Particulae in aequilibrio unidimensionali – applicatio problematum et solutionum legis primae Newtoni

^{1. Massa obiecti, m = 10 kg, fune sustentati. Invenire tensionem in fune!} g = 10 m/ ^s²

Notum:

Massa (m) = 10 kg

Acceleratio gravitatis (g) = 10 m/ ^s²

Quaesitum: Vis tensionis (T)

solution:

ΣFy ₌ 0

T – w = 0

T = w

T = mg

T = (10 kg)(10 m/s² ⁾ = 100 kg m/ ^s²

T = 30 000 Newtoni

[irp]

2. Massa rei est 10 kg. Invenire tensionem in fune... Acceleratio gravitatis = 10 m/ ^s².

Solutio

Notum:

Massa (m) = 10 kg

Acceleratio gravitatis (g) = 10 m/ ^s².

Quaesitum: Vis tensionis (T)

solution:

^{w = pondus = mg = (10 kg)(10 m/s²)}) = 100 kg m/s²

T² ₌ vis tensionis²

_{T²x =} pars x vis tensionis² = _T² cos² ⁼ 0.7 _T²

_T²y = pars y vis tensionis ² = T² _{sin 45} ° ⁼ 0.7 _T²

T² ₌ vis tensionis²

_{T²x =} pars x vis tensionis² = _T² cos² ⁼ 0.7 _T²

_T²y = pars y vis tensionis ² = T² _{sin 45} ° ⁼ 0.7 _T²

Conditio aequilibrii ΣF = 0.

axis y:

ΣFy ₌ 0

T₁y ₊ T₂y _– w = 0

0.7T ₁ + 0.7T ₂ – 100 = 0

0.7T₁ + 0.7T₂ _{= 100} —– aequatio 1

axis x:

ΣFx ₌ 0

T²x – _T¹x = ₀

0.7T ₂ – 0.7T ₁ = 0

0.7T² = _0.7T¹​

T² = T₁ _—– aequatio ₂

Magnitudinem T₁ _determina :

0.7T₁ ₊ 0.7T₁ = ₁₀₀

1.4T ₁ = 100

T² = 1₁₀₀ / 1.4

T₁ = 71.4 _Newtoni

T₁ = T₂ _ergo T₂ ₌ 71.4 _Newtoni

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1. Particulae in aequilibrio unidimensionali
2. Particulae in aequilibrio bidimensionali
3. Aequilibrium corporum funibus et trochleis connexorum
4. Aequilibrium corporum in plano inclinato