Unit 2 mm9400 ver 1.1(2014)

Equilibrium Conditions 2 - 1
Statics & Dynamics (MM9400) Version 1.1
2. EQUILIBRIUM CONDITIONS
Objectives: At the end of this unit, students should be able to:
 Understand the concept of "Force"
 Define a vector quantity.
 Distinguish between a vector and a scalar quantity.
 Define "Resultant Force."
 Understand the Addition of Forces
 Resolve a force into 2 components analytically.
 Determine analytically the resultant of several forces using rectangular
components.
 Understand the Moment of a Force
 Define the moment of a force about a point.
 Calculate the moment of a given force about a point.
 Sum up the moments due to several forces acting on body about a point.

2 - 2 Equilibrium Conditions
Version 1.1 Statics & Dynamics (MM9400)
 Understand the concept of Free Body Diagrams
 Define a Free Body Diagram (FBD).
 Know the nature of different types of forces, namely, push, pull (e.g. tension),
weight and reaction.
 Know the reactions present at various supports, e.g. rollers, pin-joints, fixed-ends.
 Sketch FBDs of two-dimensional mechanical systems in equilibrium.
 Understand the Equilibrium of coplanar, concurrent force systems
 State the conditions of equilibrium for concurrent forces.
 Determine if a body is in equilibrium.
 Solve Statics problems involving concurrent force systems.
 Understand the Equilibrium of coplanar, non-concurrent force systems
 State the conditions of equilibrium for non-concurrent forces.
 Determine if a body is in equilibrium.
 Solve Statics problems involving non-concurrent force systems.

2.1 Vector Quantity
A force is a vector quantity that has magnitude(size) and direction. The direction is
indicated by an angle  measured from a reference plane and the sense is shown by
the arrow head.
Fig. 2.1 shows a force vector lying in the plane of the paper with a magnitude of
5 N acting at an angle of counter-clockwise from the positive x-axis.
If several forces lie on the same plane, they are said to be co-planar. Fig. 2.2
shows three sets of co-planar forces, each on a different plane.
Fig. 2.2
Fig. 2.1
5 N

2.2 Resolution of a Force
Resolution of a force means “splitting or resolving” it into two or more component
forces. It is the reverse process of adding two given forces together.
The most common way to resolve a force is to split it into two mutually
perpendicular components, also known as rectangular components. See Fig. 2.3 a
(Parallelogram method) and Fig. 2.3 b (Triangle method)
Therefore F2 = F sin and F1 = F cos
where F1 and F2 are the rectangular components of F.
Example 2.1
Find the components for the given 100 N force along the x- and y- axes.

F
F1
F2
F2
F1
F


F
F1
F2
F2
F1
F

Fig. 2.3 a
Fig. 2.3 b
Fx = F cos = 100 cos 30° = 86.60 N 
Fy = F sin = 100 sin 30° = 50 N 


F
F
F
F
triangleshadedtheFrom
cos;sin
,
12
 

F = 100 N
Fx
Fy
x
y

Example 2.2
Find the magnitude of the rectangular components of the 100 N force along the x-
and y- axes and indicate their directions with respect to the horizontal.
TUTORIAL
2.1 Resolve the given forces into the specified directions.
2.2 Resolve the 300 N pulling force (tension) into rectangular components in
(a) the x and y directions.
(b) the vertical and horizontal directions
Answers :
2.1 (a) Fn = 433 N
Ft = 250 N
2.1 (b) Fx = 64.71 N
Fy = 241.5 N
2.2 (a) Fx = 281.9 N
Fy = 102.6 N
2.2 (b) Fv = 172.1 N 
Fh = 245.7 N 

F = 100 N
y
x
30° Horizontal
30°
weight
500 N
n
t
(a)
75°
250 N
y
x
(b)
15°
20°
Tension
300 N
y
x
v
h

2.3 Resultant Force (Analytical Method)
We can use rectangular components, i.e., the x- and y- components, to determine
the resultant of any number of forces.
This method is illustrated using two concurrent
forces, F1 and F2 (Fig. 2.4a). Concurrent means the
lines of action of the forces meet at a point.
1. Resolve the given forces into x- and y-components (see Fig. 2.4b & Fig. 2.4c).
2. Sum up all x-components (i.e.  Fx ), taking into account their directions
If  direction is assigned as positive, then  Fx = F2 cos  - F1 cos .
3. Sum up all y-components ( Fy ), taking into account their directions.
If  direction is assigned as positive, then  Fy = F1 sin  + F2 sin .
4. The magnitude of the Resultant FR (see Fig. 2.4d) is determined using
Pythagoras' Theorem:
2
y
2
xR )F()F(F 
5. The direction Resultant FR makes an acute angle 
with the x-axis:
x
y
F
F
θtan




