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CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
We have studied the application of forces on particles.
What if do we have a rigid body having a finite size?
If we have a rigid body having finite size, there may be different forces acting on the rigid
body at different points of application. Therefore:
The objective of this section is:
Replace a given system of forces by a simpler equivalent system acting on a rigid body.
In order to achieve this: we need to learn the followings:
Moment of a force about a point
Vector product, cross product
Principles of moments (Varignon’s theorem)
Principle of transmissibility
Moment of a force along an axis
Moment of a couple
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
Principle of Transmissibility
This principle states that the conditions of equilibrium or motion of a rigid body remain
unchanged if a force acting at a given point of the rigid body is replaced by a force of the
same magnitude and direction but acting at a different point provided that they have the
same line of action.
This principle is based on experimental evidence and cannot be proven by statics. It can
be derived from the study of dynamics of rigid bodies.
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
It should be noted that the principle of transmissibility does not apply in the
analysis of deformations and internal forces
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
MOMENT OF A FORCE ABOUT A POINT
The moment of a force about a point provides a measure of tendency of rotation
about a fixed axis along a point. It is sometimes known as TORQUE. Moment is
vector.
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
Now consider a Fy acting on the same point. No moment is produced. If the line of action
of the force passes through the point 0, no moment is produced.
Now consider a Fz acting on the same point.
Magnitude: M0= Fd
[N.m] Moment arm (perpendicular distance
from 0 to the line of action F)
Fz
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
DIRECTION: we will use the right hand rule:
Fingers follow the sense of rotation. Right hand thumb points
along the direction of the moment (moment axis).
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
A 400N force is applied to the frame and
θ=20 . Find the moment of the force at A
HW: Please work on this example.
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
SOLUTION:
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
Let’s resolve F into its rectangular components:
Fx=400N cos 20 =375.9N
Fy=400N sin 20 =136.8N
Let’s assume + (usually counter-clockwise)
MA=136.8N 3m+375.9N 2m
MA=1162.2Nm (ANSWER)
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
MOMENTS IN 3-D:
Moment of a force about point O is defined as:
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
VECTOR CROSS PRODUCT:
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
GENERALIZING:
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
PRINCIPLES OF MOMENTS: (VARIGNON’S THEOREM)
The moment of a force about a point is equal to the sum of the moments of the force’s
components about the point.
Using the distribution law.
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
Let’s use the cross product for the example problem.
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
Moment of a Force about an axis:
We know that when the moment of a force is computed about a point, the moment will
be perpendicular to the plane containing the force and the moment arm. In some cases,
we need to find the component of this moment along a specified axis.
A force is applied to the tool to open a
gas valve. Find the magnitude of the
moment of this force about the z axis.
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
Moment of a Couple:
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
Equivalent Couples:
All three couples 1,2,3 have the
same moment M.
Couples which have the same
moment are called equivalent
couples. They have the same
effect on the rigid body.
Equivalent couples must be either
in the same or parallel lines.
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
When a couple acts on a rigid body, it does not matter where the two forces forming the
couple act or what magnitude and direction they have. The only thing that counts is the
moment of the couple. (magnitude + direction)
Resultant of Couple Moment:
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
EXAMPLE:
Determine the net moment due to the two couples
(d=8ft)
Let’s resolve the forces into x and y axis.
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
HW: Please work on this question:
Find the couple moment acting on the rod in
Cartesian vector notation:
CHAPTER 2: EQUILIBRIUM OF RIGID BODIES
Same result would have been obtained if one had found individual moments of two
forces about any point.