Download Polygon of Forces

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Topic
Equilibrium of Forces and Moments
“Mechanical Engineering Science, 2nd Edition” by J.
Hannah & M. J. Hillier, Longman
Chapter 1 – Statics, Chapter 2 - Moment
_____________________________________________________________________
Reference:
Force and Newton's First Law
If the resultant force acting on a particle is zero, the particle will remain at rest (if
originally at rest) or will move with constant speed in a straight line (if originally in
motion).
To understand and use the First Law it is helpful to first develop a concept of a state
of Equilibrium. Equilibrium exists when all the forces on a particle are in balance.
The velocity of a particle does not change, if the particle is in Equilibrium.
Interpretations of the First Law
i) A body is in Equilibrium if it moves with constant velocity. A body at rest is a
special case of constant velocity i.e. v = 0 = constant.
ii) For a body to be in Equilibrium the resultant force (meaning the vector addition of
all the forces) acting on the body must be zero.
iii) A Force can be defined as 'that which tends to cause a particle to accelerate',
assuming that the force is not in Equilibrium with other forces acting on the body.
Note: A force cannot be seen only the effect of a force on a body may be seen.
Force Units
The S.I. unit of force is the Newton abbreviated as N. For example 217 N. For most
engineering applications forces are of the order of N or kN.
Force Vectors - Components in Polar and Rectangular coordinates
A Force is a vector quantity since it has direction as well as magnitude.
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fy
Fy
F

Fx
fx
Figure 1
In the x-y plane the force vector F in the above diagram can be resolved into its
components in the x and y directions (rectangular coordinates), where,
F = Fx  Fy ;
where Fx , and Fy are the components of F in the x and y directions.
The vector F can also be represented in terms of components in polar coordinates,
F and  . The components in rectangular (x,y) coordinates can be expressed in terms
of the components in polar coordinates as follows:
Fx = F.cos; and
Fy  F.sin;
and the inverse relationship as:
 Fy 
2
2
F = Fx  Fy and   tan -1  
 Fx 
Addition of Force Vectors and the Resultant Force
Figure 2 shows forces F1 and F2 acting at a point (on a particle). The vector R is
called the Resultant and is equal to the vector sum (addition) of force vectors F1 and
F2. The resultant force can be found by calculation (Algebraic) method or graphical
(triangle of force) method.
Force F1
Force F2
Particle
Note
Resultant R
means "equivalent to"
Figure 2
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Algebraic Method
The vectors can be added algebraically by first resolving the forces into their
rectangular components in the x and y directions, adding the components and then
converting back into polar coordinates.
F2, y
F2
2
F1,x
F2, x
1
F1
F1,y
Figure 3
R x  F1,x  F2 ,x  F1 cos 1  F2 cos  2
Ry  F1, y  F2, y   F1 sin 1  F2 sin  2
R  Rx  Ry
2
2
 Ry 
  tan 1  
 Rx 
Triangle of Forces Method
In this method the vectors are drawn so that the tail of one vector joins the head of the
preceding vector. The length of each force vector is proportional to the magnitude of
the force it represents and is drawn parallel to the direction of the force.
The vector R is then drawn to close the triangle. The order in which the vectors are
drawn is not important as long as they are drawn head to tail. Figure 4 shows F1
drawn first and Figure 5 shows F2 drawn first.
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Fy
Fy
F2
R
F1
R
F2
F1
Fx
Fx
Figure 4
Figure 5
Equilibrium of Forces on a Particle (Equilibrium of Concurrent Forces)
F1
Equilibrant E
Equilibrant E
Force
F2
Resultant R
Figure 6
From Newton's First Law for a particle to be in equilibrium there can be no resultant
force on the particle. The Equilibrant (the force which puts a system of forces into
equilibrium) must therefore be equal and opposite to the Resultant R to reduce the
force on the body to zero.
Equilibrant = - Resultant
E  R
R
Fy
E
F2
F1
Fx
Figure 7
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The equilibrant may be shown by reversing the direction of the arrow of the resultant.
