Ch4 Equation 3d

1
3 EQUILIBRIUM

• System isolation and
• equilibrium conditions

2
Objectives Students must be able to 1

• Course Objective
• State the conditions of equilibrium, draw free
  body diagrams (FBDs), analyse and solve problems
  involving rigid bodies in equilibrium.
• Chapter Objectives
• Analyse objects (particles and rigid bodies) in
  equilibriums
• Classify problems in equilibrium into SD and SI
  categories

3
Equilibrium Definition
stationary

• An object is in equilibrium when it is
  stationary or in steady translation

relative to an inertial reference frame.
with Constant velocity
moving
Whether object in stationary (moving in steady
translation) or not, depends on reference
frame.
really equilibrium?
Equilibrium
Newtonian Mechanics Inertial Reference Frame
- Earth Frame - Central Universe Frame
Centrifugal acceleration
Earth
At universe
4
3/1 Introduction
A

• - Equilibrium is the most important subject in
  statics.
• In statics, we deal primarily with bodies at
  rest.
• (i.e. they are in the state of equilibrium).

A
- More precisely, when a body is in equilibrium,
the wrench resultant of all forces acting on it
is zero i.e.

• These requirements are necessary and sufficient
  conditions for equilibrium i.e.

If is true body in equilibrium

If body in equilibrium is true

• From the Newtons second law of motion, a body
  that moves with constant velocity, rotates with
  constant angular velocity i.e. zero
  acceleration, can also be treated as in a state
  of equilibrium.

5
Equilibrium in 2D
- All physical bodies are inherently 3D, but
many may be treated as 2D e.g. when all forces
are on the same plane.
3/2 Mechanical system isolation

• Before we apply the equilibrium conditions
• we need to know what force or couple are
  involved.

Isolate body

• Draw Free Body Diagram (FBD)
• FBD is used to isolate body (or bodies / system
  of bodies) so that force/couple acting on it can
  be identified.

6
Free Body Diagram (FBD)

• FBD is the sketch of the body under consideration
  that is isolated from all other bodies or
  surroundings.
• The isolation of body clearly separate cause and
  effects of loads on the body.
• A thorough understanding of FBD is most vital for
  solving problems.

7
Construction of FBD
1) Pick body/combination of bodies to be
isolated
2) Isolating the body. Draw complete external
boundary of the isolated body
Free Body Diagram
3) Add all forces and moments (including that
are applied by the removed surrounding)
150 cm
50 cm
4) Indicate a coordinate system
5) Indicate necessary dimensions
Most important step is solving problems in
mechanics. If an FBD is not drawn (when it
is needed), you will get no credit ( 0 point )
for the whole problem!!!!
150 cm
50 cm
8
Note on drawing FBD

• Free Body Diagram
• Establish the x, y, z axes in any suitable
  orientation.
• Label all the known and unknown force magnitudes
  and directions on the diagram
• The sense of a force having an unknown magnitude
  can be assumed.

150 cm
50 cm

• Use different colours in diagrams
• Body outline - blue
• Load (force and couple) - red
• Miscellaneous (dimension, angle, etc.) - black

9
Equilibrium Solving Procedure

• Formulate problems from physical situations.
• (Simplify problems by making appropriate
  assumptions)
• Draw the free body diagram (FBD) of objects under
  consideration

• State the condition of equilibrium

• Substitute variables from the FBD into the
  equilibrium equations
• Substitute the numbers and solve for solutions
• Delay substitute numbers
• Use appropriate significant figures
• Technical judgment and engineering sense
• Try to predict the answers
• Is the answer reasonable?

10
Equilibrium Free Body Diagram (FBD)

• FBD is the sketch of the body under consideration
  that is isolated from all other bodies or
  surroundings.
• The isolation of body clearly separate cause and
  effects of loads on the body.
• A thorough understanding of FBD is most vital for
  solving problems.

11
Equilibrium FBD Construction
Equilibrium

• Select the body to be isolated
• Draw boundary of isolated body, excluding
  supports
• Indicate a coordinate system by drawing axes
• Add all applied loads (forces and couples) on the
  isolated body.
• Add all to support reactions (forces and couples)
  represent the supports that were removed.
• Beware of loads or support reactions with
  specific directions due to physical meanings
• Add dimensions and other information that are
  required in the equilibrium equation

12
Equilibrium Help on FBD

• Establish the x, y axes in any suitable
  orientation.
• Label all the known and unknown applied load and
  support reaction magnitudes
• Beware of loads or support reactions with
  specific directions due to physical meanings
• Otherwise, directions of unknown loads and
  support reactions can be assumed.

