ME2112 (Part 1) L2 - 1

Statics
• Statics of Particles
 Parallelogram Law, Law of Cosine & Law of Sine
 Addition of Several Forces
• Equilibrium of Rigid Bodies

• Analysis of Pin-connected Structures

Truss Frames Machines

ME2112 (Part 1) L2 - 2

Addition of Several Forces

Break the three vectors into components, then add them.

FR  F1  F2  F3
 F1 x i  F1 y j  F2 x i  F2 y j  F3 x i  F3 y j
 ( F1 x  F2 x  F3 x )i  ( F1 y  F2 y  F3 y ) j
 FRx i  FRy j
FRy
  arctan , FR  FRx2  FRy2
FRx
ME2112 (Part 1) L2 - 3

Example 1.3
An eyebolt is connected to a heavy object. Three workers are pulling
the three attached ropes by forces with different magnitudes and
directions. Determine the force resultant and predict the direction of
the object movement.

F3  50 N
Use the summation of
Cartesian components. 300
x

450
250
F2  120 N
F1  80 N

ME2112 (Part 1) L2 - 4

Solution
• Calculate the Cartesian components of each force
F1x  F1 sin 250  33.81 N or 33.81  Sign Convention
F1 y  F1 cos 250  72.50 N or 72.50   (+);  (+); (+)

F2 x  F2 sin 450  84.85 N 

F2 y  F2 cos 450  84.85 N or 84.85  y
F3 x  F3 cos 300  43.30 N  F3 y F3  50 N

F3 y  F3 sin 300  25 N  300

F1x F3x F2 x x
• Sum each component accordingly
450
250 F
1 y F  120 N
Rx  F1 x  F2 x  F3 x  94.34 N F1  80 N 2

F2 y
R y  F1 y  F2 y  F3 y  132.35 N
ME2112 (Part 1) L2 - 5

• Calculate the resultant’s magnitude and direction

R  Rx2  R y2  162.5 N Sign Convention

 (+);  (+); (+)
 Ry 
  tan 1  
 Rx 
 132.35 
 tan 1    54.52
0 y
 94.34 

x
54.52 0

162.5 N

ME2112 (Part 1) L2 - 6

Example 1.4 (Take-home exercise)

A trolley is subjected to two forces as shown below,
• given that   3 0 0 , determine by parallelogram law the magnitude of P
such that the resultant force exerted on the trolley is vertical;
• determine the corresponding resultant force.

20 0
α
1400 N
P
ME2112 (Part 1) L2 - 7

Hint: Direction of the resultant is vertical. So the parallelogram should be,

1400 P R
200 1400 N 20
0
 
1400 N  sin 300 sin  sin 

300
P
R? R
P?
P? R?
300 Try using the concept of summation
of Cartesian components to solve
this problem.

Once we have determined the angles β and γ, obtaining P and R would

be very straight forward.

ME2112 (Part 1) L2 - 8

Reading Quiz 1 (L1)

1) When a particle is in equilibrium, the sum of forces acting on it equals ___ .
(Choose the most appropriate answer)
A) A constant B) A positive number C) Zero
D) A negative number E) An integer

2) For a frictionless pulley and cable, tensions in the cable (T1 and T2) are
related as _____ .
A) T1 > T2
B) T1 = T2
C) T1 < T2 T1
D) T1 = T2 sin  T2
ME2112 (Part 1) L2 - 9

Reading Quiz 2 (L1)

1. Which one of the following is a scalar quantity?
A) Force B) Position
C) Mass D) Velocity

2. For vector addition, you have to use ______ law.

A) Newton’s Second
B) the arithmetic
C) Pascal’s
D) the parallelogram

ME2112 (Part 1) L2 - 10

Concept Questions (L1)

1000 N 1000 N
1000 N
(A) (B) (C)
1) Assuming you know the geometry of the ropes, you cannot determine the
forces in the cables in which system above?
2) Why?
A) The weight is too heavy.
B) The cables are too thin.
C) There are more unknowns than equations.
D) There are too few cables for a 1000 N weight.
ME2112 (Part 1) L2 - 11

Attention Quiz 1 (L1)

Select the correct FBD of particle A.

