 Дано: a = 1 м/с
m = 1 кг
R = -μVn
μ = 0.5 Н*с/м
α = 60˚
Найти: S-? после схода с лопатки толкателя
Решение: mx'' = Фe + R1 + R2 + N
Фe = ma
Va = Ve + Vr => R = -μVe - μVr
Пусть R1 = - μVe; R2 = - μVr => R=R1+R2
mx'' = macos60 + μVecos60 - μVr Ve = at
x'' = 0.5 + 0.25t - 0.5x'
Пусть x'=P => x''=P'
P' + 0.5P = 0.5 + 0.25t
P=UV => P'=U'V + V'U
U'V + V'U + 0.5UV = 0.5 + 0.25t
U'V + U(V' + 0.5V) = 0.5 + 0.25t
U' + 0.5V = 0
dV/V = -0.5dt
lnV = -0.5t
V = e-0.5t
U'V = 0.5 + 0.25t => U'e-0.5t = 0.5 + 0.25t
dU = [(0.5+0.25t)/(e-0.5t)]dt => U = ∫ 0.5e1/2tdt + ∫ 0.25t*e0.5tdt =>
|U = t dV = e0.5tdt | dU = dt V = 2e0.5t|
U = e0.5t + 0.5t*e0.5t - 0.5∫e0.5tdt =>
U = 0.5t*e0.5t
P = UV = 0.5t*e0.5t*e-0.5t + C => P = 0.5t + Ce-0.5t
x' = 0.5t + C*e-0.5t н.у. x'=0, t=0 => C=0 =>
1∫0 dx = t∫0 0.5tdt => 1 = 0.5*t2/2 =>
t2 = 4 => t = 2 с =>
Ve = x' = 0.5*2 = 1 м/с
Ve = at = 2 м/с
x: Ve - Vrcos60 = 1.5
y: Vrcos30 = √(3)/2 =>
VA = √(1.52 + 3/4) = √ (9/4+3/4) = √ (3) м/с
maa = -R
dVa/dt = -μdρ/dt
0∫√(3)dVa = -0.5S∫0dρ =>
- √(3) = -0.5ρ =>
ρ = 3.46 м
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