13
Equilibrium On FBD Analyses
Equilibrium

• Objective To find support reactions
• Apply the equations of equilibrium
• Load components are positive if they are directed
  along a positive direction, and vice versa
• It is possible to assume positive directions for
  unknown forces and moments. If the solution
  yields a negative result, the actual load
  direction is opposite of that shown in the FBD.

14
Contents
the heart (?) of Statics

• Equilibrium of Objects
• Particles (2D 3D)
• 2D Rigid Bodies
• 3D Rigid Bodies

Rigid Bodies

• SD and SI Problems

Ideal particle can not rotate. (no couple acting
on it)
Particle
15
Particles FBD construction

• To construct a complete FBD of a particle
• Select the particle to be isolated
• Draw the particle as a point
• Indicate a coordinate system
• Add all active forces/moments (weight, etc.)
• Add all support reactions (e.g. tension in the
  tangential direction of a cable, tensile and
  compressive forces in a compressed and stretched
  springs)

16
Particles Equilibrium Analyses 2

• Equations of Equilibrium
• Apply the equations of equilibrium
• Components are positive if they are directed
  along a positive axis, and negative if they are
  directed along a negative axis.
• Assume the directions of unknown forces in the
  positive x, and y axes. If the solution yields a
  negative result, this indicates the sense of the
  force is the reverse of that shown on the FBD.

17
Particles Equilibrium in 3D
Particle Equilibrium
z
y
x
18
H-Ex3-11) The sphere has a mass of 6 kg and is
supported as shown.Draw a free-body diagram of
the sphere, the cord CE, and the knot at C.
19
How many unknowns?
How many Equations?
Action-reaction pair - use same symbol -
opposite direction
20
Determine the tension in cables AB and AD for
equilibrium of the 250-kg engine shown.
21
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Example Hibbeler Ex 3-3 1
2unknown, 2 Eqs (at this stage)
If the sack at A has a weight of 20 lb, determine
the weight of the sack at B and the force in each
cord needed to hold the system in the equilibrium
position shown.
3unknown, 2 Eqs (at this stage)
?
23
Recommended
FBD of E
FBD of C
24
Example Hibbeler Ex 3-5 1
A 90-N load is suspended from the hook. The load
is supported by two cables and a spring having a
stiffness k 500 N/m. Determine the force in the
cables and the stretch of the spring for
equilibrium. Cable AD lies in the x-y plane and
cable AC lies in the x-z plane.
25
3D particle Equilibrium
How many unknowns, how many equations?
(no FBD, no score)
perfect answer sheet
26
Example Hibbeler Ex 3-7 1
Determine the force developed in each cable used
to support the 40-kN (4 tonne) crate shown.
3D particle Equilibrium
27
Hibbeler Ex 3-7 2
Particle Equilibrium
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Example Hibbeler Ex 3-7 4
Particle Equilibrium
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Equilibrium of 2D Rigid Bodies
2D Equilibrium

• Use similar analyses as the particles
• Additional consideration
• Action forces in supports/constraints
• Free-body diagram (FBD) of 2D rigid bodies
• Equilibrium equations (scalar form) for rigid
  bodies
• Two-force and three-force members

32
Force Reaction (2D)
To write an FBD, first, you will need to know
what kind of force we will get when eliminating
the environment/surrounding.
1) Flexible cable, belt, chain, or rope
?
?
always away from the body
T
33
2) Smooth surfaces
- Contact force at contact point normal to the
surface/contact plane
only this direction
- always compressive
3) Rough surfaces
Not always this direction