In the triangle of forces method this makes all the arrows on the vector diagram point
in the same direction around the triangle. This then provides a test for equilibrium. If
all the forces in a triangle of forces diagram are in the same direction then the
forces must be in Equilibrium.
If there are more than three forces the triangle becomes a polygon of forces.
If the diagram does not close then there is a net resultant force i.e. the body (particle)
is not in equilibrium and will tend to accelerate. (i.e. F = ma ,)
Equilibrium of Components of Force Vectors for Coplanar and Concurrent
forces
Coplanar forces: force vectors being in one plane- two dimensional system (e.g.
forces in x-y plane, no forces in z direction)
Concurrent forces: all forces vectors acting at one point or one particle
The rectangular components Fx and Fy of the force vectors which are in equilibrium
are independent variables. Therefore, if the sum of the vectors add up to zero, then
the sum of their components must also add up to zero. Hence,
for equilibrium
i=n
 Fi  0; it follows that
i=n
 Fx, i  0; and
i=n
F
i=0
i=0
i=0
y, i
 0;
for example, for a particle under the action of three forces to be in equilibrium,
Fx,1  Fx,2  Fx,3  0;
Fy,1  Fy,2  Fy,3  0;
In other words for equilibrium all the components of forces in the x direction must
add up to zero and all the components in the y direction must add up to zero.
Polygon of Forces
The same methods can be applied to multiple forces as to three forces but instead of a
triangle of forces the diagram becomes a polygon. Consider a particle in equilibrium
with five forces acting on it.
F3
F2
F5
F1
F4
Figure 8
F3
F2
F5
F1
F4
Figure 9
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Figure 8 shows the forces acting on a particle which is not a very practical case.
Figure 9 shows a more realistic case with the forces acting on a rigid body but the
result is the same provided that all the forces intersect at a common point. (i.e.
Concurrent)
Particle
is a material body whose linear dimensions are small enough to be
irrelevant.
Rigid Body is a body that does not deform (change shape) as a result of the forces
acting on it.
Figure 10 shows the polygon of forces. If any of the five forces were not present the
body would not be in equilibrium and the polygon of forces would not close. For
example suppose force F5 is removed, then there would be a net resultant R (Figure
11) acting on the body, equal and opposite to the force F5.
F1
F1
F2
F2
F4
F4
Fy
F3
F5
F3
y
R
Fx
Figure 10
x
Figure 11
In the latter case the unbalanced force or resultant has the same magnitude but
opposite direction to the force F5. Any one of the five forces can be considered as the
equilibrant to the remaining four forces.
R = -F5
Or equilibrant F5 = - Resultant R
Reactions, External Forces and Free Body Diagrams
Newton's Third Law
The forces of action and reaction between bodies in contact have the same magnitude,
but opposite in direction..
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Hammer
BANG!
Solid Surface
Figure 12
Free Body Diagram of Hammer
Hammer
Free Body
Boundary
Force on Hammer caused by Surface
Force on Surface caused by Hammer
Solid Surface
Free Body diagram of Surface
Figure 13
Figure 12 shows a hammer striking a solid surface and Figure 13 shows how the
forces between the hammer and the surface can be viewed separately.
Isolating one body or group of bodies and considering the external forces on the body
is referred to as a free body diagram (the body is free of other connected bodies).
An imaginary boundary is drawn around the free body to show which parts and forces
are to be included in the analysis.
The ‘Force on the Surface caused by the Hammer’ is an Action Force due to the
hammer and the ‘Force on the Hammer caused by the Surface’ is the Reaction Force
on the hammer.
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
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In most problems, Reaction Forces are those forces produced on a body by the fixed
points supporting the body. In the above case the ‘Force on the Hammer caused by
the Surface.
Another example is the weight of a body supported by a solid surface.
W
Free Body
W
Boundary
N Reaction
Wa Action
Figure 14
Consider a body of weight W supported by a solid surface. The body is in
equilibrium since the body's weight W, is balanced by an equal and opposite reaction
N. Mathematically, W = N.
Similarly there will be a resultant action Wa of the body on the solid surface. In the
right hand diagram the dotted box encloses what is known as a free body diagram and
shows all the external forces that act on a body, both actions and reactions.
