A 1.4-kg ball is hanging from the end of a rope. The rope hangs at an angle 27° from the vertical when a 16 m/s horizontal wind is blowing. If the wind's force on the rope is negligible, what drag force in Newtons does the wind exert on the ball?
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This is a force equilibrium. We need to find balanced forces and set up an equation to find the unknown. In this case, we want to know the drag force of the wind, D, which is operating in the horizontal direction. It pushes the ball until the "weight" of the ball balances the drag force. I put weight in quotes because it's not the full weight (mg) of the ball being balanced - rather, it's the horizontal component of the TENSION force in the rope suspending the ball.
So we need an expression for this horizontal tension force. If you sketch out the ball on a rope at an angle 27° to the vertical, you'll find the horizontal component is equal to mgsin27°. This balances the drag force exactly, so now we have an equation:
\[D=mg \sin27°\]\[D=(1.4)(9.8) \sin 27°\]\[D=6.23\]
The drag force is 6.23 N. Notice it didn't matter what the speed of the wind was!