F2
F2 sin 
F2 cos 

F1
F1 cos 
F1 sin 
 
F1
F2
Fig. 2.4 a
Fig. 2.4 b Fig. 2.4 c
Fig. 2.4 d
Fx
Fy

FR

The vector sum, or Resultant Force, FR, is a single force which has the same effect
as several forces combined. Each force can produce motion in the direction in
which it acts, but if several forces are acting at the same time, the direction of
motion will be in the direction of the resultant force.
Example 2.3
Two perpendicular forces of 30 N and 40 N are acting on a particle as shown below.
Find the resultant of the two forces.
Example 2.4
Find the resultant of the concurrent forces shown below using an analytical method.
Solution : Take  and  directions as positive
30°
50 N
60 N
Being vector quantities, the forces to be added
can be arranged to form a right-angled triangle.
The Resultant force is then represented by the
hypotenuse.
Its magnitude is determined using Pythagoras’
Theorem: FR = 22
3040  = 50 N
Its direction is defined by calculating the acute
angle  that the resultant force makes with the
horizontal side as shown.
Thus, Resultant force has magnitude FR = 50 N
and direction :
37°
FR
30N
40 N
30 N
40 N
FR

60cos30
60sin30
50

Example 2.5
Find the resultant force for the given co-planar concurrent forces using an analytical
method.
Example 2.6
Find the resultant force for the given co-planar forces.
50 N
30°
60 N
100 N
45°
80 N
40°
30°
60 N
100 N
45°
50 N

TUTORIAL
2.3 A bracket is shown with 2 forces acting on it. Determine the resultant force
acting on the bracket. (See Example 2.4)
2.4 For each of the coplanar concurrent system of forces shown, find the resultant
force:
(a)
(b)
( Ans : 16.99 N 70.1 ° )
(Ans : 6.136 kN, 16.1°)
( Ans : 7.108 kN 85.77 ° )
45°
2.4 kN
4.2 kN
8 N
50°
60°
45°
15 N 12 N
75°
20°
3 kN
2 kN
4 kN5 kN
30°

(c)
2.5 A vertical force of magnitude 400 N is required to drive a pole into the ground.
If two forces, each of magnitude 250 N, can be applied by means of cables as
shown below, determine analytically what the angle  should be.
(Ans: 36.9 o
)
30°
45°
30°
75°
20 kN
6 kN
16 kN
10 kN
(Ans : 19.95 kN 75.0° )
vertical
 
250 N 250 N

2.4 Moment of a Force
The moment of a force produces or tends to produce rotation about an axis normal to
the plane in which the force acts. A moment arises when we push a revolving door,
tighten a nut with a spanner or turn the steering wheel of a car. There may be no
motion, but we can still determine the moment if the force tends to cause rotation.
The moment of a force about a point is defined as the product of the force and
the perpendicular distance of the line of action of the force from that point.
(Since we are dealing with coplanar forces, a point actually represents an axis normal
to the plane in which the forces are acting.)
Moment of F about O, Mo = F.d
Moments are indicated as clockwise () or anticlockwise () sense, depending on
the direction in which the force tends to rotate the body about the given point or axis.
Example 2.7
Determine the moment of the 100 N force about points A, B, C and D.
M
M
M
M
D
C
B
A




M
M
M
M
D
C
B
A




F
dO
100 N
A B C
D
2 m 2 m 2 m
Fig. a
Fig. b
A B D2 m 2 m 2 m
30°
C
100 N
100cos30
0
100sin30’s line of action
passes through A, B, C &
D, hence no moment
about these points.
Alternative solution for Fig. b
Using the given force, perpendicular
distance from its line of action to the
point of consideration needs to be
calculated.
A B D2 m 2 m
2 m
30°
C
100 N
line of action of the
100 N force
dA
dB
dC

2.5 Addition of Moments
If more than one force or its components act on a body, the resultant moment ( M ),
is the algebraic sum of all the moments acting about the same point. If clockwise
moment is taken as positive then anticlockwise will be negative, or vice versa.
Example 2.8
Determine the resultant moment about various points, neglecting the weight of the
structure. (Resolve the force where necessary).
MA ( for 100N force) = _________________
MA ( for 25N force) = _________________
MA (Resultant) = _________________
MD ( for 100N force) = _________________
MD ( for 25N force) = _________________
MD (Resultant) = _________________
MD ( for 100N force) = _________________
MD ( for 25N force) = _________________
MD (Resultant) = _________________