Summation of the forces over a free body diagram will determine the resultant force
acting on the body which will be zero if the body is in equilibrium.
System of Particles or Bodies
Two or more bodies or particles connected together are referred to as a system of
bodies or particles.
External Forces
External forces are all the forces acting on a body defined as a free body or free
system of bodies, including the actions due to other bodies and the reactions due to
supports.
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Transmissibility of Force
Statics and dynamics of rigid bodies are concerned only with the external forces
applied to a body and it is assumed that a body does not deform. It also assumes that
it does not matter at which point on the body, along the line of action of the force, the
force is applied. This is known as the principle of transmissibility of force.
F
F
F
Figure 15
Loads are forces that are applied to bodies or systems of bodies.
For example, gravity acting on a mass will produce a load called the weight of a
body; pressure inside a hydraulic piston will produce a load on the piston; wind acting
on a building will produce a load on the building; water flowing from the nozzle of a
turbine will produce a load on the turbine blade, etc.
Reactions at points supporting bodies are a consequence of the loads applied to a
body and the equilibrium of a body.
Tensile and Compressive Forces
When a body is in equilibrium under the action of two forces (which must be equal
and opposite if the body is in equilibrium) the forces can either,
a) Push on the body which is called a compressive force; or,
b) Pull on a body which is called a tensile force.
Reaction
Action
Reaction
Action
Tensile Force
Compressive Force
Figure 16
The body is said to be in tension or in compression.
Procedure for drawing a free-body diagram
Since we must account for all the forces acting on the particle, the importance of
drawing a free-body diagram before applying the equation of equilibrium to the
solution of a problem cannot be over-emphasized. To construct a free-body diagram,
the following three steps are necessary.
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
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Step 1: Imagine the particle to be isolated or cut “free from its surroundings. Draw or
sketch its outlined shape.
Step 2: Indicate on this sketch all the forces that act on the particle. These forces can
be applied surface forces, reaction forces and/or force of attraction.
Step 3: The forces that known should be labeled with their proper magnitudes and
directions. Letters are used to represent the magnitudes and directions of forces that
unknown.
Example 1
Fp
Supported
pulley
Pulling
Force
Fs
F1
Spring
F1
Spring
Pulling
Force
Mass
Fs
Mass
mg
Example 2
Ceiling Support
Ring
Weight = 10 N
Consider a weight connected to a ring
that is hanging from the ceiling. Draw
a free body diagram for the weight
and the ring showing all external
forces on the free body. Also show
the force on the ceiling due to the
weight.
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Ceiling Support
Action Load = 10 N
Reaction Force = 10 N
Ring
Weight = 10 N
Free Body boundary
Gravity Load =10 N
Example 3
The diagram shows a barge been towed by two tugs using tow ropes connected to a
ring on the barge’s bow. The tow rope from tug No. 1, exerts a force of 10 kN on the
barge in line with the rope and the tow rope from tug No. 2 exerts a force of 30 kN.
tow rope
Tug No.1
25
Barge
60
Ring
.
Tug No.2
tow rope
a)
Draw a free body diagram of the ring showing the reaction force due to the barge
in terms of its two components.
b) Find the resultant force acting on the barge and its components in the x and y
directions.
c)
Find the reaction force on the ring due to the barge.
d) Find the resultant R of the two forces in tow ropes No.1 & No. 2 from the
components in the x and y directions.
Solution
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R x  10 cos 25  30 cos 60  24.06 kN
R y  10 sin 25  30 sin 60  2175
. kN
R  24.06 2  2175
. 2  32.43 kN
2175
.
  tan 1
 42.1
24.06
The equilibrant E   R
E = 32.43 kN
42.1
42.1
R = 32.43 kN
The resultant R is the sum of the actions of the tow ropes on the barge. The
equilibrant E is the reaction of the barge to the tow ropes.
Try yourself by graphical method:
a) Draw a triangle of forces with scale of 1cm = 5 kN and find the reaction force
on the ring due to the barge and its components in the x and y directions.
b) Draw a diagram showing the total reaction force on the ring as well as the tow
rope forces;
c) What is the resultant force on the barge.
d) Compare your result with the calculation method.