A
30 40
100 N

A F1 F2
A) B)
30 40°
100 N
A
F F1 F2
C) 30° D) 30° 40°
A
A
100 N 100 N

ME2112 (Part 1) L2 - 12

Attention Quiz 2 (L1)

1. Resolve F along x and y axes and write it in
vector form. F = { ___________ } N y
x
A) 80 cos (30°) i – 80 sin (30°) j
B) 80 sin (30°) i + 80 cos (30°) j
30°
C) 80 sin (30°) i – 80 cos (30°) j F = 80 N
D) 80 cos (30°) i + 80 sin (30°) j
2. Determine the magnitude of the resultant (F1 + F2) force in N
when F1 = { 10 i + 20 j } N and F2 = { 20 i + 20 j } N .
A) 30 N B) 40 N C) 50 N
D) 60 N E) 70 N
ME2112 (Part 1) L2 - 13

Free Body Diagram

A free-body diagram (FBD) is a sketch of an object or a connected group of
objects, modeled as a single particle/rigid body that is completely isolated
from its environment or surrounding bodies and represents the interactions of
its environment by appropriate external forces (and moments).
Drawing Free Body Diagram
1. Identify the body from structure that will be isolated.
2. Isolate the body by “cutting” through the structure such that the body can be
removed from the surroundings.
3. Draw sketch of the removed body and include its relative dimensions and
angles for convenience.
4. Draw all forces (external, internal, self-weight, reactions) acting on body and
label them.
5. Choose an appropriate coordinate system.

ME2112 (Part 1) L2 - 14

External and Internal Forces W

W To calculate the tensile forces in

the cables, consider a free body T3
y
diagram of the joint A as shown T3
in Fig. (b).
A
W: External forces x
A
T1 , T2and T3 are internal forces
T1 T1 T2 T2

W
W
(a) (b)
ME2112 (Part 1) L2 - 15

Construct a FBD for the object shown

W B
B
C C 
R R

R R 
A 3R/2 N
A 3R
P
2R cos 

(a) A uniform bar resting on a (b) Free-body diagram of AB

semi-circular hole

ME2112 (Part 1) L2 - 16

Example 1.5 The figure below shows a floating crane in operation. Draw a
free body diagram of the crane boom. Assume frictionless pulley.

Crane boom

www.shipspotting.com
ME2112 (Part 1) L2 - 17

Solution.
Idealized Model

From the idealized model, we can start by isolating the load and
the cables attached to the lifted weight.
T2 T1

The magnitude of T1 and T2 depends on

the location of the load’s C.G.
W

ME2112 (Part 1) L2 - 18

Free body diagram for the crane boom. T1

T1
Since the pulleys are frictionless, cable
forces will remain the same after going
T1 T1
through pulleys. T2

Axial force P acts on the supporting brace T2

member (2-force member).
T1 T2 T1
Ry
Finally, the reaction forces of the
pinned connection.
Rx

W
ME2112 (Part 1) L2 - 19

Example 1.6 Draw a free body diagram of the crane boom and main
body of the mobile crane. Use the idealized model given and assume a
frictionless pulley.

Idealized Model

ME2112 (Part 1) L2 - 20

Solution.
Idealized Model Crane boom FBD

T1

Wcrane Ry T2
W T1

Rx
N
Crane main body FBD
Load to cable FBD T 1 Ry
T2
T1
Wcrane Rx

W
N
 ME2112 (Part 1) L2 - 21

Procedures for Problem Solving

1. Particle in Equilibrium
 Under coplanar forces (2D), there are two independent relations among these forces
 Use these to determine 2 unknowns such as magnitude and direction of 1 force or magnitude of
2 forces.  Fx  0 ,  Fy  0
 If more than 2 unknowns, then 1 or more additional relations must be provided.
2. Drawing FBD
 First step in solution process. It is a key to solve the problems. No problem can be solved
without drawing FBD correctly.
 Diagram shows particle & all forces (know & unknown) acting on it.
 Indicate dimensions or angles (convenience).
3. Case 1: If only 3 forces are involved
 Best to draw these forces in tip-to-tail fashion to form a force triangle.
 Triangle can be solved by trigonometry.
4. Case 2: If more than 3 forces are involved
 Best to use an analytic approach.
 Select x and y axes and resolve each force in x and y components.
 Apply equations of equilibrium to obtain 2 unknowns

