Zvidimbu zviri muchimiro chemativi maviri - kushandiswa kwematambudziko emutemo wekutanga waNewton nemhinduro dzawo

1. Find the tension force T₁, T₂, uye T₃. Ignore cord’s uremu.

mhinduro

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

Nyorera Mutemo wekutanga waNewton on the object :

ΣF_y = 0

T₁ – w = 0

T₁ = w = m g

T₁ = (5kg)(9.8m/s²)

T₁ = 49 kg m/s²

T₁ = 49N

Apply Newton’s first law on the cord :

∑F_x = 0

T_3x - T _2x = 0

T₃ kusvika 30^o - T₂ kusvika 40^o = 0

0.87 T₃ – 0.77 T₂ = 0

0.87 T₃ = 0.77 T₂

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

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∑F_y = 0

T_3y +T_2y - T_1y = 0

T₃ pasina 30^o +T₂ pasina 40^o - T₁ = 0

0.5 T₃ + 0.64 T₂ – 49 N = 0 ———- Equation 2

Substituting T₂ in the equation 2 into the equation 2 :

0.5 T₃ + 0.64 (1.1 T₃) – 49 N = 0

0.5 T₃ + 0.70 T₃ - 49 = 0

1.2 T₃ - 49 = 0

1.2 T₃ = 49

T₃ = 49/1.2

T₃ = 41N

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T₂ = 1.1 T₃

T₂ = (1.1)(40.8 N)

T₂ = 45N

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