Science Applied Department (ASD) Centre for Foundation Studies and Extension Education (FOSEE)
PPH0075 Physics 1 Foundation in Engineering ONLINE NOTES
Lecture Notes Chapter 4: Static Equilibrium
FOSEE , MULTIMEDIA UNIVERSITY (436821-T) MELAKA CAMPUS, JALAN AYER KEROH LAMA, 75450 MELAKA, MALAYSIA. Tel [phone] Fax [phone] URL: http://fosee.mmu.edu.my/~asd/
PPH0075
Physics 1
Chapter 4
4.1 The Equilibrium of forces Static Equilibrium -object at rest Translational Equilibrium - object moves with constant v
4.2 The conditions of equilibrium 1st condition F = 0, Fx = 0 and Fy = 0 2nd condition =0 about any axis -the tendency to rotate about an axis
= r F ( Nm) = l F ( Nm)
Static Equilibrium
4.3 The center of mass and the center of gravity Center of mass- the point that moves in the same path when the body rotates
Xcm=
m1 x1 + m2 x2 + ......... + mN x N m1 + m2 + ........ + mN m y + m2 y 2 + ......... + m N y N Ycm= 1 1 m1 + m2 + ........ + m N
Center of gravity- the geometric point, determined by experiment method
4.4 Moment of inertia Moment of inertia - resistance of an object to change its rotational motion about an axis Moment of inertia , I = m1r12 + m2r22 + m3r32 +.+ mnrn2
n
=
1
mi ri 2
______________________________________________________________________________________ ASD2008/09 1/ 8
PPH0075
Physics 1
Chapter 4
CHAPTER 4 : STATIC EQUILIBRIUM
CONTENTS 4.1 The Equilibrium of forces 4.2 The conditions of equilibrium 4.3 The center of mass and center of gravity 4.4 Moment of Inertia , I 4.1 The Equilibrium of forces
The term equilibrium implies either that the object is at rest (called static equilibrium) or that is center of mass moves with constant velocity (called translational equilibrium). We deal here only with the former, which are referred to as objects in static equilibrium. In studying the equilibrium, we need to know two important parameters, i.e. Force (F) and Torque (
)
4.2 The conditions of equilibrium 4.2.1 Force,F The weight of an object (w) is the force with which gravity pulls down upon it. The tension in a string (T) is the force with which the string pulls upon the object to which it is attached The friction force (f) is a tangential force on a surface that opposes the sliding of the surface across an adjacent surface. The friction force is parallel to the surface and opposite in direction to its sliding motion. The normal force (N or FN) on a surface supported by a second surface is the component of the supporting force that is perpendicular to the surface being supported. 4.2.2 The First condition of equilibrium The vector sum of the forces acting on the particle must be zero, i.e. F = 0, Fx = 0 and Fy = 0
N
W=mg
Figure 1
______________________________________________________________________________________ ASD2008/09 2/ 8
PPH0075
Physics 1
Chapter 4
An object at rest on a table (Fig 1), has two forces acting on it, the force of gravity(weight,W) and the normal force(N) the table exerts on it. Since the net force is zero , F = ma ( a=0) F=0 N-W=0 N=W The upward force exerted by the table must be equal in magnitude to the force of gravity acting downward. Such a body is said to be in equilibrium. Exercise: The object shown in Fig 2 has weight W=60 N, is in equilibrium. If 1 = 53o and 37o , find the magnitudes of T1 and T2.
T1
2 1
2 =
T2
Figure 2
4.2.3 Torque
F=2 N
F=2 N
Figure 3
In Figure 3 , two forces acting on rod are equal in magnitude and opposite in direction , so they have a vector sum of zero; they balance each other.Although
______________________________________________________________________________________ ASD2008/09 3/ 8
PPH0075
Physics 1
Chapter 4
F = 0 in this case , common experience tells us that the rod will not remain at rest ,. Instead , it will begin to rotate. The tendency of a force to rotate an object about some axis is measured by a quantity called torque
/ moment of force.