ME2112 (Part 1) L2 - 22

Example 1.7 Determine the mass of each of the two cylinders if they
cause a sag of s  0.5 m when suspended from the rings at A and B. Note
that s = 0 when the cylinders are removed.
ME2112 (Part 1) L2 - 23

Solution: Since the spring system is symmetric, we only consider the FBD
at point A. Next, we compute the tension in AC, using the given geometry.
The tension in AC is given by
TAC  100 N/m( 2.828  2.5)  32.84 N

Equilibrium condition at point A

   Fy  0 : 32.84 sin 45o  m(9.81)  0
g
m  2.37 kg
2m
y
TAC
TAB 2.5 m 2m
45o 2.828 m
1.5 m
A x
45o
W 2m

ME2112 (Part 1) L2 - 24

Example 1.8 If the total length of cable ABC is 2.5 meters, determine x and
tension in the cable when the system is in equilibrium. Neglect the pulley
dimension and assume a frictionless pulley.

1.5 m
C
0.3 m A x

25 kg
ME2112 (Part 1) L2 - 25

Solution: Frictionless pulley, draw FBD at point B.

y
+
 Fx  0 left  right   g
T T
 left  right
(25)(9.81)
+
F y 0 T
2 sin
B x

T cos  left  T cos  right

25 kg
Total length of cable ABC is 2.5 meters,   left   right  

x 1.5  x 1.5 m
  2.5
cos  cos  0.3 m A x C
1.5
cos   ,   53.13 B
2.5 θ θ

( 25)(9.81) x 1.5  x
Therefore, T   153.3  153 N cos  
AB

BC
2 sin 53.13o
25 kg

ME2112 (Part 1) L2 - 26

To determine x,
1.5-2x 1.5 m
D E C
0.3 m A x

B
 
1.5-x 1.5-x

  53.13

0 .3 25 kg
tan 53.13o   1.333
1 .5  2 x
x  0.638 m
 ME2112 (Part 1) L2 - 27

Equilibrium of Rigid Bodies

• Introduction
• Moment of a Force & a Couple
• Drawing Free Body Diagram
• Equilibrium of 2-Force Member
• Equilibrium of 3-Force Member
• Solving 2-D Equilibrium Problems

ME2112 (Part 1) L2 - 28

Introduction
 Treatment of body as a single particle is not always possible. In
general, the size of the body and the specific points of application of
the forces must be considered. In addition, the body does not
translate and rotate.

 Most bodies in elementary mechanics are assumed to be rigid, i.e.

the actual deformations are small and do not affect condition of
equilibrium of the body. Thus the body is called a rigid body.

 A rigid body is an idealized body of finite size in which deformation

(shape or volume) is neglected under applied external forces.
 ME2112 (Part 1) L2 - 29

Learning Outcome

1) Identify Support Reactions

2) Draw a Free Body Diagram
3) Apply Equations of Equilibrium
4) Recognize two-force members
5) Recognize three-force members

For a given load on the platform,

how can we determine the forces at
the joint A and the force in the link
(cylinder) BC?

ME2112 (Part 1) L2 - 30

Applications

Ball-and-socket joints and journal bearings are often used in

mechanical systems. To design the joints or bearings, the support
reactions at these joints and the loads must be determined.
ME2112 (Part 1) L2 - 31

A 200 kg platform is suspended off an

oil rig. How do we determine the force
reactions at the joints and the forces in
the cables?
How are the idealized model and the free
body diagram (FBD) used to do this?

Two smooth pipes, each having a

mass of 300 kg, are supported by
the tines of the tractor fork
attachment. How can we
determine all the reactive forces?

Again, how can we make use of an idealized model and a

FBD to answer this question?

ME2112 (Part 1) L2 - 32

The aircraft wing is modeled as a tapered cantilever beam. The aerodynamic

load is approximately as static, varying linearly along the wing length. Two
concentrated loads from engine weights, and two moments due to changing
engine torques. (a) Aircraft; (b) wing model.

How can we determine the support reactions at A?

ME2112 (Part 1) L2 - 33

A reinforced concrete pier is used to

support bridge deck. The pier is under
shear force and bending moment. B
How can we determine the support A traffic light support structure is a cantilever
reactions at A & B? Again, how can beam under uniform distributed load and point
we make use of an idealized model loads. The weight of beam is modeled as w per
and a FBD to answer this question? unit length and traffic light as a point load.

