Force Analysis of Mechanisms(1)

Centrifugal Force, Inertia Force and Inertia Torque

Centrifugal Force

An object traveling in a circle behaves as if it is experiencing the outward force and this force is known as centrifugal force depends on the mass of the object, the speed of rotation and the distance from center. The more massive the object then the greater the force, the greater the speed of the object then the greater the force and the greater the distance from the center then the greater the force. Consider a stone being whirled round on string. The only one real force acting on stone is the tension in the string. There are no other forces acting on stone so there is a net force on stone.

Fig: Centrifugal force(F)

F=mv²/r

Inertia Force

Inertia is resistance to any type of physical object to any change in its state of motion this includes changes to its speed, direction or state of rest. It is the tendency of objects to keep moving in a straight line at constant velocity.

An inertial force is a force which resists a change in velocity of an object. It is equal to and in the opposite direction of the applied force as well as with a resistive force. This concept is based on Newton's Laws of Motion including the Law of Inertia and Action-Reaction Law. Inertial force may be examined both when you apply a force on an object and when a force is applied on you.

The mass of a body determines the momentum(p) of the body at given velocity(v).

p = mv

The factor m is referred to as inertial mass.

But mass as related to the inertia of a body can also be defined by the formula.

F=ma

Here F is the force, m is inertial mass and a is acceleration.

An imaginary force is supposed to act upon an accelerated body equal in magnitude and opposite in direction to the resultant of the real forces. Proof of inertial force can see clearly that a body cannot provide itself with motion it cannot be forced from rest except from by some foreign action. From there it stands that if a body receives motion irrespective of what that might be it cannot on its own according to neither accelerate nor slow this movement.

Inertia Torque

Torque also known as the moment of force and is the rotational analog of force. This word origin from Latin word torquere meaning "to twist". In the same way that a force is necessary to change a particle state of motion. A torque is required to change a particle or object's state of rotation.

In the vector form it is defined as

\[\overrightarrow \tau = \overrightarrow r *\overrightarrow F \]

whereτ is torque vector,F is force vector and is the position vector of the point where the force is applied relative to axis of rotation. The direction of torque is always perpendicular to the plane in which it is applied. Hence for two dimensional rotation this can be simplified to

\[\overrightarrow \tau = \overrightarrow r .\overrightarrow F \sin \varphi ..................................(\iota )\]

where [\tau \]is the distance between the axis of rotation and point at which the force is applied is magnitude of the force and is the angle between the position vector of point at which the force is applied i.e. relative to the axis of rotation and the direction in which the force is applied. The direction of this torque is perpendicular to plane of rotation. Eq. (i) shows that the torque is maximum when the force is applied perpendicular to the line joining the point at which the force is applied and the axis of rotation.

The moment of inertia otherwise known as the angular mass or rotational inertia of a rigid body determines torque required for a desired angular acceleration about a rotational axis. When a torque is applied to an object it begins to rotate with acceleration inversely proportional to its moment of inertia. This relation can be as Newton's Second Law for rotation. The moment of inertia is the rotational mass and the torque is rotational force. Angular motion obeys Newton's First Law. A practical way to calculate the magnitude of the torque is to first determine the lever arm and then multiply it times applied force. The lever arm is the perpendicular distance from the axis of rotation to line of action of force. and the magnitude of the torque is

τ = N m=Iα

Methods of Force Analysis – Introduction

A machine system is considered to be a system of an arbitrary group of bodies (links) that will be considered rigid. We involved with different types of forces in such systems.

a) Reaction Forces: Its commonly called joint forces in machine systems since the action and reaction between the bodies involved will be through contacting kinematic elements of links that form a joint. The joint forces are along the direction for which the degree of freedom is restricted. For e.g. in constrained motion direction. In the joint forces due to action and reaction. There are two links that are involved. As a convention it generally denote the joint force Fij (Mij for the couple) as that force acting on link j due to reaction of link i. Hence from Newton's third law: Fij = -Fij

Free Body Diagram

In the machine systems there are various rigid bodies. A free-body diagram is a sketch of a rigid body or group of rigid bodies in which all forces and moments acting on the rigid body are shown. Note that a free-body can involve one link or several parts of a machine or the whole machine. In the free-body diagram both the known forces and unknown forces are sketched. This diagram is need not be on a certain scale, but neatness in such diagrams is required for a clear understanding. As an example consider a rigid body, i as shown. It is connected to other rigid bodies. In free body diagram of link i. This body is isolated from other rigid bodies and all the forces (reaction forces, Fki, Fji, Mji ; external forces Fli’, Fli “) are shown. If the magnitude and/or the direction of a force is known it is indicated.

STATIC EQUILIBRIUM

From Newton's First Law, a body is in static equilibrium if the resultant of all forces excluding the inertia forces acting on a rigid body is zero. This condition results in two vector equations:

\[\sum \overrightarrow F = 0\;\;\;\;\;\;\;\;\sum \overrightarrow M = 0\]

In space, these two vector equations yield six scalar equations:

\[\begin{array}{l}
\sum {F_X} = 0\;\;\;\;\;,\;\;\;\sum {F_Y} = 0\;\;\;,\;\;\;\sum {F_Z} = 0\\
\sum {M_X} = 0\;\;\;\;\;,\;\;\;\sum {M_Y} = 0\;\;\;\;,\;\;\;\;\sum {M_Z} = 0\;\;\;\;
\end{array}\]

In case of coplanar force systems, there are three scalar equations:

\[\sum {F_X} = 0\;\;\;\;\;,\;\;\;\sum {F_Y} = 0\;\;\;,\;\;\sum {M_Z} = 0\;\]

Usually the subscript z in M_Z is dropped in coplanar force systems and M implies the moment perpendicular to the plane of forces. Static equilibrium equations in space will permit the determination of six unknowns in spatial systems and the three unknowns in planar systems. In the coplanar force systems under which the body acted on is in equilibrium, we can have the following force systems:

1) Two-force member: A rigid body acted on by two forces is in static equilibrium only when two forces are collinear and equal in magnitude but in opposite sense . If the point of application of the two forces are known points A and B. The direction of these forces will be along the line joining the points of application.

2) Two force and one-moment member: A rigid body acted on by two forces and a moment is in static equilibrium only when two forces form a couple whose moment is equal in magnitude but in opposite sense to the applied moment.

3) Three-force member: A rigid body acted on by three forces is in equilibrium only if the forces are concurrent intersect at one point. Let the force FA be completely specified in diagram. And the line of action of FB and the point of application of FC is known i.e. In plane there are three equilibrium equations and one must have three unknowns to be able to solve the equilibrium equations. When moment equilibrium equation is written for the sum of moments about the point of intersection of the line of action of FA and FB (point O), since MO=0, the moment of FC about O must be zero, or the line of action of force FC must pass through the point O. These magnitudes of the forces can then be determined from the force and moment equilibrium equations.

References:
1. H.H. Mabie and C. F. Reinholtz, “Mechanism and Dynamics of Machinery”, Wiley.
2. J.S. Rao & R.V. Dukkipati Mechanisms and Machine Theory, New Age International (P) Limited..
3. J.E. Shigley and J.J. Uicker, Jr., “ Theory of Machines and Mechanisms”, McGraw Hill.
4. B. Paul, “Kinematics and Dynamics of Planar Machinery”, Prentice Hall.
5. C. E. Wilson, J.P. Sadler and W.J. Michels, “Kinematics and Dynamics of Machinery”, Harper Row.