horizontally, with intensity

`F=12 N`
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):

`T \cos \alpha -F=0`

`T \sin \alpha -W=`
By dividing the second equation by the first one, we get

`\tan \alpha = (W)/(F)= (24.5 N)/(12 N)=2.04`
From which we find

`\alpha = 63.8 ^(\circ)`
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:

`T= (F)/(\cos \alpha)= (12 N)/(\cos 63.8^(\circ))=27.2 N`