# 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/s^{2}

__Known :__

Mass (m) = 10 kg

^{2}

__Wanted :__ The tension force (T)

__Solution :__

ΣF_{y} = 0

T – w = 0

T = w

T = m g

T = (10 kg)(10 m/s^{2}) = 100 kg m/s^{2}

T = 100 Newton

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

Solution

__Known :__

Mass (m) = 10 kg

Acceleration due to gravity (g) = 10 m/s^{2}.

__Wanted :__ The tension force (T)

__Solution :__

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

^{2}

T_{1} = the tension force 1

_{T1x} = the x-component of the tension force 1 = T_{1} cos 45^{o} = 0.7 T_{1}

_{T1y} = the y-component of the tension force 2 = T_{1} sin 45^{o} = 0.7 T_{1}

T_{2} = the tension force 2

_{T2x} = the x-component of the tension force 2 = T_{2} cos 45^{o} = 0.7 T

_{2}

_{T2y} = the y-component of the tension force 2 = T_{2} sin 45^{o} = 0.7 T_{2}

__The equilibrium condition ____ΣF = 0.__

y axis :

ΣF_{y} = 0

T_{1y} + T_{2y} – w = 0

0.7T_{1} + 0.7T_{2} – 100 = 0

0.7T_{1} + 0.7T_{2} = 100 —– equation 1

x axis :

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} —– equation 2

Determine magnitude of T_{1} :

0.7T_{1 }+ 0.7T_{1 }= 100

1.4T_{1} = 100

T_{1} = 100 / 1.4

T_{1 }= 71.4 Newton

T_{1} = T_{2} so T_{2} = 71.4 Newton

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- Particles in one-dimensional equilibrium
- Particles in two-dimensional equilibrium
- Equilibrium of bodies connected by cord and pulley
- Equilibrium of bodies on inclined plane