F
r r O l line of force / action
Figure 4 In general , we can write the torque about a given axis as =lF l - lever arm : the shortest length from the pivot to the line of force / action .
= lever arm, l force, x F
r
: a vector directed from the point of pivot to the point of application of force :an angle between
F and r
From Fig. 4 we find that l = r sin , therefore
=lF
( Nm) = (r sin ) F
= rF sin
Note
+ ve torque : force tends to produce ccw rotation about the axis - ve torque : force tends to produce cw rotation about the axis
______________________________________________________________________________________ ASD2008/09 4/ 8
PPH0075
Physics 1
Chapter 4
Example: Refer to Fig. 4. If =300 , r = 15 cm and F=0.5 N , find the torque due to force F. Solution :
=lF
= + 0.15 x 0.5 x sin 300 =+0.0375 Nm +ve sign shows that the force tends to rotate the rod in ccw direction.
4.2.4The second condition for equilibrium
= rF sin
The sum of the torques of all forces acting on the body, with respect to some specified axis, must be zero, i.e.
= 0 about any axis
Exercise: The horizontal board shown in Fig.5 is uniform and weighs 200 N. Find the tension in the two ropes supporting it when a weight W=500N is suspended as shown.
0.75L W
0.25L
Figure 5
4.3 The center of mass and center of gravity
If a body rotates, or there are several bodies that move relative to each one another, there is one point that moves in the same path that a particle would if subjected to the same net force.This point is called the center of mass The position of the center of mass is defined in the followinfg way. Suppose an object or system of objects consists of N particles having masses m1 , m2 ,m3,., mN. If the coordinates of these particles are x1 , x2, x3,.,xN respectively, then the x coordinate of thecenter of mass is defined by XCM=
m1 x 1 + m 2 x 2 +.........+ m N x N m1 + m 2 +........+ m N
______________________________________________________________________________________ ASD2008/09 5/ 8
PPH0075
Physics 1
Chapter 4
N 1 N 1
mi xi mi
N 1
N 1
mi x i M
=
=
Similar expression can be written for y coordinates of the center of mass.
mi y i M
YCM =
Exercise: Find the center of mass of the three particles in Figure 6
y
0
b
a
x
a
Figure 6
The center of gravity
The center of gravity(CG) of an object is that point at which the force of gravity can be considered to act.For uniform symetrically shaped bodies,the CG is at the geometric center. There is a conceptual difference between CG and CM , but for practical purposes they are generally at the same point. It is often easier to determine the CM or CG of an object experimentally.
A CG A B
Figure 7(a)
C B
D
Figure 7(b)
______________________________________________________________________________________ ASD2008/09 6/ 8
PPH0075
Physics 1
Chapter 4
When the object hangs at at rest from point A as shown in Fig 7(a) , the pull of gravity must act along the vertical line AB.When the object hangs at rest as shown in 7(b), the pull of gravity must act along the vertical line CD.The CG of the object will be at the intersection of the two lines. Note :CM and CG of an object is at the same point only if the object is small or the object is in a uniform gravitational field that is the value of g is constant.
4.4 Moment of Inertia , I
The resistance of an object to change its rotational motion about an axis is known as moment of inertia of the object about that axis. Moment of inertia depends on the mass of the object and the distribution of the mass from the axis of rotation. Moment of inertia, I = mr2
axis m6 r6 m1 r1 m5 r5 r2 m4
r4 r3 m3 m2
If the object consists of n particles of mass m1 , m2 , m3,.., mn at a distance of r1, r2,r3,..,rn from the axis of rotation , then the moment of inertia of the object is given by I = m1r12 + m2r22 + m3r32 +.+ mnrn2
n
=
1
m i ri2
Exercise:
______________________________________________________________________________________ ASD2008/09 7/ 8
PPH0075
Physics 1
Chapter 4
Find the moment of inertia of the system shown in Fig.8 about the axis OO' and axis PP'.Assume the mass of the rods connecting the masses to be zero.
O m
a P 4m a a a 2m P'
3m O'
Figure .8
The End
______________________________________________________________________________________ ASD2008/09 8/ 8