Apple Picking

This weekend I went to an apple orchard. Last year, I got to pick off the apples directly from the tree but this year they were all on the ground. 

I began to think of this situation in terms of forces, and how the tension force holding up the apple may decrease as it gets colder or with time (a lot of biological assumptions are being made). Assuming the apple is 0.15kg, what is the tension force at which the apple will fall?

Forces	x	y
FT	-	FT
FG	-	-mg
Total	-	may

may=(0.15kg)(0m/s2)=0N

0N=FT-mg

FT=mg=1.5N

Once the tension force is less than the force of gravity, the apple will fall from the tree. 

Though in reality, there are additional forces acting on the apple. In this next scenario, I will consider wind (6.5 N) and an angled tension force (30o). What is the tension force at which the apple will fall? (If the wind accelerates the apple by 0.5 m/s).

Forces	x	y
FT	-FTsin(30o)	FTcos(30o)
FG	-	-mg
FW	FW	-
Total	max	may

FW-FTsin(30o)=max

6.5N-FT(0.5)=(0.15kg)(0.5m/s)

FT=12.9 N

The tension force of the apple is much greater with an angled tension and the force of wind. This conceptually makes sense because the force of tension is working against the additional force of wind. For the apple to fall, Fw > FTsin(30o), and/or FG > FTcos(30o).