Ch. 3: Equilibrium
3.0 Outline
 Mechanical System Isolation (FBD)
123
123
125
2-D Systems
 Equilibrium Conditions
144
3-D Systems
 Equilibrium Conditions
174
3.0 Outline
Ch. 3: Equilibrium
124
3.0 Outline
When a body is in equilibrium, the resultant on the body
is zero.
And if the resultant on a body is zero, the body is in
equilibrium.
So,
∑F = 0
∑M = 0
is the necessary and sufficient conditions for equilibrium.
3.0 Outline
Ch. 3: Equilibrium
125
3.1 Mechanical System Isolation (FBD)
Free-Body Diagram (FBD) is the most important first
step in the mechanics problems. It defines clearly the
interested system to be analyzed. It represents all
forces which act on the system. The system may be
rigid, nonrigid, or their combinations. The system
may be in fluid, gaseous, solid, or their combinations.
FBD represents the isolated / combination of bodies as
a single body. Corresponding indicated forces may
be
1. Contact force with other bodies that are removed
virtually.
2. Body force such as gravitational or magnetic
attraction forces.
3.1 FBD
Ch. 3: Equilibrium
126
3.1 FBD
Ch. 3: Equilibrium
127
Remarks
1. Force by flexible cable is always a tension. Weight
of the cable may be significant.
2. Smooth surface ideally cannot support the tangential
or frictional force. Contact force of the rough surface
may not necessarily be normal to the tangential surface.
3. Roller, rocker, smooth guide, or slider ideally
eliminate the frictional force. That is the supports
cannot provide the resistance to motion
in the tangential direction.
4. Pin connection provides support force in any direction
normal to the pin axis. If the joint is not free to turn,
a resisting couple may also be supported.
3.1 FBD
Ch. 3: Equilibrium
128
3.1 FBD
Ch. 3: Equilibrium
129
Remarks
5. The built-in / fixed support of the beam is capable of
supporting the axial force, the shear force, and
the bending moment.
6. Gravitational force is a kind of distributed non-contact
force. The resultant single force is the weight acted
through C.M. towards the center of the earth.
7. Remote action force has the same overall effects
on a rigid body as direct contact force of equal
magnitude and direction.
8. On the FBD, the force exerted on the body
to be isolated by the body to be removed is indicated.
9. Sense of the force exerted on the FBD by
the removed bodies opposes the movement which
would occur if those bodies were removed.
3.1 FBD
Ch. 3: Equilibrium
130
Remarks
10. If the correct sense cannot be known at first place,
the sense of the scalar component is arbitrarily
assigned. Upon computation, a negative algebraic
sign indicates that the correct sense is opposite
to that assigned.
3.1 FBD
Ch. 3: Equilibrium
131
Construction of FBD
1. Make decision which body or system is to be isolated.
That system will usually involve the unknown quantities.
2. Draw complete external boundary of the system
to completely isolate it from all other contacting
or attracting bodies.
3. All forces that act on the isolated body by the removed
contacting and attracting bodies are represented
on the isolated body diagram. Forces should be
indicated by vector arrows, each with its magnitude,
direction, and sense. Consistency of the unknowns
must be carried throughout the calculation.
4. Assign the convenient coordinate axes.
Only after the FBD is completed should the governing
equations be applied.
3.1 FBD
Ch. 3: Equilibrium
132
3.1 FBD
Ch. 3: Equilibrium
133
Note
1. Include as much as possible the system in FBD
while the unknowns are still being revealed.
2. Internal forces to a rigid assembly of members do not
influence the values of the external reactions. And so
the external response of the mechanism as a whole
would be unchanged.
3. Include the weights of the members on FBD.
4. Try to get the correct sense of unknown vectors by
visualizing the motion of the whole system when
the supports are pretended to disappear. The correct
sense will oppose the motion’s direction.
5. Follow the action of force prototypes in determining
the forces acted by the removed bodies.
3.1 FBD
Ch. 3: Equilibrium
134
3.1 FBD
Ch. 3: Equilibrium
Ax
135
Ay
MO
Ox
Oy
Bx
Ax
Ay
3.1 FBD
Ch. 3: Equilibrium
136
3.1 FBD
Ch. 3: Equilibrium
137
F
F
By
Ax
MA
Ax
3.1 FBD
Ch. 3: Equilibrium
138
3.1 FBD
Ch. 3: Equilibrium
139
3.1 FBD
Ch. 3: Equilibrium
1.
140
y
T
x
mg
F
N
mg
2.
On verge of being rolled over
means the normal force N = 0
P
y
R
N=0
T
x
3.1 FBD
Ch. 3: Equilibrium
T
3.
141
y
Rx
x
Ry
L
4.
y
N
AX
mg
mOg
x
Ay
3.1 FBD
Ch. 3: Equilibrium
mg
5.
T
O
142
y
x
F
∑M
N
O
=0
R
6.
y
AX
BX
x
Ay
By
3.1 FBD
Ch. 3: Equilibrium
143
7.
T
y
AX
x
Ay
8.
mg
BX
y
T
By
x
AX
Ay
L
3.1 FBD
Ch. 3: Equilibrium
144
3.2 2-D Equilibrium Conditions
A body is in equilibrium if all forces and moments
applied to it are in balance. In scalar form,
=
∑ Fx 0=
