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HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Engineering Mechanics
STATICS
Equilibrium of a Particle
LECTURE 3
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Condition of the Equilibrium of a ParticleA particle is in equilibrium provided
• it is at rest if originally at rest or
• has a constant velocity if originally in motion
To maintain equilibrium, it is necessary to satisfy Newton’s first law of motion, which requires that the resultant force acting on a particle to be equal to zero.
This condition may be stated mathematically asnF = 0
The above equation is necessary and sufficient
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Condition of the Equilibrium of a ParticleAdditionally, if the particle is moving
⇒ Newton's second Law: ΣF = ma
But to satisfy equilibrium
ΣF = 0 ⇒ ma = 0 ⇒ a = 0
⇒ particle has constant velocity
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
The Free-Body Diagram
To apply the equation of equilibrium;All the known and unknown forces (nF ) must be considered
The best way to do that is through drawing the particle’s
FREE BODY DIAGRAM
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
The Free-Body Diagram
Procedure:
1. Draw outline shape of the particle to be isolated
2. Show all forces acting on the particle
3. Identify each force
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Springs• With linear elastic springs,
deformation is linearly proportional to the applied force
• The elasticity of the spring is defined by means of the spring stiffness k
F = ks
s = l – lo
If s > 0F must “pull”
If s < 0 F must “push”
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Cables and Pulleys
Remarks:
• Weights of Cables are not to be considered
• Cables support only Tension in the direction of the Cable
• For any θ the cable is subjected to a constant tension T throughout its length
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Example 3-1
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Coplanar Force System
If a particle is subjected to a system of forces that lie in the x-y plane
For the above vector equation to be satisfied then
These scalar equations must be satisfied
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Scalar Notation
• In the above example, when drawing the FBD, we assumed the senseof the unknown F is to the right.
• But the equation resulted in F = -10 N, which indicates that F must have a sense to the left to hold the particle in equilibrium
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Example 3-2
FBD
Equilibrium Equations
From the above Equilibrium Equations, then
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Example 3-3
FBD
Equilibrium Equations
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Example 3-3
FBD
FBDNewton’s 3rd Law
Principle of Action and Reaction
Equilibrium Equations
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-16/1
The cylinder D has a mass of 20 kg. If a force of F=100N is applied horizontally to the ring at A, determine the largest dimension d so that the force in cable AC is zero.
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-16/2Free-Body diagram of ring A:• Force from cable AC (FAC=0)• Force from cable AB (FAB)• Weight of cylinder D {W = 20(9.81) = 196.2N}• Force F = 100N
F = 100N x
y
W = 20(9.81) = 196.2N
FAB
θ⇒
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-16/3
)(
)(coscos
2196.2inθ0196.2inθ
0
1100θ0100θ
0:mEquilibriu of Equations
=⇒=−⇒
=↑+
=⇒=+−⇒
=⎯→⎯
∑
∑+
sFsF
F
FF
F
AB
AB
y
AB
AB
x
F = 100Nx
y
W = 20(9.81) = 196.2N
FAB
θ
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-16/4
NF
FF
sFF
AB
AB
AB
AB
AB
2220cos62.99
100(1)in Substitute
9962θ9621θ
100196.2
θθ
12
2196.2inθ1100θ
.
..tancossin
)()(
)()(cos
=°
=⇒
°=⇒=⇒
=⇒
=
=
F = 100Nx
y
W = 20(9.81) = 196.2N
FAB
θ
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-16/5
θ
mddd
4225199262
51θ22
d1.5θ
:geometry From9962θ
..).(tan
.)(tan
tan
.
=⇒−=⇒
−=⇒
+=
°=
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-22/1
The springs on the rope assembly are originally stretched 1 ft when θ = 0°. Determine the vertical force F that must be applied so that θ = 30°.
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-22/2Free-Body diagram of A:• Tension from cable AB (Fs)• Tension from cable AD (Fs)• Vertical Force F
Fs
x
F
Fs
30ο30ο⇒
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-22/3
lbFF
FsF
F
lbFkxF
ftxllx
ft
s
y
s
s
3390303392
0inθ2
0
33931130
3111312- stretched are springs the30θwhen
31230cos2
cosθ2BA
o
.))(sin.(
)(
.).(
..
.
=⇒=−°⇒
=−⇒
=↑+
==⇒
==−=⇒
=°=
=°
==
∑Fs
x
F
Fs
30ο30ο
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Three Dimensional Force System
⎪⎩
⎪⎨
⎧
===
⇒
=++⇒
=
∑∑∑
∑∑∑
∑
000
0:components , , their into resolved are forces theIf
0:require wemequilibriu particleFor
z
y
x
zyx
FFF
FFF
F
kjikji
00
00
00
321
321
321
=++−⇒=
=−+⇒=
=+−⇒=
∑∑∑
zzzz
yyyy
xxxx
FFFF
FFFF
FFFF
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/1
If the maximum allowable tension in cables AB and AC is 500 lb, determine the maximum height z to which the 200-lb crate can be lifted. What horizontal force F must be applied? Take y = 8 ft.