MA = _____________________________

MC = _____________________________
ME = _____________________________
MF = _____________________________
100 N
A CB4 m 2 m
30°
25 NFig. a
100 N
A CB4 m 2 m
30°
25 N
1.5 m
Fig. c
D
100 N
A CB4 m 2 m
30°
25 N
D
1.5 m
Fig. b
100 N
A
B
C
D
2 m
2 m 2 m
Fig. d
30°
F
3 m
E
3 m

Example 2.9
Determine the resultant moment about point A of the structure due to the 2 forces.
Given : AB = BC = 3 m; CD = 1.5 m.
Example 2.10
Calculate the resultant moment about point B for the rectangular plate which weighs
1 kg at G. AB = 100 mm; BC = 160 mm.
TUTORIAL
2.6 Determine the sum of moments of the given forces about the fixed end A of the
cantilever. ( Ans : 557 Nm  )
B
500 N
D
C
100 N
80°
A
AB
40°
D C
4 kN
2 kN
8 kN
AB = CD = 0.2 m
BC = 0.3 m
30°
A C
B
110° 20 N
30 N
GG

2.7 Calculate the moment of the given force about point A in the following cases,
showing the resolved components: ( Ans : 58.99 Nm  , 953.5 Nm  )
2.6 Couples
Couples are commonly encountered in engineering. A couple consists of two equal
and opposite forces having separate lines of action. We often represent a couple with
a curved arrow, i.e. () or ()
A couple has the following characteristics:
1. The resultant force of a couple is zero.
2. The moment of a couple is the product of one of the forces and the perpendicular
distance between their lines of action.
3. The moment of a couple is the same for all points in the plane of the couple.
Example 2.11
Calculate the moment of the
couple applied to the steering
wheel shown. Diameter of the
steering wheel is 45 cm.
For the die-holder shown,
calculate the moment of the
couple about points A, B, C
& D.
MA = ______________
MB = ______________
MC = ______________
MD = ______________
_______________________
800 N
38°A
B
AB = 1.2 m
BC = 0.4 m
Fig. b
C
B
Fig. a
100 N
10°
25°A
C
AB = 0.6 m
BC = 0.3 m
3 N
3 N
b
AC
B
D
a/2 a/2 F
F

Tutorial
2.8 For the piping system shown below, calculate the moment
(a) due to the couple. ( Ans : 9 Nm  )
(b) of the 20 N force about points A and B.
( Ans : 17.18 Nm , 22.18 Nm  )
2.9 Determine the sum of the moments about fulcrum B of the bell-crank due to
the applied forces and the couple.
( Ans : 18.60 Nm  )
10 N
20 N
A
60°
AB = 0.5 m
BC = 1.8 m
CD = 0.9 m
10 N
B
C
D
AB = 0.2 m
BC = 0.18 m
130 N
75°
100 N
kg
A B
C
60°
16 Nm

2.10 For the framework shown below, calculate the sum of moments of the 3 given
forces about point A. Each triangular frame is an equilateral triangle of side 1 m.
( Ans : 13.77 kNm  )
2.11 Determine the sum of moments about point A due to all forces and couple
acting on the 10 kg plate, which has its centre of gravity at G. The plate
measures 0.8 m by 1.2 m. ( Ans : 217.3 Nm  )
B
10 kN
2 kN
5 kN
A
3 kNm
40°
25°
100 N
70 N
90 N
A
30 Nm
Couple
G

2.7 Free Body Diagram
A mechanical system can be a single body or a combination of bodies. The first
important step to analyzing a system is to isolate it or its component parts. Applied
loads (forces & couple/moment) and reactions (support forces/moments) are then put
in.
Then you have what is called a Free Body Diagram(s). See illustrations below.
 Free body diagram of a single body is usually used to analyze reactions at
supports, for example Ax, Ay and M above.
 Free body diagrams of the component bodies will be required when reactions at
the mating surfaces are required. An example of this can be seen above, when Bx
and By at point B are required.
 Support reactions are represented by arrows with arbitrarily assumed
direction/sense. See table on the next page for guidance. The correct sense of the
reactions can be confirmed through calculations.
By
Bx
D1
D2
D3C1
C2
C3
C
B
D
Bx
By
Ay
Ax
M
A
B
C D
Mechanical System
Free Body Diagram of single body
W
C3 C2 C1
WD
B
Ay
Ax
M
WC
Free Body Diagrams of component bodies