∑ Fy 0 =
∑ MO 0
• The x-y coordinate system and the moment point O
can be chosen arbitrarily.
• Complete equilibrium in 2-D motion must satisfy all
three equations. However, they are independent
to each other. That is, equilibrium may only be satisfied
in some generalized coordinates.
• System in equilibrium may stay still or move with
constant velocity. In both cases, the acceleration is zero.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
145
Categories of equilibrium
Some equations are automatically satisfied and
so contribute nothing in solving the problems.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
146
Weights of the members negligible
Equilibrium of a body under the action of two force only:
The forces must be equal, opposite, and collinear.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
147
Equilibrium of a body under
the action of three force only:
The lines of action of the three
forces must be concurrent.
The only exception is when the
three forces are parallel.
The system may be reduced to the
three-force member by successive
addition of the known forces.
If all forces are concurrent, then the
equilibrium statement calls for the
closure of the polygon of forces.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
148
Alternative Equilibrium Equations
Three independent equilibrium conditions:
=
∑ Fx 0=
∑ M A 0=
∑ MB 0
(
¬ AB ⊥ x-direction
)
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
149
Alternative Equilibrium Equations
Three independent equilibrium conditions:
=
∑ M A 0=
∑ MB 0 =
∑ MC 0
A, B, and C are not on the same straight line
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
150
Constraints and Statical Determinacy
The equilibrium equations may not always solve all
unknowns in the problem. Simply put, if #unknowns
(including geometrical variables) > #equations, then we
cannot solve it. This is because the system has more
constraints than necessary to maintain the equilibruim.
This is call statically indeterminate system. Extra
equations, from force-deformation material properties,
must also be applied to solve the redundant constraints.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
151
Constraints and Statical Determinacy
mg
P
P
Ax
Q
#unknowns = 2
#equilibrium eqs. = 2
statically determinate
Ay
By
Cy
Ax
Bx
Cx
F
Bx
Ay
By
#unknowns = 4
#equilibrium eqs. = 3
statically indeterminate
#unknowns = 6
#equilibrium eqs. = 3
statically indeterminate
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
152
Adequacy of Constraints
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
153
Problem Solution
1. List known – unknown quantities, and check the
number of unknowns and the number of available
independent equations.
2. Determine the isolated system and draw FBD.
3. Assign a convenient set of coordinate systems.
Choose suitable moment centers for calculation.
4. Write down the governing equation, e.g. ∑ M O = 0 ,
before the calculation.
5. Choose the suitable method in solving the problem:
scalar, vector, or geometric approach.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/1
154
In a procedure to evaluate the strength of the
triceps muscle, a person pushes 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. State
any assumptions.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/1
155
Assumption: contraction force from biceps muscle
acts at point O
1. Known: weight of lower hand,
pushing force
Unknown: triceps force, biceps force
2. FBD: lower hand
 ∑ M=
0 
O
-T × 25-1.5g ×150 + 160 × 300
= 0
T=1832 N
 ∑ Fy = 0 
y
T
T-C-1.5g +160=0
C=1977 N
C
x
1.5g
160 N
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/2
156
1. Unknown: l,R
Known: m, b, T
2. FBD: tensioning system with cut-cable
R
T
y
x
F
equivalent tension forces at the middle pulley
F = 2Tcos30
mg
T
Three-force member with mg, F, and O
For equilibrium, three lines of action must be concurrent.
2Tbcos30
 ∑ M O= 0  F × b-mg × l = 0
∴l =
mg
 ∑ F = 0 
R= F2 + ( mg ) = 3T 2 + m 2 g 2
2
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/3
157
The exercise machine consists of a lightweight cart which is
mounted on small rollers so that it is free to move along
the inclined ramp. Two cables are attached to the cart –
one for each hand. If the hands are together so that the cables
are parallel and if each cable lies essentially in a vertical plane,
determine the force P which each hand must exert on its cable
in order to maintain an equilibrium position. The mass of the
person is 70 kg, the ramp angle is 15°, and the angleβis 18°.
In addition, calculate the force R which the ramp exerts
on the cart.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/3
Assumption: negligible rail friction
70g
T
T
15°
T
158
R
9°
9°
x’
1. Unknown: P, T, R
2. FBD: exercise machine, pulley
 ∑ Fx ' = 0 
∑ F ' = 0
y


