lec 1 - fundamental concepts, force vectors
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ES 11 Lecture 1
➔the study of the relationship among
forces and their effects on bodies.
➔the science which describes and predicts the conditions for rest and motion of bodies under the action of forces.
Mechanics
Rigid Bodies
Statics
Dynamics
Deformable Bodies
Fluids
Compressible
Incompressible
➔represents the action of one body on another
➔may be exerted by actual contact or at a distance
➔characterized by its:
➛Point of application
➛Magnitude
➛Line of action
➔represented by a vector.
➔Development of other forces
➛Reactions
➛Internal forces
➔Deformation of the body
➔Acceleration of the body
Applied Force
Applied Force
Reaction
Development of force or forces at
points of contact with other
bodies (reactions).
Development of forces within the
body itself (internal forces) A A
Applied Force
Applied Force
Deformation of the body
Applied Force
Acceleration of the body
10 N
30o
10 N 30o
➔Point of application (forces acting on the same particle have the same point of application)
➔Magnitude
➔Direction ➛Line of action (angle w.r.t. a fixed axis)
➛Sense
Treatment of bodies as particles - the shape and size of the object does not significantly affect the solution of the problems under consideration.
Rigid Bodies - the problems considered in this course are assumed to be non-deformable.
Acceleration due to gravity
g = 9.81 m/s^2 or g = 32.2 ft/s^2
Quantity SI English
Length m (meter) ft (feet)
Mass kg (kilogram) slugs
Time s (seconds) s (seconds)
Force kg m/s2 OR N (newtons)
lbs (pounds)
P
Q
A
R
A
➔Single equivalent force having the same effect as the original forces acting on the particle
Parallelogram Law
• The resultant is the diagonal of the parallelogram with the two forces as its sides
Triangle Law
• Derived from the parallelogram law
• If the two vectors are placed tip-to-tail, the resultant is the third side of the triangle
P
Q A R
P
A
Q A P
Q
P
Q A P+Q = R
Q’ QPR
BPQQPR
cos2222
• Law of cosines,
• Law of sines,
P
C
R
B
Q
A sinsinsin
➔ Vectors are defined as mathematical expressions possessing magnitude and direction, which add according to the parallelogram law.
➔Forces are sliding vectors
➔Is vector addition commutative?
➔Is vector addition associative?
➔Is vector addition commutative?
➛The addition of vector is commutative.
P
Q A
P+Q
Q’
PQQP
P
Q A
Q+P
P’
➔Adding more than 2 vectors
If the vectors are coplanar, the resultant may be obtained by using the polygon rule for the addition of vectors – arrange the given vectors in a tip-to-tail fashion and connect the tail of the first vector with the tip of the last one
P
Q A
Q
S
R
P
Q A
Q
S
R
P+Q
➔Is vector addition associative?
➛Vector addition is associative
P
A
Q
S
R
P+Q
P
A
Q
S
R
Q+S
(P+Q)+S
P+ (Q+S)
P+Q+S = (P+Q)+S = P+ (Q+S)
PPP 2
Pn
- have the same direction as P with magnitude Pn
P 1.5P -2P
P
P
- Product nP of a positive integer n and a vector P
•Concurrent forces: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces.
The two forces act on a bolt at
A. Determine their resultant.
SOLUTION:
• Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant.
• Trigonometric solution - Apply the triangle rule.
From the Law of Cosines,
155cosN60N402N60N40
cos222
222 BPQQPR
N73.97R
From the Law of Sines,
A
A
R
QBA
R
B
Q
A
20
04.15N73.97
N60155sin
sinsin
sinsin
04.35
25
Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle if the resultant of the two forces applied to the hook support is to be horizontal, (b) the corresponding magnitude of R.
50 N
25O
P
26
50 N
25O
P
R
R
O sin
35
25sin
50
sin
(a) Determining &
OO
14.3735
25sin50sin 1
OO 25180
O86.117
NR 22.73
27
50 N
25O
P
R
R
O sin
35
25sin
50
sin
(b) Determining R
sin
sin50R
Two forces are applied as shown to a hook support. Knowing that the magnitude of the resultant of the two forces is a 50 – N horizontal force, determine the value of α for which the applied force 2 is minimum.
50 N
25O
P
Force 1
Force 2
➔It has been shown that the resultant of forces acting at the same point (concurrent forces) can be found.
➔In the same way, a given force F can be resolved into components.
29
➔Vector force components: two or more force vectors which, together, have the same effect as a single force vector.