Table Showing Reactions for Various Support Types
Type of Support Direction of reaction No. of unknowns
1. Rollers
1 unknown reaction.
Direction is normal to
the supporting surface.
2. Smooth surfaces
3. Cables/chain 1 unknown tension F
pointing away from the
body.
4. Rigid links or bars
1 unknown reaction.
Direction along the link,
either in compression or
tension.
50°
rigid link
R
R
R
R
50°
R
50o
50o
R
R
F
F
F
F

Type of Support Direction of reaction No. of unknowns
5. Pinned or Hinged
2 unknown rectangular
components of reaction.
Directions are not
restricted to just vertical
& horizontal
6. Rough Surfaces
2 unknown rectangular
Directions are normal
and tangential to the
supporting surface.
7. Fixed or Built-in Support 2 unknown rectangular
1 unknown fixing
(resisting) moment M
Ry
Rx
Rx
Ry
M
Rx
Ry
Rx
Ry
Rx
Ry
Ry
Rx

TUTORIAL (Sketching Free Body Diagrams)
2.12 Complete the free body diagram of the body to be isolated:
Body to isolate Mechanical system FBD of body
a)
Bell
crank
b)
Structure
ABCD
c)
Beam
d)
Crate
e)
Beam
&
Pulley
B
110°
30° 20 N
30 N
A
C
AB
40°
D C
4 kN
2 kN 8 kN
AB = CD = 0.2 m; BC = 0.3 m
800 N
38°A
B
AB = 0.8 m; BC = 0.4 m; CD = 0.4 m
C
D70°
50°
Rough ground
Smoothwall
AB = 5 m
BC = 2 m
Weight mg at G
G
A
B
C
15 °
3 kN
30 ° cable
A
O
T
pulley
beam
ABD C
A
B
C
D
B
A
C

2.8 Equilibrium Conditions
For a body to be in a state of balance or equilibrium, all forces and moments acting
on it are in balance. In other words, resultant forces as well as moments equal zero.
 F = 0 and M = 0
For a 3-dimensional or spatial force system, the above conditions must be applied
with respect to x, y and z directions which are mutually perpendicular.
For a 2-dimensional or coplanar force system, the following conditions are
necessary and sufficient for complete equilibrium.


where subscript “o” refers to any point on or off the body.
Note : Only the first 2 conditions need to be used when considering the equilibrium
of a point where all forces meet i.e. concurrent force system.
Example 2.12 ( Equilibrium of coplanar concurrent forces )
Figure shows concurrent forces acting on a truss joint. With the help of a FBD of the
joint, determine the magnitude of the forces A and B for equilibrium.
( Hint: Resolve all forces along the directions of A & B )
10 kN
30°
5 kN
2 kN
B
A
 Fx = 0
Fy = 0
Mo = 0
B
A
10
5cos30
5sin30
2

Example 2.13
Three cables supporting a weight D form a
concurrent force system at B. Sketch the relevant
FBDs and determine the force in each cable.
Example 2.14
For the magnified joint B of the plane truss shown, find the forces in the members AB
& BE. The forces in the 2 members are assumed as indicated (i.e. under tension)
5 kg
30°
A
45°
C
B
D
B
5 kN
FAB
FBE
5.4375 kN
33.69°
36.87°
D
C
F
A
B
E
5 kN 8 kN
AF = 3 m; FE = 6 m; FB = 4 m
D
FBD
mg
FBCcos30
FBCsin30FABsin45
FABcos45
FBD
B
B

TUTORIAL (Equilibrium of Coplanar, Concurrent Forces)
2.13 Force A = 3.5 kN and B = 2 kN are applied to the gusset plate shown below.
Knowing that the plate is in equilibrium, determine the magnitude of forces C
and D.
( 1.73 kN , 2.5 kN )
2.14 The members of a plane truss are connected to a gusset plate shown below.
Find the magnitude of the force T and P to maintain equilibrium.
(See Example 2.12)
( 13.3 kN , 10.2 kN )
A = 3.5 kN D
C
B = 2 kN
30 °
O
T
O
5 kN
30° O
8 kN
T
P
45 °