70gsin15 - Tcos9 = 0
R-70gcos15-Tsin9 = 0
T = 179.9 N
R = 691 N
2P
x’
T - 4Pcos9 = 0
P = 45.5 N
2P
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
159
P. 3/4 A uniform ring of mass m and radius r carries
an eccentric mass mo at a radius b and is in
an equilibrium position on the incline, which
makes an angleαwith the horizontal. If the
contacting surfaces are rough enough to
prevent slipping, write the expression for the
angleθwhich defines the equilibrium position.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/4
160
1. Unknown: F, N, θ
2. FBD: ring+eccentric mass
mg
mog
x’
F
N
 ∑ M O 0 
=
 ∑ Fx ' = 0 
Fr - m o gbsinθ = 0
m o gbsinθ
r
r 

m 
∴θ = sin -1  1 +
sin
α


b
m
o




∴F =
F - ( m o + m ) gsinα = 0
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/5
161
The hook wrench or pin spanner is used to
turn shafts and collars. If a moment of 80 Nm
is required to turn the 200 mm diameter collar
about its center O under the action of the applied
force P, determine the contact force R on the
smooth surface at A. Engagement of the pin
at B may be considered to occur at the periphery
of the collar.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/5
R
162
B
80 Nm
NA
y
shaft & hook as one system
 ∑ M O =
0 
80-P × 0.375 = 0
∴ P = 213.3 N
x
Three - force member
 ∑
=
M B 0 
N A × 0.1sin 60 − P × ( 0.375 + 0.1cos60
=
) 0
N A = 1047 N
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/6
163
The small crane is mounted on one side of the
bed of a pickup truck. For the positionθ=40°,
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.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/6
110
164
D
340
C
d
O
α
360
α
B
geometry at BCDO
−1  360 + 340sin 40 − 110 cos 40 
°
tan
=

 56.2
 340 cos 40 + 110sin 40 
d = 360cosα = 200 mm
α
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
165
P. 3/6
y
x
120g
C
Ox
O
Oy
Three - force member
O
=
∑ M O 0 120g × ( 785 + 340 ) cos 40 − C × d = 0 C = 5063 N
F
p = 2 = 2.58 MPa
πr
 ∑ Fx =
0 
O x − Ccosα = 0
Ox =
2820 N
 ∑ Fy =
0 
- O y − 120g + Csinα = 0
Oy =
3030 N
O = O 2x + O 2y = 4140 N
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/7
166
The rubber-tired tractor shown has a mass of 13.5 Mg with
the C.M. at G and is used for pushing 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 wheels is 80 percent
of the normal force under that wheel. Also find the total normal
reaction NB under the rear pair of wheels at B.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
y’
P. 3/7
167
13500g
x’
0.8NA
NA
 ∑ Fx ' 0 
=
P - 0.8N A − 0.8N B + 13500g ×
∑
=
Fy' 0 