➔How many components could a force be resolved into?
➛Infinite number of possible components
➔For a force resolved into two components,
32
15 . 0 °
40 . 0 °
45 . 0 °
a
b
P Given P is 800 N, determine the components of the force in a and b axes.
4045180
33
15 . 0 °
40 . 0 °
45 . 0 °
a
b
P 15.0O
25.0O P=800N
95O
40O
45O
AF
BF
A
O
B
OO
FF
40sin45sin
800
95sin
NFB 85.567
NFA 19.516
➔A force vector may be resolved into perpendicular components
j F F y y ˆ
j
i i F F x x ˆ
F
q
y
x
i j and - unit vectors of
magnitude 1 directed along the
positive x and y axes, respectively.
xF yF - vector components of
FFx , Fy – scalar components of
F
- may be positive or negative
depending upon the sense of xF
and yF
- the absolute values are equal
to the magnitude of the
component forces and xF yF
qcosFFx q sinFFy
F = Fxi +Fy j
0 35
0 145 q
N F 800
x F
y F FX = -655.3 N i
FY = 458.9 N j
2 - 37
SQPR
• Wish to find the resultant of 3 or more concurrent forces,
jSQPiSQP
jSiSjQiQjPiPjRiR
yyyxxx
yxyxyxyx
• Resolve each force into rectangular components
x
xxxx
F
SQPR
• The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces.
y
yyyy
F
SQPR
x
yyx
R
RRRR 122 tan q
• To find the resultant magnitude and direction,
38
tan q =
x
y
F
F
F = 22
yx FF q
F
x F
y F
x
y
qR
R
x R
y R
x
y
xx FR
yy FR x
y
RR
Rqtan
Determine the resultant of the three forces below
39
25o
45o
350 N
800 N
600 N
60o
y
x
40
25o
45o
350 N
800 N
600 N
60o
y
x
RX = F x = 350 cos 25O + 800 cos 70O -600 cos 60O
RX = 317.2 + 273.6 – 300 = 290.8 N
RY = F y = 350 sin 25 + 800 sin 70O +600 sin 60O
RY = 147.9 + 751 + 519.6 = 1419.3 N
F= 290.8 N i +1419.3 N j
Resultant, F
NF 14493.14198.290 22 O4.788.290
3.1419tan 1 q
F = 1449 N 78.4O
The resultant of concurrent forces acting on a particle in space will also act at the same particle.
Only the magnitude and direction are to be determined.
y
z
x
o
F1
F2
F3
Note: In this illustration, the magnitude and direction of all of the forces are given.
From the force polygon (warped), the resultant can be drawn from the tail of the first force to the head of the last force.
The magnitude and direction of the resultant can be computed using successive use of the triangle law.
y
z
x
o
F1
F2 F3
z q
x q
y q
R
Note: The sine and cosine laws are hard to implement because usually the given angles are absolute.
The rectangular components of a force can be determined easily depending on the given characteristics of the force.
Given the Magnitude and Two Angles
y
z
x
o
F
zq
yq
y
z
x
o
F
zq
xyq
xyF
zF
zxy FF qsin
zz FF qcos
xyzxyxyy FFF qqq cossincos
xyzxyxyx FFF qqq sinsinsin
Given the Magnitude and Two Angles
y
z
x
o
F
zq
xyq
xyF
zF
zxy FF qsin
y
z
x
o
xyq
xyF
zF
xF
yF
zz FF qcos
Given the Magnitude and Two Angles
y
z
x
o
xyq
xyF
zF
xF
yF xyzxyxyy
xyzxyxyx
FFF
FFF
qqq
qqq
cossincos
sinsinsin
222
ˆˆˆ
FzFyFxF
kFzjFyiFxF
In vector form,
Given the Magnitude and Three Absolute Angles
Fx = Fcos qx Fy = F cos qy Fz = Fcos qz
where cos qx, cos qy
and cos qz are direction cosines F = Fxi + Fyj + Fzk
y
z
x
o
x q
F
y q
z q
A force of 500N forms angles of 600, 450 and 1200, respectively, with the x,y and z axes. Find the components Fx, Fy and Fz of the force. Find also the vector representation of the force.