• A rough surface can produce a tangential force
  (F, friction) as well as a normal force (N)
• direction of F depend on situations (chapter 6)

only this direction
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4) Roller supports
only this direction
5) Freely sliding guide
The vector N may be up or down depend on problem.
If not known, you may assume any of the two.
After further calculation, if N is , correct
sense was assumed. If negative, N goes the
other way.
not always this direction
M
M existence due to its bending resistance
35
6) Pin connection
not always this direction
As a general rule, if a support prevents
translation of a body in a given direction, then
a force is developed on the body in the opposite
direction. Similarly, if rotation is prevented,
a couple is exerted on the body.
7) Built-in or fixed support
not always this direction
Weld
A
A
36
8) Gravitational attraction
Resultant of the gravitational attraction is the
weight W mg and act toward center of the earth
passing through the center mass G
m
9) Spring action
x
F
Normal distance
For linear springs , F kx
37
Equilibrium
You may assume either case. The sign will
indicate its sense of direction later.
38
Equilibrium construction of a FBD
2D Equilibrium

• To construct a complete FBD for a 2D rigid bodies
• determine which body is to be isolated
• draw external boundary of isolated body
• indicate a coordinate system (axes)
• add all loads (forces and couples, be they
  applied or support)
• No FBDs ? Cannot apply equilibrium conditions
• ? NO SCORES

39
F1
F3
F2
mass m
A
mass m
P
Correct?
A
B
No internal force is shown in FBD
m
not always mg
A
B
Write FBD before allowing any movements of
vectors (sliding/free)
40
3/A Correct the incomplete FBD
for all figures below
41
3/B Correct the wrong or incomplete FBD
for all figures below
If rough surface
42
3/B Correct the wrong or incomplete FBD
for all figures below
43
We will choose problem 1 and 4 as samples
System Isolation FBD
Equilibrium
44
_ with the incline
45
Example Hibbeler Ex 5-1 1
2D Equilibrium
Draw the free-body diagram of the uniform beam
shown. The beam has a mass of 100 kg.
46
Example Hibbeler Ex 5-1 2
2D Equilibrium
47
Example Hibbeler Ex 5-3 1
2D Equilibrium
Two smooth pipes, each having a mass of 300 kg,
are supported by the forks of the tractor. Draw
the FBDs for each pipe and both pipes together.
48
2D Equilibrium
49
Example Hibbeler Ex 5-3 2
2D Equilibrium
50
Example Hibbeler Ex 5-4 1
2D Equilibrium
Draw the FBD of the unloaded platform that is
suspended off the edge of the oil rig shown. The
platform has a mass of 200 kg.
51
Example Hibbeler Ex 5-4 2
2D Equilibrium
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3/3 Equilibrium Conditions

• We can find the wrench resultant force

P
P
equilibrium
equilibrium
P

• Equilibrium conditions (3D) of the body is
  equivalents to

(for some point P )
We can prove this!
(for all point O)
- For co-planer forces only (2D), eq conditions
is equivalents to
?Fx 0 ?Fy 0 ?Mo 0 (for all
point O) (2D)
?Fx 0 ?Fy 0 ?Mo 0 (for some
point O) (2D)
55
3/3 Equilibrium Conditions
P

• We can find the wrench resultant force

P
P
P
if in equilibrium
O

• Equilibrium conditions (3D) of the body is
  equivalents to

( moment for some specific point P )
We can prove this!
?Fx 0 ?Fy 0 ?Mo 0 (for some
point O) (2D)
56
3/3 Equilibrium Conditions

• We can find the resultant force and couple at
  any point in the body

Easy - Why?
No force to cause moment effect
O
O
Can couple exist?

• Equilibrium conditions (3D) of the body is
  equivalents to

( moment for all point )
(trivially easy)
We can prove this (later) !
?Fx 0 ?Fy 0 ?Mo 0 (for some
point O) (2D)
57
Equilibrium and its Independent Equations
Equilibrium (no tendency to initiate
translation, no tendency to initiate rotation)
A
A
equilibrium
A
A
any O
(trivially easy)
(some point A)
(any point O)
Important Meaning
A
2D
A
Only at most 3 independent Equations
If is true body in equilibrium

If body in equilibrium is true
A

• From the Newtons second law of motion, a body
  that moves with constant velocity, rotates with
  constant angular velocity i.e. zero
  acceleration, can also be treated as in a state
  of equilibrium.