ME2112 (Part 1) L2 - 34

External and Internal Forces

• Forces acting on rigid bodies
are divided into two groups:
- External forces
- Internal forces
• External forces are shown in a FBD.
If unopposed, each external force can
impart a motion of translation or
rotation or both.
• Internal forces, such as the force
between each wheel and the axel it is
mounted on, are never shown on a FBD.
ME2112 (Part 1) L2 - 35

 Reaction: When two bodies in contact, there is a pair of Reaction

Forces between them. They are always equal in magnitude but
opposite in direction (Newton’s third law)

 Weight: Weight, W = mg, is the force exerted on a body by gravity

where m is the mass of the object, g gravitational acceleration.

 Tension: Tension is the force transmitted through a cable when it

is pulled tight by forces acting from opposite ends. The tension force
is always directed along the length of the cable and directed away from the
body.

ME2112 (Part 1) L2 - 36

Conditions for Rigid-Body Equilibrium

In contrast to the forces on a particle, the forces on a
rigid-body are not usually concurrent and may cause
rotation of the body (due to the moments created by
the forces).

Forces on a particle
For a rigid body to be in equilibrium, the net force
as well as the net moment about any arbitrary point
O must be equal to zero.

F  0 (no translation)
and
M O  0 (no rotation)
Forces on a rigid body
ME2112 (Part 1) L2 - 37

Moment of a Force & a Couple

Moment of a Force
The moment of a force about a point provides a measure of the
tendency for rotation (sometimes called a torque).

ME2112 (Part 1) L2 - 38

In a 2-D case, the magnitude of the moment is M O  Fd

As shown, d is the perpendicular distance from point O to the line of

action of the force.

In 2-D, the direction of MO is either clockwise (CW) or counter-

clockwise (CCW), depending on the tendency for rotation.
ME2112 (Part 1) L2 - 39

Moment of a force about a point is equal to the sum of the moments of the
force components about the same point.

Fy
x F
Fx

y Based on principle of moments,

Mo
+ M o   Fx y  Fy x
O

ME2112 (Part 1) L2 - 40

Example 1.9 Determine the moment of force F about point O.

Solve the same problem with more than one method.

30
O

A
70

F= 5 kN
ME2112 (Part 1) L2 - 41

Solution I
We can adopt the conventional
x-y axes and find the x and 3 cos 30◦
y components of the force F. O
30
y 3 sin 30◦
Fx  F cos 70 o

Fx
Fy  F sin 70o A 70 x

Moment about point O,

+ MO  Fx (3sin 30o )  Fy (3cos30o ) Fy F= 5 kN
 ( F cos70o )(3sin 30o )  ( F sin 70o )(3cos30o )
 (5 cos70o )(3sin 30o )  (5sin 70o )(3cos30o )
 9.642 kNm

ME2112 (Part 1) L2 - 42

Solution II
We can also set the x-y axes to be
parallel and perpendicular to O 30
beam axis.
y
Fx  F cos 40 o

30
Fy  F sin 40o 40

20
Fx x
Moment about point O, Fy
F= 5 kN

+ MO   Fy (3)  ( F sin 40o )(3)  (5sin 40o )(3)

 9.642 kNm
ME2112 (Part 1) L2 - 43

A torque or moment of 12 Nꞏm is required to rotate the wheel.

Why does one of the two grips of the wheel above require less force to rotate
the wheel?

ME2112 (Part 1) L2 - 44

Moment of a Couple
A couple is a pair of equal and opposite, non-collinear parallel forces.
F
d1  M  Fd
d
A d2
F B

About A, M A  Fd About B, M B  Fd1  Fd 2  Fd

The moment of a couple is the same about any point in the plane of the couple. The
moment of a couple is often called a moment, since this is the net effect of a couple.
It may be denoted by M .
Since the moment of a couple depends only on the distance between the forces, the
moment of a couple is a free vector. It can be moved anywhere on the body and
have the same external effect on the body.
ME2112 (Part 1) L2 - 45

Equivalent of Force-Couple
A single force F acting on a point B in the cantilever beam may be considered as a
force F acting on the point A plus a couple or a moment M about the point A.

F F F
d d

A B = A B
(i) F (ii)

F
M = Fd
= d

A B
(iii)
Condition (i) and (iii) are equivalent as far as the external effects are concerned.
(External effects : the system’s rotation/translation and reaction forces)
They are not equivalent in terms of internal forces.