2.15 A street lamp of mass 15 kg is supported at mid-point between two poles by a
single cable ABC. If the length of the cable is 20 m and the deflection BD at
mid-point is 0.2 m, sketch the free body diagram of point B and determine the
force in the cable.
( 3.68 kN )
2.16 A 100 N force pulls at point B of a cable ABC as shown. Sketch the free body
diagram of point B and determine the tensions in AB and BC. (See example
2.14)
(75.25 N, 96.45 N )
D C
B
A
A40°C
B60°
100 N
15°

2.17 Shown below is a simple framework consisting of light metal rods pinned
together at their ends. At one joint A hangs a chandelier of mass 25 kg.
Calculate the forces acting in AB & AC pinned at the joint A. Assume that both
rods are in tension. (See Example 2.14)
(381.5 N ; – 292.3 N )
40°
A
B
C

Worked example ( Equilibrium of non-concurrent forces)
For the beam shown,
a) sketch a free body diagram of the
beam ABC, given that its weight is
negligible.
b) determine the reactions on the beam at
roller A and pin-joint B.
Solution:
a) Indicate the dimensions of the body
for computing the components and
moments of forces.
b) When applying equilibrium conditions, note that MC = 0 will result in 2
unknowns ( RA & By ) in the equation.
* Summing moments about the point where an unknown acts will
enable the other unknown to be found.
40sin50o
(0.3) – RA(0.2) = 0
(Roller reaction) RA = 45.96 N 


YB – RA – 40sin50 = 0
YB – 45.96 – 30.64 = 0
YB = 76.60 N 
XB – 40cos50 = 0
XB = 25.71 N 
RB =
2
B
2
B XY  = 22
76.6025.71  = 80.8 N
B = tan-1
(
B
B
X
Y
) = tan-1
(
71.25
60.76
) = 71.4°
A
B C
40 N
50 °
50°
0.2 m
A CB
0.3 m
40 N
XB
YB
RA
MB = 0
Fy = 0
Fx = 0

Example 2.15
The 500 mm x 400 mm plate is subjected to forces as shown. Prove that the plate is in
equilibrium. Neglect the weight of the plate.
Example 2.16
Determine the reactions at A and B for the light beam shown below.
(The reactions at the supports keep the beam at rest)
500 mm
400 mm
50 N
40 N
30 N80 N
60 N
100 N
A
45° C D
2 m3 m2 m
100 N600 N
B A C D B

60°
A
B
C
25°
D
100 N
AB = 1.5 m
BC = 1.5 m
Example 2.17
For the piping system shown below,
a) sketch a free body diagram of the piping.
b) calculate reactions at the fixed end A.
Example 2.18
An awning structure is supported as shown. It weighs 500 N and the centre of gravity
is at point B. A load of 100 N hangs from C. Determine
a) the tension in the tie rod BD.
b) the reaction at the pin support A.
10 N
20 N
A
60°
AB = 0.5 m
BC = 1.8 m
CD = 0.9 m
10 N
B
C
D
FBD
FBD
A
B
C
D

Example 2.19
A light beam and pulley system shown below is at rest. Determine
a) the reaction acting at the centre B of the pulley.
b) the reactions at A, B and C on the beam.
Solution: Pulley has to be isolated from the pin support at B. FBDs of the separate
bodies are thus required. Note the opposing sense of the component
reactions at B on both the bodies.
a)
b)
A
B
D
30°
1.5 m1.5 m
600 N
CB
A CB

TUTORIAL (Equilibrium of Coplanar, Non-concurrent, Forces)
2.18 Calculate the support reactions for the beams/structure shown below. Neglect
the mass of the beams.
i)
ii)
iii)
AB = CD = 0.2 m
BC = 0.3 m
AB
40 °
D C
4 kN
2 kN 8 kN
Ans ( 5.5 kN 73.8 o
; 557 Nm  )
Ans ( 645.7 N 88.12° ; 54.91 Nm  )
5 kN
19 kN
B
2 m2 m2 m
Ans ( 11 kN  , 3 kN  )
B
65°A
D C
100 N
500 N
50 N
AB = 4.5 m
BC = 1.5 m
CD = 1.5 m

2.19 Calculate the tension in the chain at B and the reaction at the pin support A.
( 1268.4 N, 1083.2 N , 6.779° )
2.20 An advertising sign of mass 10 kg is supported on a light beam AB as shown.
Determine, with the aid of a free body diagram of AB:
a) cable tension at B.
b) magnitude and direction of the reaction at A. ( 125 N ; 114 N , 18.3o
)
800 N
38°A
B
AB = 0.8 m; BC = 0.4 m; CD = 0.4 m
C
D70°
30°
B A
10 kg
1.4 m0.8 m