A
58
Equilibrium Eqn. for 2D Rigid Bodies
2D Equilibrium
Scalar Form
59
2D Equilibrium Procedure 1
2D Equilibrium

• Free-Body Diagram
• Establish the x, y axes in any suitable
  orientation.
• Draw an outlined shape of the body.
• Show all the forces and couple moments acting on
  the body.
• Label all the loadings and specify their
  directions relative to the x, y axes. The sense
  of a force or couple moment having an unknown
  magnitude but known line of action can be
  assumed.
• Indicate the dimensions of the body necessary for
  computing the moments of forces.

60
2D Equilibrium Procedure 2
2D Equilibrium

• Equations of Equilibrium
• When applying the force equilibrium equations,
• orient the x and y axes along lines that will
  provide the simplest resolution of the forces
  into their x and y components.
• If the solution of the equilibrium equations
  yields a negative scalar for a force or couple
  moment magnitude, this indicate that the sense is
  opposite to that which was assumed on the FBD.

61
2D Equilibrium Procedure 3
2D Equilibrium

• Equations of Equilibrium
• Apply the moment equation of equilibrium about a
  point O that lies at the intersection of the
  lines of action of two unknown forces.
• In this way, the moments of these unknowns are
  zero about O, and a direct solution for the third
  unknown can be determined.

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H5.7) Find the tension in the cord at C and the
horizontal and vertical component of reactions at
pin A. The pulley is frictionless.
The body is in equilibrium.
?Fx 0 ?Fy 0 ?MAnyPoint 0
T are the same at both side?

If couple also applies at pully,T are the same in
both side?
If pulley rotates, T are the same in both side ?
with constant angular velocity?
64
Independent Equations
y
Whatever 2D-equilibrium conditions you used,
at most 3 unknowns (scalar) may be found.
x
Body in equilibrium
not independent equations!
Hey , why you cannot get anymore, you can take
another points to take moment, and get new
equation?
?Fx 0 ?Fy 0 ?Mo 0 (for some point
O)
However, sometimes you can not get those 3
unknowns in 2D problem.
?Fx 0 ?Fy 0 ?Mo 0 (for all point O)
65
H5.7) Find the tension in the cord at C and the
horizontal and vertical component of reactions at
pin A. The pulley is frictionless.
The body is in equilibrium.
?Fx 0 ?Fy 0 ?MAnyPoint 0
Not independent Equations!

66
Example Hibbeler Ex 5-6 1
2D Equilibrium
Determine the horizontal and vertical components
of reaction for the beam loaded as shown.
Neglect the weight of the beam in the calculation.
67
Example Hibbeler Ex 5-6 2
2D Equilibrium
Find Ay, Bx, By
68
Example Hibbeler Ex 5-6 3
2D Equilibrium
69
Example Hibbeler Ex 5-6 4
2D Equilibrium
70
Example Hibbeler Ex 5-8 1
2D Equilibrium
The link shown is pin-connected at A and rests
against a smooth support at B. Compute the
horizontal and vertical components of reactions
at the pin A.
71
Example Hibbeler Ex 5-8 2
2D Equilibrium
72
Example Hibbeler Ex 5-8 3
2D Equilibrium
73
3/57 200-kg beam has a C.G. at G. The 80-kg man
is exerted a 300-N force on the rope. Calculate
the force reaction at A which is a weld pin.
3 unknowns, 3 Equations. You can solve it.
mg 80(9.81)
300 N
T
mg
mg?
N
300 N
200(9.81)
C
N
300N
T
300N?
T300N
80(9.81)
80(9.81)
200(9.81)
4 unknowns, 3 Eq. You cant solve it?
200(9.81)
300 N
You can solve it ! Use FBD of rod
T
300N
C
N is internal force.
tension is internal force.
74
3/57 200-kg beam has a C.G. at G. The 80-kg man
is exerted a 300-N force on the rope. Calculate
the force reaction at A which is a weld pin.
80(9.81)
80(9.81)
Suns Gravitational force
200(9.81)
C
tension is internal force.
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Draw the FBD of the foot lever shown.
why?
77
?Fx 0 ?Fy 0 ?Mo 0
Two-Force Member In Equilibrium
To keep the body in equillibrium, the second
force must
Body in Equilibrium
body in equilibrium
bodies in equilibrium
78
?Fx 0 ?Fy 0 ?Mo 0
Three-Force Member for Equilibrium
To keep the body in equillibrium, the second
force must
1)
(co-planer)
2)
Concurrent
OR Parallel (with proper moment arm)
Parallel (with proper moment arm)
concurrent
4 forces need to be concurrent or parallel to
make body in equilibrium?
O
?M 0
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Special Member 2-Force Member 1
2D Equilibrium
Definition
The object is a two-force member when subjected
to two equivalent forces acting at different
points.
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Special Member 2-Force Member 2
2D Equilibrium

• If a two-force member is in equilibrium, the two
  forces
• are equal in magnitude.
• are opposite in direction.
• have the same line of action.
• There are no couple moments.