ME2112 (Part 1) L2 - 46

Example 1.10 The Figure below shows a typical traffic light in Singapore.
If each traffic light weighs 20 kg and the self weight of pole is negligible,
replace the loading with equivalent load and moment at point A.

4m

5m

2.5 m

A
Idealized model
ME2112 (Part 1) L2 - 47

Solution:
4m 4m Principle of
D D transmissibility
C C of force

W
5m B = 5m B

W 2.5 m 2.5 m

A A 784.8 Nm
392.4 N
Force equilibrium
Moment equilibrium
+
F x 0
+ M A  W 4   (9.81  20 )( 4)
+ F y  W  W  2  (9.81  20)  784 .8 Nm
 392.4 N

ME2112 (Part 1) L2 - 48

READING QUIZ
F = 12 N
1. What is the moment of the 12 N force about point A
(MA)?
A) 3 Nꞏm B) 36 Nꞏm C) 12 Nꞏm
d=3m
D) (12/3) Nꞏm E) 7 Nꞏm • A

2. The moment of force F about point O is

defined as MO = ___________ .
A) r × F B) F × r
C) r • F D) r * F
ME2112 (Part 1) L2 - 49

Reduction of a system of forces to one force and a couple

---- equivalent force-couple system at a given point O

When several forces and couple moments act on a

body, we can move each force and its associated
couple moment to a common point O.
Now we can add all the forces and couple
moments together and find one resultant force-
couple moment pair.

n
FR   Fi
i
n n
( M R )O   ( MO )i   M   (ri  Fi )   M
i i

ME2112 (Part 1) L2 - 50

WR = W1 + W2
(MR)o = W1 d1 + W2 d2

If the force system lies in the x-y plane (a 2-D case), then the reduced
equivalent system can be obtained using the following three scalar equations.

FRx   Fx
FR y   Fy
( M R )O   M O   M
ME2112 (Part 1) L2 - 51

Example 1.11 Consider the following canopy beam. If the suspend

tension is 6 kN, replace the loadings with a single force and determine
where its line of action crosses the beam, measured from A.

C
60

0.3 m
5 kN 10 kN B
A

1m 1m 1m
www.mapes.com

ME2112 (Part 1) L2 - 52

Solution: Plan
(i) y
• Calculate the force resultant x 0.3 m
and y components from (i), 5 kN 10 kN 60 B
i.e. ∑Fx and ∑Fy A x
• The idea is that the moment 1m 1m 1m
resultant about A is equal
between (i) and (ii).
• By equating both systems, we (ii)
can obtain d.
∑Fy
• Finally, we can calculate the
resultant of Fx and Fy as ∑Fx
A
well as its direction.
d
ME2112 (Part 1) L2 - 53

(i) Calculate the force resultant’s x and y components and the moment resultant
about point A.
y 6sin 600
0.3 m
5 kN 10 kN 60 B 6 cos 600 60 B
A x
1m 1m 1m 6cos 60o (0.3)

+ F x  6cos60o  3 kN
+ F y  5 10  6sin 60o  9.8 kN

+ M A( i )  5(1)  10(2)  6sin 60o (3)  6cos 60o (0.3)

 8.51 kNm

ME2112 (Part 1) L2 - 54

(ii) Calculate the moment resultant about A by using the force resultants
calculated in (i).

∑Fy 9.8 kN
y
∑Fx 3 kN
A A
x
d d

+ M A( ii )  9.8(d )
Equate moment resultants from (i) and (ii), we have

M A(i )   M A(ii )
8.51 =  9.8(d )
8.51
d  0.87 m
9.8
ME2112 (Part 1) L2 - 55

(iii) Calculate the resultant force and its direction

9.8 kN
10.3 kN
y
3 kN 73
A A
x
d 0.87 m

FR  (3) 2  ( 9.8) 2  10.3 kN

 9.8 
  arctan    73
o

 3 

ME2112 (Part 1) L2 - 56

Equations of Equilibrium
External forces and moments acting on a rigid body may be reduced to a
force-couple system at some arbitrary point O. When the force and the
couple are both equal to Zero, the rigid body is said to be in Equilibrium.
n n
FR   Fi  0; ( M R )O   (ri  Fi )   M  0
i i

 F  0;  F
x y  0; F  0
z

 M  0;  M
x y  0; M  0 z

Note: 6 equations to solve 6 unknowns

For 2-D rigid body contained in the x-y plane, we have

F x  0; F y  0; M O 0
ME2112 (Part 1) L2 - 57

Process of Solving Rigid Body Problems

Actual physical system Idealized model FBD of platform

For analysing an actual physical system, first we need to create an idealized model.
Then we need to draw a FBD showing all the external (active & reactive) forces.
Finally, we need to apply the equations of equilibrium to solve for any unknowns.

