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

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

Known :

Mass (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s²

Wanted : The tension force (T)

Solution :

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.

Solution

Known :

Mass (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s².

Wanted : The tension force (T)

Solution :

^{w = weight = 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 sin 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 sin 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

[wpdm_package id=’486′]

1. Particles in one-dimensional equilibrium
2. Particles in two-dimensional equilibrium
3. Equilibrium of bodies connected by cord and pulley
4. Equilibrium of bodies on inclined plane