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/2
y = 8ft⇒
Free-Body diagram of A:• Tension from cable AB (FAB = 500 lb)• Tension from cable AC (FAC = 500 lb)• Weight of the crate (W = 200 lb) • Force F
A
FAB = 500 lb
200 lb
F
FAC = 500 lb
x
z
y
5 ft
4 ft
8 ft
5 ft
4 ft
8 ft
(4-z) ft
x’y’
z’
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/3
lb489
4500489
4000489
2500
500489
4489
84895
489485
485
axes as x'-y'-z'origin, asA Consider
222
222
2222
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
−+
−+
−+−
−+
−=⇒
=
−+
−+
−+−
−+
−=
−+=−+−+−=
−+−−=
k)(
)(j)(
i)(
F
uF
k)(
)(j)(
i)(
u
)()()()(
k)(jir
AB
ABAB
AB
AB
zz
zz
zz
zz
zzr
z
AB
FAC = 500 lb
FAB = 500 lb
200 lb
Fx
y
z
5 ft5 ft
4 ft
4 ft
8 ft
8 ft
(4-z) ft
x’y’
z’
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/4
{ }
{ }jF
kW
k)(
)(j)(
i)(
F
FF
k)(
)(j)(
i)(
F
AC
ACAB
AB
F
zz
zz
zz
zz
=
−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
−+
−+
−+−
−+=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
−+
−+
−+−
−+
−=
lb200
lb489
4500489
4000489
2500
thuslsymmetrica are and
lb489
4500489
4000489
2500
222
222
FAC = 500 lb
FAB = 500 lb
200 lb
Fx
y
z
5 ft5 ft
4 ft
4 ft
8 ft
8 ft
(4-z) ft
x’y’
z’
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/5
( )
( ))(
)()(
)()(
1489
8000
0489
8000
0489
4000489
4000
0
0489
2500489
2500
0
2
2
22
22
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+=⇒
=+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+
−⇒
=+−+
−−+
−⇒
=
=−+
+−+
−⇒
=
∑
∑
zF
Fz
Fzz
F
zz
F
y
x
y = 8ft
FAC = 500 lb
FAB = 500 lb
200 lb
Fx
y
z
5 ft5 ft
4 ft
4 ft
8 ft
8 ft
(4-z) ft
x’y’
z’
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/6
( )
( )
( ))()(
)(
)()(
)()(
)(
2489
41000200
0200489
41000
0200489
4500489
4500
0
1489
8000
2
2
22
2
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+
−=⇒
=−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+
−⇒
=−−+
−+
−+
−⇒
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+=
∑
z
z
z
z
zz
zz
F
zF
z
y = 8ft
FAC = 500 lb
FAB = 500 lb
200 lb
Fx
y
z
5 ft5 ft
4 ft
4 ft
8 ft
8 ft
(4-z) ft
x’y’
z’
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/7
( )
( )
Fz
F
z
z
zF
16004
8z-4200
(1)by (2) Dividing
2489
41000200
1489
8000
2
2
=−⇒
=⇒
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+
−=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+=
)(
)()(
)(
y = 8ft
FAC = 500 lb
FAB = 500 lb
200 lb
Fx
y
z
5 ft5 ft
4 ft
4 ft
8 ft
8 ft
(4-z) ft
x’y’
z’
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/8
( )
( )
22
2
2
2
1600898000
160089
18000
489
18000
489
8000(1) From
16004
⎟⎠⎞
⎜⎝⎛+=⎟
⎠⎞
⎜⎝⎛⇒
⎟⎠⎞
⎜⎝⎛+
=⇒
−+=⇒
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−+=
=−
FF
F
Fz
F
zF
Fz)(
y = 8ft
FAC = 500 lb
FAB = 500 lb
200 lb
Fx
y
z
5 ft5 ft
4 ft
4 ft
8 ft
8 ft
(4-z) ft
x’y’
z’
HCU – Dept. of Mechtronics Eng’gDr.-Ing. Saleh Chehade
Problem 3-59/9
ftzF
z
lbF
FF
FF
0729314
931831
160016004
83189
10146
8910146
8916008000
1600898000
7
27
2
22
22
..
.
.
.
=−=⇒
===−
=×
=⇒
=×⇒
=−
⇒
⎟⎠⎞
⎜⎝⎛+=⎟
⎠⎞
⎜⎝⎛
y = 8ft
FAC = 500 lb
FAB = 500 lb
200 lb
Fx
y
z
5 ft5 ft
4 ft
4 ft
8 ft
8 ft
(4-z) ft
x’y’
z’