# 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 :

ΣFy = 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 45o = 0.7 T1

T1y = the y-component of the tension force 2 = T1 sin 45o = 0.7 T1

T2 = the tension force 2

T2x = the x-component of the tension force 2 = T2 cos 45o = 0.7 T2

T2y = the y-component of the tension force 2 = T2 sin 45o = 0.7 T2

The equilibrium condition ΣF = 0.

y axis :

ΣFy = 0

T1y + T2y – w = 0

0.7T1 + 0.7T2 – 100 = 0

0.7T1 + 0.7T2 = 100 —– equation 1

x axis :

ΣFx = 0

T2x – T1x = 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. 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

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