Moments and Couples
The Definition of a Moment
When a force acts on a body (not in equilibrium) in addition to tending to cause it to
translate (move in a straight line), it can cause it to rotate. The force multiplied by the
perpendicular distance to the point of rotation is called a Moment. Moment is a
measure of the turning effect about a point (or an axis).
Moment = Force X Perpendicular Distance
M  Fd
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F
A
A
d
Figure 17 (a)
M=Fxd
Figure 17 (b)
Figure 17 (a) shows a force F acting at a perpendicular distance d from a point A
which creates a moment M = F.d. The distance d is called the moment arm. Moments
are shown diagrammatically by a curved arrow as in Figure 17 (b). If the force acts at
a point B which is a distance r from A then the moment must be obtained from the
perpendicular distance d and not the distance r.
F
A
r

B
d
M = F.d
= F.r.cos 
Figure 18
Moment = Force X Perpendicular Distance
M  F  r cos
The value of the moment M is said to be obtained by taking the moment of the force
F about the point A or briefly by taking moments about A.
Note that the in the above examples the body is not in equilibrium hence the force F
tends to accelerate the body in translation and rotation.
Moment Vector
Moment is the cross product of the force vector and the distance vector. Therefore
moment is also a vector. Its magnitude is equal to the product of force and the
perpendicular distance from a point (or axis). The direction of the moment vector is
perpendicular to the plane containing the force and distance vectors or along the
direction of the axis of rotation.
In two dimensional problems, the direction of the moment vector can be described
completely by its sense, that is clockwise or anti-clockwise. Summation of moments
in a single plane then reduces to the algebraic sum, where a +ve is assigned arbitrarily
to one sense or the other.
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Resultant (force and moment) of a system of forces
Figure 19 (a) shows an arbitrary body subjected to a number of forces F1, F2, F3. The
combined effect, or resultant, R, of these three forces is determined by the vector sum,
so that
R = F1 + F2 + F3
These can be evaluated graphically by scale drawing, or analytically by resolving
each force into components, say x and y, summing the components in each direction
to give
Rx = F1x + F2x + F3x
Ry = F1y + F2y + F3y
and then recombining to find the resultant force. Figure 19 (b) shows clearly that the
two methods are equivalent but, whichever is used, only the magnitude and direction
of the resultant can be found. The line of action of the resultant is still unknown.
l3
O
Rx
F3
l2
l1
l
R
F2
F 3y
F3
R
Ry
F2
F1
F1
F 1y
F 1x
(a)
F 3x
F 2y
F 2x
(b)
Figure 19 Resultant of a system of forces
The line of action can be obtained by equating the sum of moments of F1, F2 and F3
about some arbitrarily chosen point, to the moment of the resultant, R, about O.
Taking anti-clockwise positive, this gives
Mo = F1l1+ F2l2 + F3l3 = Rl
where l is the moment arm length of the line of action of R from O. Hence the
position of R is determined (Figure 19a).
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In some cases the vector diagram, Figure 19b, may close by itself, so that R = 0;
however, this does not necessarily mean that the sum of the moments is also zero.
Consider the special case of two equal and opposite forces, F, distance d apart, as
shown in Figure 20. Clearly there is no resultant force, but taking moments about O,
clockwise positive, gives
Mo = F(d + l) - Fl = Fd
O
l
F
d
F
Figure 20
A system such as this is known as a couple. Note that the moment of a couple is the
same about every point in its plane.
Example 4
The diagrams show a body subjected to four forces; calculate the total (resultant)
moment on the body.
a)
300 mm
15 N
30 N
170 mm
30 N
A
100 mm 15 N
Taking moments about the corner A
+
M  30  017
.  15  0.3  30  0.05  15  01
.
 30  0120
.
