Answer
(a) The magnitude of the wind's force exerted on the tire is $0.21~W$
(b) The tension in the rope is $1.02~W$
Work Step by Step
(a) The vertical component of the tension in the rope is equal in magnitude to the weight.
$W = T~cos~\theta$
The horizontal component of the tension is equal in magnitude to the force of the wind $F_w$:
$F_w = T~sin~\theta$
We can divide the second equation by the first equation:
$\frac{F_w}{W} = \frac{sin~\theta}{cos~\theta}$
$F_w = W~tan~\theta$
$F_w = W~tan~12^{\circ}$
$F_w = 0.21~W$
The magnitude of the wind's force exerted on the tire is $0.21~W$
(b) We can find the tension in the rope:
$T~cos~\theta = W$
$T = \frac{W}{cos~\theta}$
$T = \frac{W}{cos~12^{\circ}}$
$T = 1.02~W$
The tension in the rope is $1.02~W$