N A + N B − 13500g ×
0.8NB
NB
15
= 0
152 + 1002
100
= 0
15 + 1002
100
15
 ∑ M A =
0 
N B × 1.8 − P × 0.6 -13500g ×
× 1.2 − 13500g ×
× 0.825 =
0
152 + 1002
152 + 1002
=
=
N A 6.3
kN, N B 124.7 kN, P = 85.1 kN
2
alternative equations:
 ∑ M A 0 =
 ∑ M B 0=
  ∑ Fx ' 0 
=
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
168
P. 3/8 Pulley A delivers a steady torque (moment) of
100 Nm to a pump through its shaft at C. The
tension in the lower side of the belt is 600 N.
The driving motor B has a mass of 100 kg and
rotates clockwise. Determine the magnitude R
of the force on the supporting pin at O.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
169
mg
P. 3/8
T
100 Nm by load
600 N
y
 ∑ M C = 0 
x
100g
T
( 600 - T ) × 0.225 − 100 =
 ∑ M
=
0 
D
∴ T = 155.6 N
0
O y × 0.25 − 600 × ( 0.2 − 0.075 ) − 100g × 0.125
- T × 0.075 - Tcos30 × 0.2 + Tsin30 × 0.125 = 0
O y = 906 N
600 N ∑ Fx = 0 
Tcos30 + 600 - O x = 0
∴ O x = 734.7 N
O = O 2x + O 2y =
1.17 kN
D
Ox
P
Oy
 ∑=
Fy 0 
Tsin30 -100g - P + =
Oy 0
∴ P = 2.8 N
spring compressed to resist rotation of the body
=
 ∑ M D 0=
  ∑ Fx 0  =
 ∑ M O 0 
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/9
170
When setting the anchor so that it will dig into the sandy bottom,
the engine of the 40 Mg cruiser with C.G. at G is run in reverse
to produce a horizontal thrust T of 2 kN. If the anchor chain
makes an angle of 60°with the horizontal, determine the
forward shift b of the center of buoyancy from its position when
the boat is floating free. The center of buoyancy is the point
through which the resultant of the buoyant force passes.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
171
y
P. 3/9
40000g
x
b
B
x
A
free floating (no thrust, tension): buoyancy force = weight, acting at C.G.
backward motion: new buoyancy force acting at new position
to maintain equilibrium
 ∑ Fx
 ∑ Fy
0 
Acos60 - 2000 = 0
0 
B - 40000g - Asin60 = 0
∴ A = 4 kN
∴ B = 395864 N
 ∑=
M A 0  40000g × 8 - 2000 × 3 - Bx = 0
b = 8 - x = 85.2 mm
∴ x = 7.915 m
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/10
172
A special jig for turning large concrete pipe sections (shown
dotted) consists of an 80 Mg sector mounted on a line of rollers
at A and a line of rollers at B. One of the rollers at B is a gear
which meshes with a ring of gear teeth on the sector so as to
turn the sector about its geometric center O. When α= 0,
a counterclockwise torque of 2460 Nm must be applied to the
gear at B to keep the assembly from rotating. When α = 30,
a clockwise torque of 4680 Nm is required to prevent rotation.
Locate the mass center G of the jig by calculating r and θ.
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
173
F2
P. 3/10
2460 Nm
F1
 ∑ M B = 0 
4680 Nm
α = 0° : 2460 - F1 × 0.24= 0, F1 = 10250 N
α = 30° : - 4680 + F2 × 0.24= 0,
F2= 19500 N
80000g
 ∑ M O = 0 
α = 0° : 80000g × rcosθ − 10250 × 5 = 0
α = 30° : -80000g × rcos (180-30-θ ) + 19500 × 5 = 0
=
r 367 mm,=
θ 79.8°
y
F
NA
x
NB
3.2 2-D Eqilibrium Conditions
Ch. 3: Equilibrium
174
3.3 3-D Equilibrium Conditions
A body is in equilibrium if all forces and moments
applied to it are in balance. In scalar form,
=
∑ Fx 0=
∑ Fy 0=
∑ Fz 0
=
∑ M Ox 0 =
∑ M Oy 0=
∑ M Oz 0
• The x-y-z coordinate system and the moment point O
can be chosen arbitrarily.
• Complete equilibrium in 3-D motion must satisfy all
six equations. However, they are independent
to each other. That is, equilibrium may only be satisfied
in some generalized coordinates.
• System in equilibrium may stay still or move with
constant velocity. In both cases, the acceleration is zero.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
175
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
176
Categories of equilibrium
Some equations are automatically satisfied and
so contribute nothing in solving the problems.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
177
Constraints and Statical Determinacy
The equilibrium equations may not always solve all
unknowns in the problem. Simply put, if #unknowns
(including geometrical variables) > #equations, then we
cannot solve it. This is because the system has more
constraints than necessary to maintain the equilibrium.
This is call statically indeterminate system. Extra
equations, from force-deformation material properties,
must also be applied to solve the redundant constraints.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
178
Adequacy of Constraints
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
179
P. 3/11 The light right angle boom which supports the
400 kg cylinder is supported by three cables
and a ball-and-socket joint at O attached to the
vertical x-y surface. Determine the reactions
at O and the cable tensions.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/11
TAC
180
TBD
O
TBE
400g
n AC =
−0.408i + 0.408 j − 0.816k
=
n BD 0.707 j − 0.707k
n BE =
−k , n OE =
i
n OB = 0.6i + 0.8k
n OD = 0.6i + 0.8 j
 ∑ M OB = 0  to find TAC
( −0.75i ) × ( −400gj) + ( 2k ) × TACn AC  n OB= 0 ⇒ TAC= 4808.8 N
 ∑ M OD = 0  to find TBE
( 2k ) × TACn AC + ( 0.75i + 2k ) × ( −400gj) + (1.5i ) × TBE n BE  n OD= 0 ⇒ TBE= 654 N
 ∑ M OE = 0  to find TBD
( 2 j) × TBDn BD + ( 0.75i + 2k ) × ( −400gj) + ( 2k ) × TACn AC  n OE =0 ⇒ TBD =2775.1 N
 O x 1962=
N, O y 0=
N, O z 6540 N
 ∑ F = 0=
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/12
181
The 600 kg industrial door is a uniform rectangular panel which
rolls along the fixed rail D on its hanger-mounted wheels A and B.
The door is maintained in a vertical plane by the floor-mounted
guide roller C, which bears against the bottom edge. For the
position shown compute the horizontal side thrust on each of
the wheels A and B, which must be accounted for in the design
of the brackets.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/12
182
Bx
Ax
Bz
Az
600g
NC
 ∑ M AB= 0  600g × 0.15 − N C × 3= 0 ⇒ N C= 294.3 N
 ∑ M Az= 0  N C × 0.6 − Bx × 3= 0 ⇒ Bx= 58.86 N
 ∑ Fx = 0 
A x + Bx − N C = 0 ⇒ A x = 235.44 N
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
183
P. 3/13 The smooth homogeneous sphere rests in the
120°groove and bears against the end plate
which is normal to the direction of the groove.
Determine the angle θ, measured from the
horizontal, for which the reaction on each side
of the groove equals the force supported by
the end plate.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/13
184
Projection onto two orthogonal planes
z
mgcosθ
mg
z
θ
y
N2
x
N1 30°
Nr
N1cos30+N2cos30
 ∑ =
=
Fy 0  N
N=
N
1
2
=
 ∑ Fz 0=
 mgcosθ 2Ncos30
 ∑ Fx 0=
 N r mgsinθ
=
if N r = N,
tanθ =1/ 2 cos 30
⇒ θ =30°, N = mg/2
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
185
P. 3/14 The mass center of the 30 kg door is in the
center of the panel. If the weight of the door is
supported entirely by the lower hinge A, calculate
the magnitude of the total force supported by
the hinge at B.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
186
z
P. 3/14
Bx
By
Ax
30g
y
Ay
30g
x
 ∑ Fx = 0   ∑ M A y = 0  30g × 0.36 − Bx ×1.5 = 0, Bx = A x = 70.6 N


