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

1. Find the tension force T₁T₂Et T₃. Ignore cord’s massa.

Solutio

(a) Free-body diagram for object (b) Free-body diagram for cord

applicare Prima lex Newtoni on the object :

ΣF_y = 0

T₁ – w = 0

T₁ = w = m g

T₁ = (5 kg)(9.8 m/s²)

T₁ = 49 kg m/s²

T₁ = 49 N

Apply Newton’s first law on the cord :

ΣF_x = 0

T_3x - T _2x = 0

T₃ XXX cos^o - T₂ XXX cos^o = 0

T 0.87₃ – 0.77 T₂ = 0

T 0.87₃ = 0.77 T₂

T₂ = 0.87 T₃ / 0.77 = 1.1 T₃ ———- Equation 1

-

ΣF_y = 0

T_3y T +_2y - T_1y = 0

T₃ peccatum 30^o T +₂ peccatum 40^o - T₁ = 0

T 0.5₃ + 0.64 T₂ – 49 N = 0 ———- Equation 2

Substituting T₂ in the equation 2 into the equation 2 :

T 0.5₃ + 0.64 (1.1 T₃) – 49 N = 0

T 0.5₃ + 0.70 T₃ - III XI =

T 1.2₃ - III XI =

T 1.2₃ = 49

T₃ = 49 / 1.2

T₃ = 41 N

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T₂ = 1.1 T₃

T₂ = (1.1)(40.8 N)

T₂ = 45 N

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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