# Calculation of Static Equilibrium: Case Studies and Examples

A body is said to be in equilibrium when its neither in a state of motion nor its state of energy changes over a period of time. An undisturbed object continues to remain in its state of equilibrium. ScienceStruck explains with examples how to compute static equilibrium.

ScienceStruck Staff

Last Updated: Jul 16, 2017

Balancing Act

For calculation of static equilibrium of an object, its mass is considered to be concentrated at the center.

Equal and Opposite Forces

A body under the influence of equal and opposite forces is not always in equilibrium. Consider the following two cases.

Consider case II in the diagram above. In this diagram the object O experiences upward as well as downward force on its edges. The object is not in equilibrium.

What is Static Equilibrium

Examples of Balanced Forces

**Book placed on a table**

A book placed on the table experiences two forces, the upward force of the table and downward force of gravity. These forces are in opposite directions and have the same magnitude thus, balancing each other. Hence, the book is considered to be in a state of equilibrium.

**Weight Hung by multiple strings**

In the diagram given below, an object is hung by three strings. To understand its state we need to calculate the forces acting upon it. If the net sum of all the forces acting on the object is zero, it implies that the object is in equilibrium.

How to Calculate Static Equilibrium

**Problem I:**Consider a point upon which three forces with magnitudes 4N, 8.2N, and 9.5N act at angles 150o, 65o, 270o respectively. Calculate the net force.

Ax = 4N * cos150

^{o}= 4 * 0.87 = 3.48N leftwards

Ay = 4N * sin150

^{o}= 4 * 0.5 = 2N upwards

Bx = 8.2 * cos65

^{o}= 8.2 * 0.423 = 3.47N rightwards

By = 8.2 * sin65

^{o}= 8.2 * 0.91 = 7.5N upwards

Cx = 9.5 * cos270

^{o}= 9.5 * 0 = 0N

Cy = 9.5 * sin270

^{o}= 9.5 * 1 = 9.5N downwards

Upward direction = 2+7.5 = 9.5N

Downward direction = 9.5N

The magnitude of the upward force is equal to the magnitude of downward force. Thus, these forces cancel each other.

Summation of forces working in:

Rightward direction = 3.47N

Leftward direction = 3.48N

The magnitude of the rightward force is equal to the magnitude of leftward force. Thus, these forces cancel each other.

Static Equilibrium of a Rigid Body (Motionless)

**CASE I**

If the vector sum of all the forces acting on the object is zero, implies that object is in translational equilibrium. In other words net external force should be zero.

S(F) = 0

**CASE II**

If the sum of torques acting externally on the object is zero, implies that the object is in rotational equilibrium. In other words net external torque should be zero.

S(t) = 0

**SFx = 0, SFy = 0, and Stz = 0**

When both the above conditions are satisfied implies that the rigid body is in static equilibrium.

**Problem II:**Consider the system in the given figure where in two masses are suspended by strings. This system is in the state of static equilibrium. Calculate Ta, Tb, Tc, and the angle Θ.

**Solution:**Remember that the sum of forces and, torque is always equal to zero.

Consider the object suspended on the left side of the system.

Finding the summation of forces along the x axis.

-Ta sin 30

-Ta sin 30

^{o}+ Tb = 0**(Equation I)**Finding the summation of forces along the y axis.

Ta cos30

Ta cos30

^{o}- 50N = 0**(Equation II)**Consider the object suspended on the right side of the system.

Finding the summation of forces along the x axis.

-Tb + Tc sinΘ = 0

-Tb + Tc sinΘ = 0

**(Equation III)**Finding the summation of forces along the y axis.

Tc cosΘ - 40N = 0

Tc cosΘ - 40N = 0

**(Equation IV)**Ta cos30

^{o}= 50N (Substituting the value of cos30

^{o}which is 0.866.)

Ta = 50N/cos30

^{o}= 50/0.866 = 57.74N

On solving equation I:

Tb = Ta sin30

^{o}(Substituting the value of sin30

^{o}which is 0.5.)

Tb = 57.74*0.5 = 28.87N

On solving equation III:

Tc SinΘ = Tb

(Substituting the value of Tb in equation III)

Tc sinΘ = 28.87N -------

**Equation V**

Tc cosΘ = 40N -------

**Equation VI**

Dividing equation V by equation VI.

tanΘ = Tc sinΘ/Tc cosΘ

tanΘ = 28.87/40

tanΘ = 0.72175

Θ = tan-1(0.72175)

**Θ = 35.82o**

Finding the value of Tc (Substituting the value of Tb and Θ in equation III)

Tc sin35.82 = 28.87

Tc = 28.87/0.95

Tc = 30.39N

**Ta = 57.74N, Tb = 28.87N, Tc = 30.39N, Θ = 35.82**

^{o}