 ∑ Fy = 0   ∑ M A x = 0  By ×1.5 − 30g × 0.9 = 0, By = A y = 176.6 N
B=
B2x + B2y = 190.2 N
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/15
187
One of the three landing pads for the Mars Viking lander is
shown in the figure with its approximate dimensions. The mass
of the lander is 600 kg. Compute the force in each leg when
the lander is resting on a horizontal surface on Mars. Assume
equal support by the pads and consult Table D/2 in Appendix D
as needed.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
FDC
P. 3/15
188
TCB
g=3.73 m/s2
TCA
200g
n DC =
0.35i − 0.936k , n CA =
−0.7664i + 0.418 j + 0.4877k
 ∑ M BA = 0  to find FDC
( 0.85k + 0.1i ) × FDCn DC − 200g × 0.55j  j = 0
⇒ FDC =
1049.1 N
 ∑ Fx = 0  and symmetry about x-z plane
FDCn DC i − 2TCA × 0.7664 =0 ⇒ TCA =TCB =239.5 N
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/16
189
The uniform 15 kg plate is welded to the vertical shaft, which
is supported by bearings A and B. Calculate the magnitude
of the force supported by bearing B during application of the
120 Nm couple to the shaft. The cable from C to D prevents
the plate and shaft from turning, and the weight of the
assembly is carried entirely by bearing A.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/16
190
Ay
15g
z
A
By x
Bx
y
15g
T
x
n DC =
−0.95i − 0.316 j
 ∑ M Oz = 0  120 + 0.6i × Tn DC k = 0, T = 632.9 N
 ∑ M A = 0  − Bx × 0.2 + 15g × 0.3 + Tx × 0.68= 0, Bx= 2265 N
y