The ability to recognize 2-force members is
important in the analyses of structures.
81
Special Member 2-Force Member 3
2D Equilibrium
Example
82
Special Member 2-Force Member 4
2D Equilibrium
Example
83
Special Member 3-Force Member 1
2D Equilibrium
Definition

• The object is a three-force member when subjected
  to
• three equivalent forces acting at different
  points.
• If a three force member is in equilibrium, the
  three
• forces are
• coplanar and
• either parallel or concurrent.

84
Special Member 3-Force Member 2
2D Equilibrium
Concurrent Forces
Parallel Forces
85
Draw the FBD of the foot lever shown.
why?
Two force in Equilibrium
You should write the FBD of 2-force (in
equilibrium) like this.
86
The lever ABC is pin-supported at A and connected
to a short link BD as shown. If the weight of the
members is negligible, determine the force of the
pin on the lever at A.
4 unknowns, 3 eq.
Hibbeler Ex 5-13
2 unknowns, 3 eq.
Two Force Members
87
Solve by using 3-force member Concept
Three Forces are concurrent.
Hibbeler Ex 5-13
O
0.5 m
0.50.2 0.7 m
Assume Directions
0.5-0.1 0.4 m
2 force members
3-Unknowns, 3-Eqs
88
The lever ABC is pin-supported at A and connected
to a short link BD as shown. If the weight of the
members is negligible, determine the force of the
pin on the lever at A.
2 force members
0.5 m
O
Ans
89
Any two-force members or three-force members
here?
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91
3/27 In a procedure to evaluate the strength of
the triceps muscle, a person pushed down on a
load cell with the palm of his hand as indicated
in the figure. If the load-cell reading is 160 N,
determine the vertical tensile force F generated
by the triceps muscle. The mass of the lower arm
is 1.5 kg with mass center at G.
92
Free Body Diagram
If what you want o know is Oy ?
If what you want o know is Ox ?
0.15
0.15
0.25
1.5(9.81)
3 Unknown, 3 Equations.
You can solve it.
By easier way.
Because what we want to know is the force T of
triceps. Taking summation of moments at point O
will eliminate Ox and Oy which are unknowns out,
and the calculation will be easier.

Ans
93
Technique Finding unknown force only in 1 step
Concrete slab with mass of 25,000 kg is
pulled by cable of tension P. Determine the
tension T in the horizontal cable using only 1
equilibrium equation.
Given ? 60
D
A
6 m
Key
Here we have 3 unknowns (P,R,T). To use only
1equilibrium eq. To determine T, we need to take
moment at the point where the other unknown
forces (R,P) passes.
94
D
W mg
A
6 m
ANS
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96

• Alternative Equilibrium equations
• Recall In general, you have these three
  equilibrium equations
• (for some point O) ?Fx 0
  ?Fy 0 ?Mo 0 (E)

- Alternatively, you may use either
?MA 0
?Fx 0
?MB 0
?Mc 0
?MB 0
?MA 0
A B C not on the same straight line
line AB not _ to the x direction
R0
R0
not _
B
B
x
A
A
C
97
Calculate the magnitude of the force at pin A.
400 N
500 N
80 N-m
D
How many unknowns ?
We have 3 unknowns of
Ay
Ax
By
98
Calculate the magnitude of the force at pin A.
400 N
500 N
we can determine Ay from

80 N-m
we can determine Ax from

Thus, the resultant of force at pin A
ANS
99
Constraints and Statical determinacy
need to know when a problem can be solved or
what force/couple can be found.