Idealized Free Body Equilibrium

Model Diagram Equations

ME2112 (Part 1) L2 - 58

Drawing Free-Body Diagram

Example of FBD

Idealized model Free-body diagram (FBD)

1. Draw an outlined shape. Imagine the body to be isolated or “free” from its constraints
and draw its outlined shape.
2. Show all the external forces and couple moments. These typically include: a) applied
loads, b) support reactions, and c) the weight of the body.
3. Label loads and dimensions on the FBD: All known forces and couple moments
should be labeled with their magnitudes and directions. For the unknown forces and
couple moments, use letters like Ax, Ay, MA. Indicate any necessary dimensions.
ME2112 (Part 1) L2 - 59

Example of FBD
Given: The operator applies a vertical force to the
pedal so that the spring is stretched 37.5 mm and the
force in the short link at B is 100 N.
Draw: An idealized model and FBD of the foot pedal.

F  ks  ( 4 N/mm)(37.5 mm)  150 N

ME2112 (Part 1) L2 - 60

Support Reactions for 2-D Structure

If a support prevents translation
of a body in a particular direction,
then the support exerts a force on
the body in that direction.
If a rotation is prevented, then the
support exerts a moment on the
body.
Reactions in 2-D structures can be
divided 3 type of groups:

1) Reactions equivalent to a force

with known line of action.
ME2112 (Part 1) L2 - 61

Supports Reactions for 2-D Structure (Continued)

2) Reactions equivalent to a
force of unknown
direction and magnitude.

3) Reactions equivalent to
a force of unknown
direction and magnitude
and a couple of
unknown magnitude.

Rocker support Cable support Fixed support

Pin support Pin support

ME2112 (Part 1) L2 - 62

Practice of FBD

A B
150 kN

B is the most correct, though C is also

correct. A & D are incorrect; why?
Choose the most C 150 kN
D
150 kN
correct FBD for the
150 kN

frame support
shown.
ME2112 (Part 1) L2 - 63

READING QUIZ 1 (L2)

1. In statics, a couple is defined as __________ separated by a
perpendicular distance.
A) two forces in the same direction
B) two forces of equal magnitude
C) two forces of equal magnitude acting in the same direction
D) two forces of equal magnitude acting in opposite directions

2. The moment of a couple is called a _________ vector.

A) Free B) Spinning
C) Fixed D) Sliding

ME2112 (Part 1) L2 - 64

READING QUIZ 2 (L2)

1. A general system of forces and couple moments acting on a rigid body
can be reduced to a ___ .
A) single force
B) single moment
C) single force and two moments
D) single force and a single moment

2. The original force and couple system and an equivalent force-couple

system have the same _____ effect on a body.
A) internal B) external
C) internal and external D) microscopic
ME2112 (Part 1) L2 - 65

ATTENTION QUIZ 1 (L2)

1. If a force of magnitude F can be applied in four different 2-D configurations
(P,Q,R, & S), select the cases resulting in the maximum and minimum torque
values on the nut. (Max, Min).
A) (Q, P) B) (R, S)
C) (P, R) D) (Q, S)

S
P
Q R
10 N 5N
3m P 2m

2. Using the CCW direction as positive, the net moment of the two forces about
point P is
A) 10 Nm B) 20 Nm C) -20 Nm
D) 40 Nm E) -40 Nm

ME2112 (Part 1) L2 - 66

ATTENTION QUIZ 2 (L2)

1. A couple is applied to the beam as shown. Its moment equals _____ Nꞏm.
A) 50 B) 60 50 N

C) 80 D) 100 1m 2m 5
3
4

2. Consider three couples acting on a body. Equivalent systems will be _______ at

different points on the body.
A) Different when located
B) The same even when located
C) Zero when located
D) None of the above.