Izinhlayiya eziku-equilibrium eyodwa - ukusetshenziswa kwezinkinga nezixazululo zomthetho wokuqala kaNewton

^{1. Isisindo sento, m = 10 kg, esisekelwe yintambo. Thola ukucindezeleka entanjeni!} g = 10 m/s ²

Kwaziwa:

Isisindo (m) = 10 kg

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Okufunwayo: Amandla okucindezela (T)

Isixazululo:

ΣF _y = 0

T – w = 0

T = w

T = mg

T = (10 kg)(10 m/s ² ) = 100 kg m/s ²

T = 100 Newton

[irp]

2. Isisindo sento singama-10 kg. Thola ukucindezeleka entanjeni….. Ukusheshisa ngenxa yamandla adonsela phansi = 10 m/s ².

Isixazululo

Kwaziwa:

Isisindo (m) = 10 kg

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ².

Okufunwayo: Amandla okucindezela (T)

Isixazululo:

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

T ₁ = amandla okucindezela 1

_{I-T 1x} = ingxenye engu-x yamandla okucindezela 1 = I-T ₁ cos 45 ^o = 0.7 I-T ₁

_{T 1y} = ingxenye ka-y yamandla okucindezela 2 = T ₁ sin 45 ^o = 0.7 T ₁

T ₂ = amandla okucindezela 2

_{I-T 2x} = ingxenye engu-x yamandla okucindezela 2 = I-T ₂ cos 45 ^o = 0.7 I-T ₂

_{T 2y} = ingxenye ka-y yamandla okucindezela 2 = T ₂ sin 45 ^o = 0.7 T ₂

Isimo sokulingana ΣF = 0.

i-axis ka-y:

ΣF _y = 0

T _1y + T _2y – w = 0

0.7T ₁ + 0.7T ₂ – 100 = 0

0.7T ₁ + 0.7T ₂ = 100 —– isibalo 1

i-axis x:

ΣF _x = 0

T _2x – T _1x = 0

0.7T ₂ – 0.7T ₁ = 0

0.7T ₂ = 0.7T ₁

T ₂ = T ₁ —– isibalo 2

Thola ubukhulu be-T ₁ :

0.7T ₁ + 0.7T ₁ = 100

1.4T ₁ = 100

I-T ₁ = 100 / 1.4

I-T ₁ = 71.4 Newton

T ₁ = T ₂ ngakho T ₂ = 71.4 amaNewton

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1. Izinhlayiya ezilingana ngokulingana okukodwa
2. Izinhlayiya ezilingana ngezindlela ezimbili
3. Ukulingana kwemizimba exhunywe ngezintambo nama-pulley
4. Ukulingana kwemizimba endizeni ethambekele