 15  0.2  6.6 Nm
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Note that the forces form two couples or pure moments 3.6 Nm and 3.0 Nm
(resultant force =0, moment is the same about any point)
Equilibrium of Moments
In previous section, the forces on a body were defined as being in equilibrium when
‘the vector sum of all the forces acting on the body was zero’. In the same way, the
sum of all the moments is zero when the body is in moment equilibrium. This can be
written mathematically as:
n
M
i
 0;
M1  M 2 .......  0
or
i 1
It can be proved that if the body is in equilibrium the sum of the moments of all the
forces on acting on a rigid body is the same for all points on the body. This means
that it does not matter at which point on a rigid body you choose for taking
moments about.
In Example 4, the bodies are in vertical and horizontal force equilibrium thus the
moment does not depend on where you take moments. In general this is not the case.
If there is a non-zero force on the body the resultant moment will depend upon where
you take moments.
Example 5
0.5 m
10 N
B
15 N
1.0 m 1.25 m
A
5N
2.5 m
3.0 m
Figure A
For the body shown in Figure A find the resultant moment and the equilibrant
moment. Note that in this case the body is not in vertical and horizontal equilibrium.
Taking moments about A, gives the sum of the moments i.e. the resultant moment
M R  10  0.5  15  10
.  5  2.5  22.5 Nm
but if moments are taken about B the result is,
+
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+
M R  10  2.5  15  0.25  5  0.5  18.75 Nm
This difference is because the body is not in vertical and horizontal equilibrium as
was the case in the previous examples. There is therefore no unique value for the
resultant moment since the value depends on where the resultant force acts.
General Equations of Equilibrium of a Plane (Two Dimensional) Rigid Body
(Non-concurrent forces)
A plane rigid body is one for which all the force vectors lie in the two dimensional x y plane . For equilibrium,
i=n
n
 Fi  0;
M
i
0
i 1
i=0
and expanding the force vector into its components in the x and y directions,
i=n
 Fx, i  0; and
i=n
 Fy, i  0;  M i  0
n
i=0
i=0
i 1
for example, with three forces acting on a rigid body,
Fx,1  Fx,2  Fx,3  0;
Fy,1  Fy,2  Fy,3  0;
M 1  M 2 .......  0
For complete equilibrium of a plane rigid body all three equations must be satisfied.
Applications of Equilibrium
Types of Beam Supports
Many engineering components can be 'approximated' or 'modeled' by a single span
beam. The calculation of stresses and deflections in a beam is often preceded by an
analysis of equilibrium. In most cases this involves obtaining the reactions i.e. the
forces and moments at the supports of the beam. Consider the following cases;
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A beam simply supported on points or knife-edges
A side view or elevation of a simply supported beam
A diagrammatic view of a simply supported beam
A beam built in to a wall
Called a 'built in' beam
or 'fixed beam'
and sometimes called a ‘Cantilever’
Diagrammatic sketch of built in beam
Types of Support and Connections
Deciding how a body or a structure is supported (connected to the ground) or
connected to an adjacent body is very important when calculating the reactions
applied to the body.(free body diagram). The three most common assumptions are:
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Force perpendicular
to surface only.
Pin Joint supported by a roller
Ry
Both parallel and
Rx
perpendicular forces
Pin Joint Fixed to the
ground
Fix Support
Ry
Two components of force and
Rx
M
a moment
Ry
Types of Loading on Beams
Loading on beams can be divided into two types,
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a)
W
Point load
W
Diagrammatic representation
W / unit length
Distrubuted load
(uniform)
b)
W / unit length
Diagrammatic representation
W / unit length
Another way of representing
Uniformly distributed loads
Distributed loading can be either uniform or non-uniform,
Non-uniformly distributed loads
c)
The point supports of a simply supported beam can be replaced by vertical reactions
at those points. Point or simple supports allow the beam to rotate at the support hence
the moment is zero at this point. Theoretically there could be a horizontal reaction if
there was a horizontal load but this is rarely the case for beams. To ensure that no
horizontal loads are implied a support which allows horizontal movement can be
used.
W
Beam with right hand support on rollers to allow free horizontal movement
The analysis of beam and other statics problems is aided by the use of free body
diagrams and the replacement of supports and loads by reaction forces and weights on
a free body diagram.
