Conditions for Equilibrium in 2D | Engineering Mechanics for Mechanical Engineering PDF Download

Equilibrium as the condition in which the resultant of all forces and moments acting on a body is zero. Stated in another way, a body is in equilibrium if all forces and moments applied to it are in balance. These requirements are contained in the vector equations of equilibrium, Eqs. 3/1, which in two dimensions may be written in scalar form as

ΣFx = 0             Σ Fy = 0 ΣMo = 0      (3/2)

The third equation represents the zero sum of the moments of all forces about any point O on or off the body. Equations 3/2 are the necessary and sufficient conditions for complete equilibrium in two dimensions. They are necessary conditions because, if they are not satisfied, there can be no force or moment balance. They are sufficient because once they are satisfied, there can be no imbalance, and equilibrium is assured. These equations show that the acceleration of the mass center of a body is proportional to the resultant force ΣF acting on the body. Consequently, if a body moves with constant velocity (zero acceleration), the resultant force on it must be zero, and the body may be treated as in a state of translational equilibrium. For complete equilibrium in two dimensions, all three of Eqs. 3/2 must hold. However, these conditions are independent requirements, and one may hold without another. Take, for example, a body which slides along a horizontal surface with increasing velocity under the action of applied forces. The force–equilibrium equations will be satisfied in the vertical direction where the acceleration is zero, but not in the horizontal direction. Also, a body, such as a flywheel, which rotates about its fixed mass center with increasing angular speed is not in rotational equilibrium, but the two force–equilibrium equations will be satisfied.

 

  Conditions for Equilibrium in 2D | Engineering Mechanics for Mechanical Engineering

In addition to Eqs. 3/2, there are two other ways to express the general conditions for the equilibrium of forces in two dimensions. The first way is illustrated in Fig. 3/6, parts (a) and (b). For the body shown in Fig. 3/6a, if ΣMA = 0, then the resultant, if it still exists, cannot be a couple, but must be a force R passing through A. If now the equation ΣFx = 0 holds, where the x-direction is arbitrary, it follows from Fig. 3/6b that the resultant force R, if it still exists, not only must pass through A, but also must be perpendicular to the x-direction as shown. Now, if ΣMB = 0, where B is any point such that the line AB is not perpendicular to the x-direction, we see that R must be zero, and thus the body is in equilibrium. Therefore, an alternative set of equilibrium equations is

ΣFx = 0  ΣMA = 0     ΣMB = 0   

where the two points A and B must not lie on a line perpendicular to the x-direction. A third formulation of the equilibrium conditions may be made for a coplanar force system. This is illustrated in Fig. 3/6, parts (c) and (d). Again, if ΣMA = 0 for any body such as that shown in Fig. 3/6c, the resultant, if any, must be a force R through A. In addition, if ΣMB 0, the resultant, if one still exists, must pass through B as shown in Fig. 3/6d. Such a force cannot exist, however, if ΣMC = 0, where C is not

Conditions for Equilibrium in 2D | Engineering Mechanics for Mechanical EngineeringConditions for Equilibrium in 2D | Engineering Mechanics for Mechanical Engineering

Conditions for Equilibrium in 2D | Engineering Mechanics for Mechanical EngineeringConditions for Equilibrium in 2D | Engineering Mechanics for Mechanical Engineering

collinear with A and B. Thus, we may write the equations of equilibrium as

ΣMA = 0     ΣMB = 0    ΣMC = 0

where A, B, and C are any three points not on the same straight line. When equilibrium equations are written which are not independent, redundant information is obtained, and a correct solution of the equations will yield 0

0. For example, for a general problem in two dimensions with three unknowns, three moment equations written about three points which lie on the same straight line are not independent. Such equations will contain duplicated information, and solution of two of them can at best determine two of the unknowns, with the third equation merely verifying the identity 0 = 0.

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FAQs on Conditions for Equilibrium in 2D - Engineering Mechanics for Mechanical Engineering

1. What are the conditions for equilibrium in 2D?
Ans. The conditions for equilibrium in 2D are: 1. The net force acting on the object must be zero. This means that the vector sum of all the forces in both the x and y directions must be zero. 2. The net torque acting on the object must be zero. This means that the vector sum of all the torques must be zero, considering the point of rotation. 3. The object must be in translational equilibrium, which means that the sum of the forces in the x direction and the sum of the forces in the y direction must both be zero. 4. The object must also be in rotational equilibrium, which means that the sum of the torques in the clockwise direction and the sum of the torques in the counterclockwise direction must both be zero. 5. The center of mass of the object must remain at rest or move at a constant velocity.
2. How is the net force calculated in 2D equilibrium problems?
Ans. In 2D equilibrium problems, the net force is calculated by adding up the forces acting on the object in both the x and y directions. This can be done by resolving each force into its x and y components using trigonometry. The sum of the x components of the forces should be equal to zero, and the sum of the y components of the forces should also be equal to zero. This ensures that the object is in translational equilibrium.
3. What is the significance of the net torque in 2D equilibrium?
Ans. The net torque in 2D equilibrium is significant because it determines whether an object is in rotational equilibrium or not. If the net torque is equal to zero, it means that the object is not experiencing any rotational acceleration and is in rotational equilibrium. This requires that the sum of the torques in the clockwise direction is equal to the sum of the torques in the counterclockwise direction. If the net torque is not zero, the object will experience rotational acceleration and will not be in equilibrium.
4. How can we determine if an object is in translational equilibrium in 2D?
Ans. To determine if an object is in translational equilibrium in 2D, we need to check if the sum of the forces in the x direction and the sum of the forces in the y direction are both equal to zero. This means that the object is not experiencing any net force in either direction, and it will remain at rest or move at a constant velocity. If the sum of the forces in either direction is not zero, the object is not in translational equilibrium.
5. Can an object be in equilibrium even if it is moving in 2D?
Ans. Yes, an object can be in equilibrium even if it is moving in 2D. This is possible if the object is moving at a constant velocity, which means that it has a net force of zero. In this case, the object is in translational equilibrium. However, it is important to note that even though the object is in equilibrium, it is still moving, indicating that there might be forces acting on it. These forces are balanced in such a way that the object's motion remains constant.
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