# Condition of Equilibrium under the Action of Non-Parallel Coplanar Forces

The conditions of equilibrium under the action of non-parallel coplanar forces are:

1. **Forces: **The vector sum of all the forces acting on a body must be zero i.e.Â

**(i) **The algebraic sum of the horizontal components must be zero âˆ‘F_{x} = 0

**(ii)** The algebraic sum of the vertical components of the forces must be zero âˆ‘F_{y} = 0

2. **Moments: **Â The algebraic sum of the moments of all the forces about any axis perpendicular to the plane of the forces must be zero.

### Example 1:

In the diagram below an object of mass 500kg is kept in equilibrium by three cables. Find the tension in each of the three cables. (g = 9.8ms^{-2})

**Solution:**

Tension in cable 3 can be found easily

T_{3} = mg

âˆ´ T_{3} = 500 x 9.8 = 4900N (in the negative direction)

To find T_{1} and T_{2}, since the object is in equilibrium âˆ‘F_{x} = 0 and âˆ‘F_{y} = 0

The horizontal component of T_{1} = T_{1X }

The vertical component of T_{1} = T_{1Y}

The horizontal component of T_{2} = T_{2X }

The vertical component of T_{2} = T_{2Y}

The horizontal component T_{1x} = T_{1} cos 30Âº (**Note:** this is moving in the negative direction)

The vertical component T_{1y} = T_{1} sin 30Âº

The horizontal component of T_{2x} = T_{2} cos 30Âº_{ }

The vertical component of T_{2y} = T_{2} sin 30Âº

**Sum of the horizontal components âˆ‘F _{x};**

= T_{2x} – T_{1x} = 0 (we subtract since T_{1x} is moving in the negative direction)

T_{2} cos 30Âº – T_{1} cos 30Âº = 0

0.9397T_{2} – 0.8660T_{1} = 0 ……(i)

**Sum of the horizontal components âˆ‘F _{y};**

= T_{2y} + T_{1y} – T_{3} = 0

T_{2} sin 30Âº + T_{1} sin 30Âº – 4900 = 0

0.3420T_{2} + 0.5T_{1} – 4900 = 0 …….(ii)

We can then find T_{1} and T_{2} by substituting.

From equation (i)

0.9397T_{2} – 0.8660T_{1} = 0

0.9397T_{2} = 0.8660T_{1}

T_{2 }= \( \frac{0.866}{0.9397}\scriptsize T_1 \)

T_{2 }= 0.9216 T_{1} ….(iii)

substitute the value of T_{2} into equation (ii)

0.3420T_{2} + 0.5T_{1} – 4900 = 0 …….(ii)

0.3420(0.9216T_{1}) + 0.5T_{1} – 4900 = 0

0.3152T_{1} + 0.5T_{1} = 4900

0.8152T_{1} = 4900

T_{1} = \( \frac{4900}{0.8152} \)

T_{1} = 6011N

To find T_{2} substitute the value of T_{1} into equation (iii)

T_{2 }= 0.9216 T_{1} ….(iii)

T_{2 }= 0.9216 x 6011

T_{2 }= 5540N

### Example 2:

A body of mass of 10kg is kept in equilibrium by two ropes as shown in the diagram below. If one rope pulls the body in the horizontal direction and the other in a direction of 40Âº with the vertical, calculate the tension in each rope. (g = 10m/s^{2})

**Solution:**

The mass of 10k is equal to a force of 100N (F = mg = 10 x 10 = 100N)

To find T_{1} and T_{2}, since the object is in equilibrium âˆ‘F_{x} = 0 and âˆ‘F_{y} = 0

The horizontal component of T_{1} = T_{1}

The vertical component of T_{1} = 0 (no vertical component)

The horizontal component of T_{2} = T_{2X }

The vertical component of T_{2} = T_{2Y}

The horizontal component T_{1x} = T_{1} (**Note:** this is moving in the negative direction)

The horizontal component of T_{2x} = T_{2} cos 50Âº_{ }

The vertical component of T_{2y} = T_{2} cos 40Âº

**Sum of the horizontal components âˆ‘F _{x};**

T_{2} cos 50Âº – T_{1} = 0

0.6427T_{2} = T_{1}

or T_{1} = 0.6428T_{2}

**Sum of the vertical components âˆ‘F _{x};**

T_{2} cos 40Âº – 100 = 0

T_{2} cos 40Âº = 100

âˆ´ T_{2} = \( \frac{100}{cos40^{\circ}}\)

T_{2} = 130.54N

but

T_{1} = 0.6428T_{2}

T_{1} = 0.6428 x 130.54

T_{1} = 83.909N

T_{1} = 83.91N

### Example 3:

A metal frame weighing 40N is hung on a nail using two ropes as shown. Find the tension in the rope.

**Solution:**

Due to symmetry, the tension in each rope will be the same. Because of this, the quickest way to solve this question is by simply writing an equilibrium equation in the y-direction: (there is no need resolving in x-dirextion)

Resolving in vertical direction, âˆ‘Fy= 0

T cos60Âº + T cos60Âº â€“ 40N = 0

2Tcos60Âº = 40N

Tcos60Âº = \( \frac{40}{2} \)

Tcos60Âº = 20

\( \scriptsize T \: \times \: \normalsize \frac{1}{2} \scriptsize = 20 \) \( \frac{T}{2} \scriptsize = 20 \)T = 40N

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