2.21 A 4 kN crate with its mass centre at G rests against a smooth wall as shown.
Determine with the aid of a free body diagram,
a) the reaction at point A of the crate.
b) magnitude and direction of the ground
reaction at B.
2.22 The flip-up table-top ABC weighs 20 N and
its CG is at the middle of AC. It is supported
by a hinge at A and a pin-ended bar at B
which is anchored at the wall.
Calculate the reactions at A and B of the
table-top.
( 219 N, 36.4o
, 274.1 N , 50o
)
AB = 0.2 m
BC = 0.4 m
60 N
50 °
A B C
50°
AB = 5 m
BC = 2 mA
B
C
G
Rough ground
Smoothwall
( 0.8782 kN ; 4.095 kN ; 77.62° )

2.23 A car park gate control has a beam CB of negligible mass and a counter weight
C of 200 kg. Calculate the tension T in the rope at B and the reaction at the
pivot A.
(435.2 N ; 2349 N, 84.7o
)
45°
60°
T
A
B
C
A
3.3 m
1 m

2.24 The resistance of the bracket to bending is tested under a 2 kN load as shown.
Determine the reactions acting on the bracket at:
a) A. ( 1.25 kN  )
b) the hinge O. (2.358 kN 58o
)
2.25 For the bell crank lever shown, determine:
a) force acting on the roller at R if the pull at P is 80 N. (128.8 N )
b) reaction at the hinge Q. (202.1 N, 11.4o
)
250 mm
400mm
2 kN
O
A
PQ = 0.3 m
QR = 0.18 m
75°
Q
P
Roller
80 N
45o
R

2.26 A load of 3 kN acts on a light beam 2.2 m long which is pivoted at O as shown
below. The part of the cable between the pulley and the beam is vertical.
a) Sketch separate free body diagrams of the beam and the pulley.
b) Find the tension F in the cable required to hold the beam
in equilibrium. (9.76 kN)
c) Determine the reaction at O on the beam. (6.91 kN 83.5o
)
d) Determine the reaction at A on the pulley. (16.9 kN , 60o
)
15 °
3 kN
30°
0.7 m
2.2 m
cable
O
F
A

2.27 A load of 50 kg is held in equilibrium by a light angle bracket ABC, pulley and
cable as shown. Assuming that all the forces are coplanar,
a) determine the tension in the cable;
b) sketch the free body diagram of the pulley alone;
c) sketch the free body diagram of the angle bracket ABC alone;
d) determine the magnitude and direction of the force exerted on the bracket
by the pin at C;
e) determine the reaction at A on the bracket.
( 490.5 N , 849.6 N 60º  , 708.0 N  )
Cable
Roller
Angle
Pulley 75 mm
500mm
300 mm
B
A
60O
50 kg
C

*2.28 Figure shows a portable mobile crane lifting a cargo weighing half a tonne. The
boom ABC weighs 50 kg and the centre of gravity is at point G. Find:
(a) force in the hydraulic cylinder BD which acts at point B of the boom.
(b) reaction at the hinge A.
Given: AC = 2.0 m
AB = 1.2 m
AG = 0.9 m
( 14.80 kN ; 10.48 kN , 45.08o
)
C
A
30°
60°
B
G
D

*2.29 A mass of 10 kg is supported by two cables DB and DE as shown in the figure.
The beam ABC has a mass of 3 kg and is held in equilibrium by a pin at A and
a frictionless roller at C.
a) Sketch a free body diagram of the point D and show that the tensions in
the cables DB and DE are 67.1 N and 52.2 N respectively.
b) Sketch a free body diagram of the uniform beam ABC and determine:
i) the reaction at C;
ii) the reaction at A.
( (i) Rc = 204.8 N, 45º (ii) 187.4 N, 17.8º )
************************
unit2.doc
frictionless
rollerE
A
D
45°
10 kg
60°
40°
60°
G
A
B
C
0.6 m
0.4 m

Unit 2 mm9400 ver 1.1(2014)

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (7)

Similar to Unit 2 mm9400 ver 1.1(2014)

Similar to Unit 2 mm9400 ver 1.1(2014) (20)

Recently uploaded

Recently uploaded (20)

Unit 2 mm9400 ver 1.1(2014)