Rob

Me neither. That would have to be some SERIOUS force! Let's investigate.

velocity
v̅ = Δs
Δt
v = ds
dt

acceleration
a̅ = Δv
Δt
a = dv
dt

equations of motion
v = v0 + at
x = x0 + v0t + ½at2
v2 = v02 + 2a(x − x0)
v̅ = ½(v + v0)

newton's 2nd law
∑ F = m a
∑ F = dp
dt

weight
W = m g dry friction ? = μN centrip. accel.
ac = v2
r
ac = − ω2 r

momentum
p = m v

impulse
J = F̅ Δt
J = ⌠
⌡ F dt

impulse-momentum
F̅ Δt = m Δv
⌠
⌡ F dt = Δp

work W
= F̅Δs cos θ
W = ⌠
⌡ F · ds

work-energy
F̅Δs cos θ = ΔE
⌠
⌡ F · ds = ΔE

kinetic energy
K = ½mv2 general p.e.
ΔU = − ⌠
⌡ F · ds
F = − ∇U

gravitational p.e.
ΔUg = mgΔh efficiency
ℰ = Wout
Ein

power
P̅ = ΔW
Δt
P̅ = F̅v cos θ
P = dW
dt
P = F · v

angular velocity
ω̅ = Δθ
Δt
ω = dθ
dt
v = ω × r

angular acceleration
α̅ = Δω
Δt
α = dω
dt
a = α × r − ω2 r

equations of rotation
ω = ω0 + αt
θ = θ0 + ω0t + ½αt2
ω2 = ω02 + 2α(θ − θ0)
ω̅ = ½(ω + ω0)

2nd law for rotation
∑ τ = I α
∑ τ = dL
dt

torque
τ = rF sin θ τ = r × F

moment of inertia
I = ∑ mr2
I = ⌠
⌡ r2 dm

rotational work
W = τ̅Δθ
W = ⌠
⌡ τ · dθ

rotational power
P = τω cos θ P = τ · ω

rotational k.e.
K = ½Iω2

angular momentum
L = mrv sin θ
L = r × p
L = I ω

universal gravitation
Fg = − Gm1m2 r̂
r2


gravitational field
g = − Gm r̂
r2

gravitational p.e.
Ug = − Gm1m2
r

gravitational potential
Vg = − Gm
r

orbital speed
v = √ Gm
r

escape speed
v = √ 2Gm
r

hooke's law
F = − k Δx elastic p.e. Us = ½kΔx2 s.h.o.
T = 2π √ m
k

simple pendulum
T = 2π √ ℓ
g

frequency
? = 1
T

angular frequency
ω = 2π? density
ρ = m
V

pressure
P = F
A

mass continuity
ρ1A1v1 = ρ2A2v2

volume continuity
A1v1 = A2v2

bernoulli's equation
P1 + ρgy1 + ½ρv12 = P2 + ρgy2 + ½ρv22

dynamic viscosity
η = F̅/A
Δvx/Δz
η = F/A
dvx/dz

kinematic viscosity
ν = η
ρ

aerodynamic drag
R = ½ρCAv2 mach number
Ma = v
c

reynolds number
Re = ρvD
η

froude number
Fr = v
√gℓ

young's modulus
F = E Δℓ
A ℓ0

shear modulus
F = G Δx
A y

bulk modulus
F = K ΔV
A V0

surface tension
γ = F
ℓ

--- NOPE! Not possible!