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.
Balancing Act
For calculation of static equilibrium of an object, its mass is considered to be concentrated at the center.
Equilibrium can be stable or unstable depending upon how long the system can remain in its state of equilibrium. A body in equilibrium that is disturbed, but comes back to equilibrium is said to be stable. Consider stirring tea in a mug. It becomes still after sometime. A body in equilibrium that is disturbed, and loses it state of equilibrium permanently is said to unstable. Consider a tennis ball balanced on a racket. It can fall if moved a little. When an object is disturbed by external forces which impart acceleration to it, it loses its stable state of equilibrium. This article will help you with static equilibrium calculations.
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 I in the diagram above. In this diagram the object O experiences upward as well as downward force on its surface. The object is in equilibrium.
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
If all the forces acting on a motionless object are balanced, it is said to be in static equilibrium. To understand the concept of static equilibrium is it important to determine the forces acting on the object and, their directions. On determining the forces, we must find the vector sum of these forces.
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
Forces act on individual points of an object. However, this fact makes the summation of forces difficult. Hence for simplifying the whole concept of equilibrium, the object’s mass is considered to be one single number rather than discrete points.
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.
Initially, try to draw a composite diagram depicting all the forces involved. For the above problem the resultant composite diagram of the forces is as follows:
Finding the components of force along the X and Y axis.
Ax = 4N * cos150o = 4 * 0.87 = 3.48N leftwards
Ay = 4N * sin150o = 4 * 0.5 = 2N upwards
Bx = 8.2 * cos65o = 8.2 * 0.423 = 3.47N rightwards
By = 8.2 * sin65o = 8.2 * 0.91 = 7.5N upwards
Cx = 9.5 * cos270o = 9.5 * 0 = 0N
Cy = 9.5 * sin270o = 9.5 * 1 = 9.5N downwards
Summation of forces working in:
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)
For a rigid body to be considered in the state of static equilibrium, following two conditions need to be followed.
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 30o + Tb = 0
(Equation I)
Finding the summation of forces along the y axis.
Ta cos30o – 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
(Equation III)
Finding the summation of forces along the y axis.
Tc cosΘ – 40N = 0
(Equation IV)
On solving equation II:
Ta cos30o = 50N (Substituting the value of cos30o which is 0.866.)
Ta = 50N/cos30o = 50/0.866 = 57.74N
On solving equation I:
Tb = Ta sin30o (Substituting the value of sin30o 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 = 35.82o
A photo frame hung on a wall, a glass kept on the table, or a bridge, do not undergo translational or linear motion at any point. Hence, we can say that they all follow the concept of static equilibrium.