1. Statically Determinate

Improper Constraint
- Not enough constraint
Unknown 3 , Independent eq 2
Unknown 3 , Independent eq 3
2) Statically Indeterminate

• Not in equilibrium

- cant get all unknowns
No solution (cant maintain moment)
Unknown 4 , Independent eq 3
unknown 3 , independent 2
100
Independent Equations
y
Whatever 2D-equilibrium conditions you used,
at most 3 unknowns (scalar) may be found.
x
Body in equilibrium
not independent equations!
Hey , why you cannot get anymore, you can take
another points to take moment, and get new
equation?
?Fx 0 ?Fy 0 ?Mo 0 (for some point
O)
However, sometimes you can not get those 3
unknowns in 2D problem.
?Fx 0 ?Fy 0 ?Mo 0 (for all point O)
Independent Equations on categories of 2D problem.
Equilibrium
1) Co-linear
x
Depend on x-axis selection
Choose this direction as x axis
?Fx 0 ?Fy 0 ?Mo 0
(Choose point O on the line of action)
101
2) Concurrent at a point
?Fx 0 ?Fy 0 ?Mo 0
y
concurrent at point O
x
O
3) Parallel
?Fx 0 ?Fy 0 ?Mo 0
Depend on x-axis selection
y
x
4) General
?Fx 0 ?Fy 0 ?Mz 0
y
x
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Note on Solving Problems
1. If we have more unknowns than the number of
independent equations, then we have a statically
indeterminate situation. We cannot solve these
problems using just statics.
2. The point (axis) which we check for the
moment, affects the simplicity of the solution.
Choosing appropriate one is the key for fast
problem solving.

1. The order in which we apply equations may affect
   the simplicity of the solution. For example,
   solving ? FX O first allows us to find the
   first unknown quickly.

x
4. If the answer for an unknown comes out as
negative number, then the sense (direction) of
the unknown force is opposite to that assumed
when starting the problem.
N
If you get Nlt0, something wrong!
104
3/48 The small crane is mounted on one side of
the bed of a pickup truck. For the position q45,
determine the magnitude of the force supported
by the pin at O and the oil pressure P against
the 50-mm-diameter piston of the hydraulic
cylinder BC.
AOBC?
Body in interest
AOC?
Because We also want to find force P exerted by
BC to the object AOC
105
Ignore mass of link CB
q45 Find P ,O
y
FBD
x
C
C
B
which dirction?
2 Force Member
// with CB
Direction // CB
Can you find a ?
F
To find C
0.11
0.34
q
C
G
O
q
E
D
F
0.36
a
H
B
Ans
106
y
q45 Find P ,O
A
FDB
0.785
x
W
F
0.11
0.34
C
O
F
Ox
q
Oy
a
B
To find reactions at O
Ans
107
3/46 It is desired that a person be able to
begin closing the van hatch from the open
position shown with a 40-N vertical force P. As a
desired exercise, determine the necessary force
in each of the two gas-pressurized struts AB. The
mass center of the 40-kg door is 37.5 mm directly
below point A. Treat the problem as
two-dimensional.
Why direction CB?
1125 mm
A
550 mm
O
600 mm
175 mm
P 40N
B
108
Find F
1125 mm
2F
y
Oy
A
2F
A
550 mm
O
G
x
G
Ox
O
600 mm
W(40)(9.81)
175 mm
Wmg
B
P 40
B
ANS
109
3/38 The hinge P fits loosely through the frame
tube, and the frame tube has a slight clearance
between the supports A and B. Determine the
reactions on the frame tube at A and B,
associated with the weight L of an 80-kg person.
Also, calculate the changes in the horizontal
reactions at C and D.
Assumption no couple reactions develop at C,D
L 80(9.81)
?
110
Find appropriate L to make Bodies in equilibrium.
known
q
known
x?
111
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112
3/52 The rubber-tired tractor shown has a mass of
13.5 Mg with the center of mass at G and is used
for pusshing or pulling heavy loads. Determine
the load P which the tractor can pull at a
constant speed of 5 km/h up the 15-percent grade
if the driving force exerted by the ground on
each of its four wheel is 80 of the normal force
under the wheel. Also find the total normal
reaction N under the rear pair of wheels at B.
Normal reaction at A and B are equal?
If not, which one should be larger?
113
FBD
Wmg13.510009.81132400 N
Ans
114
Recommended Problem
3/34 3/49 3/35 3/54 3/57