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C
A
B
W
Diagrammatic view of simply
supported beam with a
b
a
concentrated load W
W
B
A
Free body diagram replacing
R B loads and supports by forces
C
RA
w N/m
A
a
b
B
Diagrammatic view of simply
supported beam with uniformly
distributed load w
F= w(b-a)
Free body diagram replacing
loads and supports by forces
(a+b)/2
The uniformly distributed load w has been replaced by a single force acting at the
centre of the distribution of load.
Example 6
Consider a beam built into a wall as shown in Figure A. When a beam is built into or
fixed to a wall so that it cannot rotate at the wall it is called a cantilever beam. The
beam is supporting a load of weight W, at a distance d from the wall.
Wall
Built in End
or Fixed End
Point Load of weight W
d
Beam
Figure A
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Figure B shows the free body diagram of the beam with a reaction force V and
reaction moment MA at the wall A.
W
Moment Reaction MA
C
A
B
d
Reaction V
Free Body Diagram of Beam
Figure B
The reactions are the forces that put the beam in equilibrium. Therefore, from force
and moment equilibrium and taking moments about A;
 M  M  W. d  0
  F  V  W  0
+
A
y
M A  W. d
VW
(The point C on the beam is d/4 from B, what will be the value of MA by taking
moments about C. Compare your answer with the one above.)
Example 7
Find the reactions at the supports for the beam shown in the diagram.
20 kN
5 kN/m
3.5m
3.5m
7m
First draw a free body diagram, showing that the uniformly distributed load can be
considered to act at the mid point along the load and supports can be replaced by
reaction forces.
20 kN
35 kN
RB
RA
3.5m
7m
3.5m
Free Body Diagram
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W  5000  7  35 kN
Taking moments about A,
+ve
From vertical equilibrium
 R B  14  35  10.5  20  35
. 0
R B  3125
. kN
R A  R B  35  20
 R A  23.75 kN
Example 8
A steel block of weight W is placed on a beam resting on a pivot (or fulcrum). The
block is lifted by applying a force F at a distance a from the fulcrum. If the steel
block is a distance b from the fulcrum express F in terms of m, a and b.
b
a
F
b
a
F
W
W
R
Illustration
Free Body Diagram
 F = 0;
Taking moments about pivot;
Thus, F 
b
W;
a
+
+ ; R - F - W = 0
 M  Wb  Fa  0
Where the ratio
a
is referred to as the Mechanical Advantage.
b
For example if a = 30 cm and b is 10 cm the beam or lever is said to have a
mechanical advantage of 3:1 (3 to 1). This means that the force F can lift a weight W
three times greater than F. Taking moments about the pivot eliminates the reaction R
from the calculation but from the vertical equilibrium equation above:
R=F+W
Moments can be taken about any point, such as at point of application of F, hence,
Wa  b  Ra  0
Solving the two latter equations to eliminate R gives the same solution as before.
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
Dept of REFM
Example 9
Two forces are acting on a beam as shown in figure. If the beam is fixed supported at
position A and free supported at position B (that is point B can only take up vertical
force). Determine the reaction forces at the supports A and B.
(Hint: By considering the sum of force is zero at equilibrium and taking the moment
at point A and at point B.)
350N
A
B
0.15m
30o
200N
0.30 m
0.40 m
Solution:
Consider the sum of vertical forces and horizontal forces are zero, and since R Ax = 0
as point B can only take up vertical force.
RAx = 200 x sin30o = 100 N
RAy + RBy = 350 + 200 cos30o = 350 + 173.2 N
= 523.2 N
Taking moment about A,
RBy x 0.3 = 350 x 0.15 + 200 cos30o x 0.4
RBy x 0.3 = 121.8 N
RBy = 405.9 N
Therefore the reaction at point B is 405.9N upward.
RAy = 523.2 - 405.9 = 117.3 N
The reaction at point A is 117.3 N upward and 100 N to the left.
