Particles in the one-dimensional equilibrium – application of the Newton’s first law problems and solutions

^{1. Massa of an object, m = 10 kg, supported by a cord. Find the tension in the cord!} g = 10 m/s²

Magħruf:

Massa (m) = 10 kg

Aċċelerazzjoni minħabba l-gravità (g) = 10 m/s²

Meħtieġ: The tension force (T)

Soluzzjoni:

ΣF_y = 0

T – w = 0

T = w

T = m g

T = (10 kg)(10 m/s²) = 100 kg m/s²

T = 100 Newton

2. Mass of the object is 10 kg. Find the tension in the cord….. Acceleration due to gravity = 10 m/s².

Soluzzjoni

Magħruf:

Massa (m) = 10 kg

Aċċelerazzjoni minħabba l-gravità (g) = 10 m/s².

Meħtieġ: The tension force (T)

Soluzzjoni:

^{w = piż = m g = (10 kg)(10 m/s2}) = 100 kg m/s²

T₁ = the tension force 1

_T1x = the x-component of the tension force 1 = T₁ cos 45^o = 0.7 T₁

_T1y = the y-component of the tension force 2 = T₁ dnub 45^o = 0.7 T₁

T₂ = the tension force 2

_T2x = the x-component of the tension force 2 = T₂ cos 45^o = 0.7 T₂

_T2y = the y-component of the tension force 2 = T₂ dnub 45^o = 0.7 T₂

The equilibrium condition ΣF = 0.

y axis :

ΣF_y = 0

T_1y +T_2y – w = 0

0.7T₁ + 0.7T₂ - 100 = 0

0.7T₁ + 0.7T₂ = 100 —– equation 1

x axis :

ΣF_x = 0

T_2x - T_1x = 0

0.7T₂ – 0.7T₁ = 0

0.7T₂ = 0.7T₁

T₂ =T₁ —– equation 2

Determine magnitude of T₁ :

0.7T₁ + 0.7T₁ = 100

1.4T₁ = 100

T₁ = 100/1.4

T₁ = 71.4 Newton

T₁ =T₂ so T₂ = 71.4 Newton

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1. Partiċelli f'ekwilibriju unidimensjonali
2. Partiċelli f'ekwilibriju bidimensjonali
3. Ekwilibriju ta' korpi konnessi permezz ta' ħbula u taljoli
4. Ekwilibriju ta' korpi fuq pjan inklinat