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Simora [160]
3 years ago
11

Why wind makes things that is light make them fly away if wind blow them but not heavy things?

Physics
2 answers:
klemol [59]3 years ago
8 0
Light things have less density while a heavy object is more dence
DochEvi [55]3 years ago
6 0
Let's do math before we look for information. First, what is the force that keeps you anchored to the ground? This is the force of static friction, which is Fs=μmg
F
s
μ
m
g
. What is this force opposing? The force of drag from the wind pushing on you. For the velocities involved (a high Reynolds number regime), the drag is quadratic in velocity, Fd=12ρv2CdA
F
d
1
2
ρ
v
2
C
d
A
, where ρ
ρ
is the density of atmosphere, v
v
is the velocity, Cd
C
d
is a dimensionless drag coefficient, and A
A
is your body's cross-sectional area. So let's set the forces equal and solve for the velocity:

v2=2μmgρCdA
v
2
2
μ
m
g
ρ
C
d
A
We'll be very ballpark about this. The density of air is ρ≈1.2 kg/m3
ρ
1.2
kg/m
3
. I'll say your mass is 50 kg
50
kg
. Per this paper, we'll say CdA≈0.84 m2
C
d
A
0.84
m
2
. Per this thread, we'll say μ=0.4
μ
0.4
.

Putting all these numbers in gives us v≈20 m/s
v
20
m/s
, or about 45 mph. But, this is just enough to make your body move (compared to standing still on the ground). It would take at least a 70 mph wind to overcome the force of gravity, and even then, that's assuming the wind keeps pushing on you with your body turned to face it (or away from it), not sideways. Hard thing to guarantee given how the body is likely to tumble or spin.

It's hard to be exact about this sort of thing, but let's just say this: going out in this kind of storm is a bad idea. The numbers aren't clear-cut enough to say you're safe, so better safe than sorry.

shareciteimprove this answer
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drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

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1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

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b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

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Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

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