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
Dept of REFM
Resultant at A is:
RA = (117.32 + 1002)
= 154.1 N
angle  = 49.55o
117.3

100
Example 10
The diagram shows a rectangular plate suspended from a wall. Support A is a fixed
pinned support and B is pinned but can move freely in the vertical direction. Find the
reactions at A & B
55 kN
55 kN
HA
A
VA
750 mm
750 mm
HB
B
600 mm
600 mm
Illustration
Free Body Diagram
   Fy  VA  55  0
Vertical Equilibrium:

Horizontal Equilibrium


  Fx  H A  H B  0
Moment Equilibrium:taking moments about B
+
 M H
A
 0.75  55  0.6  0
Solving the three equations,
VA  55 kN ;
H A  44 kN ;
H B  44 kN
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
Dept of REFM
Tutorial 1
Find the components of the 500 N force in rectangular coordinates
y
500N
147
x
(Ans.-419.3 N, 272.3 N)
Tutorial 2
A force vector has the components -45 N and 72 N in rectangular coordinates, in the x
and y directions respectively, what are the components in polar coordinates. Draw the
vector diagram.
(Ans. 84.9 N, 122)
Tutorial 3
Find the magnitude and direction of the resultant (i.e. in polar coordinates) of the two
forces shown in the diagram,
a) Using the graphical Method
b) Using the algebraic calculation method.
750N
450N
60
45
Ans. 969N, 93.4 from the +ve x axis
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
Dept of REFM
Tutorial 4
There are three forces acting on a particle which is in equilibrium find the value of the
forces P and Q.
Hint: The inclination of P and Q is given and is fixed but the sign of P and Q may
not be as shown in the diagram.
Solve the problem by resolving forces into components.
Q
370N
15
30
45
P
(Ans. Q = -714 N, P = 714 N)
Tutorial 5
The diagram shows three forces acting on a particle
45 kN
84 kN
20
30
50
122 kN
Find the equilibrant by drawing the polygon of forces and by calculation method.
Ans: E = 61 kN at 37o to the x-axis.
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
Dept of REFM
Tutorial 6
The diagram shows four forces acting on a particle.
60 kN
100 kN
60
30
10
80 kN
110 kN
Find the equilibrant and the resultant using the polygon of forces.
21 ;
(Ans. E = 44 kN
R = 44 kN
21
)
Tutorial 7
Two weights are suspended on ropes which are tied to a ring embedded in a solid
wall. The ropes run over frictionless pulleys so it may be assumed that the tension
force in the ropes is equal on both sides of the pulley.
a) Draw a free body diagram for the ring, the weights and the pulley/rope system.
Frictionless
Pulley
Ring
30 Rope
Solid
Wall
20
50 N
Weight
Rope
Frictionless
Pulley
Weight
40 N
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
Dept of REFM
Tutorial 8
The diagram shows ship mooring ropes tied to a ring bolt
1570 N
350 N
30
30
1700 N
2550 N
What is the magnitude and direction of the resultant force on the ring bolt and the
reaction at the anchoring point. Calculate the vertical and horizontal components of
the reaction. (Ans. 2967 N, -20, -2788 N, 1015 N)
Tutorial 9
The 150 N collar may slide on a frictionless vertical rod and is connected by a rope,
that runs over a frictionless pulley, to a 170 N counter weight. Determine the value of
h for which the system is in equilibrium.
40 cm
frictionless
rod
h
170 N
Weight
150 N
Collar
(Ans. 75 cm)
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
Dept of REFM
Tutorial 10
Find the reaction forces at A and B.
100 kN
40 kN
A
B
2m
0.5m
5m
0.5m
Ans. 126 kN, 14 kN
Tutorial 11
Find the reaction force and moment at the built-in end of the cantilever shown.
20kN/m
1m
10 kN
3m
5.5 m
Ans. 30 kN, - 25 kNm
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics
Dept of REFM
Tutorial 12
The fixed crane shown in the diagram is attached at its support by a pin joint at A and
a rolling connection at B. The weight of the jib is 10 kN and acts as shown.
Calculate the reactions at A and B due to lifting a 50 kN weight.
A
4m
50 kN
10 kN
B
6m
10 m
(at A,-215 kN,60 kN, at B, 215 kN)
ENG2301 Principles of Engineering – Lecture Notes on Basic Mechanics