 ∑ M=
= 0, B=
0  By × 0.2 − Ty × 0.68
680 N
Ax
y
B=
B2x + B2y = 2635 N
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
191
P. 3/17 The uniform 900x1200 mm trap door has a
mass of 200 kg and is propped open by the light
strut AB at the angle θ= atan(4/3). Calculate
the compression FB in the strut and the force
supported by the hinge D normal to the hinge
axis. Assume that the hinges act at the extreme
ends of the lower edge.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
192
P. 3/17
z
Dz
Dx
TAB
C
200g
Dy
y
x
n AB =
−0.2857i − 0.4286 j + 0.857k
 ∑ M Cx = 0 
∑ M C = 0
y


 ∑ M=
0 
Cz
Dn =
[0.9j × TABn AB ]i − 200g × 0.45cos53.13=
0, TAB = 688 N
− 200g × 0.6 + D z ×1.2= 0, D z= 981 N
− D y ×1.2 + ( −TABn AB i ) × 0.9
= 0, D
=
147.4 N
y
D 2y + D 2z = 992 N
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/18
193
The uniform rectangular panel ABCD has a mass of 40 kg
and is hinged at its corners A and B to the fixed vertical surface.
A wire from E to D keeps edges BC and AD horizontal. Hinge A
can support thrust along the hinge axis AB, whereas hinge B
supports force normal to the hinge axis only. Find the tension
T in the wire and the magnitude B of the force supported by
hinge B.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
194
x
P. 3/18
Bz
By
TDE
z
40g
Ay
n DE =0.35i − 0.707 j + 0.61k
0 
Ax
 ∑ M=
y
Az
Ax
0.6j × 40g ( − cos 30k − sin 30i )  i
+ [1.2 j × TDE n DE ]i = 0 ⇒ TDE =
278.55 N
 ∑ M A = 0  1.2i × 40g ( − cos 30k − sin 30i )   j − 2.4Bz = 0, Bz = 169.9 N
y


 ∑
 By 0 N ∴ Bn =
169.9 N
M AE 0=
=
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/19
195
Under the action of the 40 Nm torque (couple) applied to the
vertical shaft, the restraining cable AC limits the rotation of the
arm OA and attached shaft to an angle of 60°measured from
the y-axis. The collar D fastened to the shaft prevents downward
motion of the shaft in its bearing. Calculate the bending moment
M, the compression P, and the shear force V in the shaft at
section B. (note: Bending moment, expressed as a vector,
is normal to the shaft axis, and shear force is also normal to
the shaft axis.)
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
196
P. 3/19
x
y
MBx
Vy
Vx
40 Nm
TAC
MBy
P
section the shaft at B revealing the reaction force and moment
=
n AC 0.53i + 0.38 j − 0.758k
 ∑ M z =0  40 + [ 0.18 j × TACn AC ]k =0 ⇒ TAC =419.3 N
 ∑=
Fz 0  P + TACn AC k = 0 ⇒ =
P 317.8 N
V=
Vx2 + Vy2 =
2
− P 2 = 273.5 N
TAC
=
k + 0.18 j) × TACn AC 0
 ∑ M B 0  M Bx i + M By j + 40 + ( −0.09=
=
=
M Bx 42.87
Nm, M By 20.0 Nm
∴ M b=
M 2Bx + M 2By= 47.3 Nm
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
197
P. 3/20
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
198
z
P. 3/20
y
x
FBD of reel only
 ∑ M O= 0 
y


100 × 0.15 − P × 0.3
= 0, =
P 50 N
 ∑ M
0 
=
Bx
200 × 0.2 − 100sin15 × 0.35
NB
NC
NA
58.13 N
− Psin30 × 0.075 − N C × 0.5
= 0, N=
C
 ∑ M B= 0 
y


− N A × 0.525 − N C × 0.2625 + 200 × 0.2625
+ 100 cos15 × 0.52 − 100sin15 × 0.2237
− Pcos30 × 0.635 + Psin30 × 0.1125 = 0, N A = 108.56 N
 ∑ F=
0 
z
N A + N B + 100sin15 + N C − Psin30 − 200
32.44 N
= 0, N=
B
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
199
P. 3/21 The drum and shaft are welded together and
have a mass of 50 kg with mass center at G.
The shaft is subjected to a torque (couple) of
120 Nm, and the drum is prevented from
rotating by the cord wrapped securely around
it and attached to point C. Calculate the
magnitudes of the forces supported by
bearings A and B.
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
200
P. 3/21
50g
z
Ax
Az
y
T
Bx
Bz
x
66.87°
0.15
∑ M B = 0
y


 ∑ M =
0 
Bz
 ∑ M B=
0 
x
 ∑ F=
0 
x
 ∑ Fz = 0 
A=
T × 0.15 − 120 = 0, T = 800 N
= 0, A=
Tcos66.87 × 0.36 − A x × 0.7
161.6 N
x
0.18
T
0.24
= 0, A=
50g × 0.3 + Tsin66.87 × 0.36 − A z × 0.7
588.6 N
z
A x + Bx − Tcos66.87
= 0, B=
152.6 N
x
A z + Bz − 50g − Tsin66.87
= 0, B=
637.6 N
z
A 2x + A 2z = 610.4 N, B =
B2x + Bz2 = 655.6 N
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
201
P. 3/22
3.3 3-D Eqilibrium Conditions
Ch. 3: Equilibrium
P. 3/22
double U-joint
202
z
TBC
Oy
TAD
M
Ox
Oz
x
n BC =
0.13i − 0.91j + 0.39k , n AD =
−0.48i − 0.84 j + 0.241k
50g
 ∑
 O z 0 N
=
M AB 0=
 ∑=
M z 0  (1.8i × TBCn BC )k + ( 2.1j × TADn AD )k = 0
 ∑
=
M x 0  ( 2.1j × TBCn BC )i + ( 2.1j × TADn AD )i − 50g=
× 2.1 0
=
TBC 625
=
N, TAD 1024 N
 ∑ M y= 0  M + 50gx + (1.5i × TBCn BC ) j = 0
=
M 365.66 − 490.5 x
 ∑ Fy =
0 
 ∑ Fx =
0 
O=
O y + TBCn BC  j + TADn AD  j =
0, O y =
1429 N
O x + TBCn BC i + TADn AD i =
0, O x =
410 N
O 2x + O 2y + O 2z = 1487 N
3.3 3-D Eqilibrium Conditions