Fx = F cos qx = (500N)cos 600 Fx = +250N Fy = F cos qy = (500N) cos 450 Fy = +354N Fz = F cos qz = (500N) cos 1200 Fz = -250N
y
z
x
o
F
qx qy
qz
Fx
Fy
Fz
F = 250N i + 354N j – 250N k
y
z
x
o
F
qx qy
qz
Fx
Fy
Fz
F = 250N i + 354N j – 250N k
Note: The angle a force F forms with an axis should be measured from the positive side of the axis and will always be between 0 and 1800.
kzjyixFF ˆcosˆcosˆcos qqq
let = cos qx i + cos qy j+ cos qz k
= unit vector x= cos qx
y= cos qy
z= cos qz
x2 + y
2 + z2 = 1
cos2 qx + cos2 qy + cos2 qz = 1
it follows that,
F = F The force vector is equal to the product of the magnitude of the force and the unit vector.
A force has the components Fx = 20N, Fy = -30N, Fz = 60N. Determine its magnitude F and the angles qx, qy, qz it forms with the coordinate axes.
y
z
x
o
Fx
Fy
Fz 222 FxFyFxF
NNF
NNNF
704900
)60()30()20( 222
qx = cos-1 (Fx / F) = cos-1 (20/70)
qx = 73.40
qy = cos-1 (Fy / F) = cos-1 (-30/70)
qy = 115.40
qz = cos-1 (Fz / F) = cos-1 (60/70)
qy = 31.00
y
z
x
o
Fx
Fy
Fz F
F = 70 N
Given the Magnitude and Two Points where Force Passes
222 dzdydxd
zzd
yyd
xxd
oez
oey
oex
Fd
dxFx F
d
dyFy F
d
dzFz
kFzjFyiFxF ˆˆˆ
y
z
x
o
e e e z y x E , ,
o o o z y x O , ,
F
kFzjFyiFxF ˆˆˆ
y
z
x
o
eee zyxE ,,
eee zyxO ,,
F
FxFy
Fz
d
dxx qcos
d
dyy qcos
d
dzz qcos
Determine the vector representation of the given force.
z
x
O
2.4m E
1.5m
F=1.6kN
1.2m
1.5m
y
➔EXAMPLE 1-8
z
x
O
2.4m E
1.5m
F=1.6kN
1.2m
1.5m
y
O(1.2, 1.5, 0.0) E(0.0, 2.4, 1.5)
dx = 0.0 -1.2 = -1.2 dy = 2.4 -1.5 = +0.9 dz = 1.5 - 0.0 = +1.5
2225.19.02.1 d
d = 2.12m
kNkNFx 905.0)6.1(12.2
2.1
kNkNFy 679.0)6.1(12.2
9.0
kNkNFz 131.1)6.1(12.2
5.1
kkNjkNikNF ˆ131.1ˆ679.0ˆ905.0
Fx
Fz
z
x
O
E
y
Fy
The resultant R of two or more forces in space will
be determined by summing their rectangular
components. Graphical or trigonometric methods are
generally not practical in the case of forces in space.
FR FzFyjFxiRzkRyiRxi
kFzjFyiFx
FxRx FyRy FzRz
222 RzRyRxR
R
Rxx qcos
R
Ryy qcos
R
Rzz qcos
Determine graphically, the magnitude and direction of the resultant of the two forces using (a) Parallelogram law and (b) the triangle rule.
900 N
600 N
30o
45o
A parallelogram with sides equal to 900 N and 600 N is drawn to scale as shown.
From the scaled drawing of the forces, the resultant is
R 1400 N 900 N
600 N
30o
45o
R
q q 46o
Note: The triangle rule may also be used. Join the forces in a tip to tail fashion and measure the magnitude and direction of the resultant.
For the magnitude of R, using the cosine law:
oR 135cos6009002600900 222
NR 13916.1390
For angle q, using sine law:
sin
600
135sin
O
R
900 N
600 N
30o
135o
R
q
OO 8.1730 qThe angle of the resultant:
OO
8.171391
135sin600sin 1
Sample Problem 2.10 : (2.89) A frame ABC is supported in part by cable DBE which passes through a frictionless ring at B. Knowing that the tension in the cable is 385 N, determine the components of the force exerted by the cable on the support D.
400mm
210mm 280mm
510mm
480mm 600mm
x z
A
B
C
D E
O(0, 510, 280)
E(480, 0, 600)
dx = xE – xO = 480-0=480
dy = yE – yO = 0-510=-510
dz = zE – zO = 600-280=320
d = 770 mm
400mm
210mm
280mm
510mm
480mm
600mm
x z
A
B
C
D E
TDB = 385N
TDBY
TDBX
TDBZ
E(480, 0, 600)
O(0, 510, 280)
NTd
dT DB
x
DBX 240385770
480
NTd
dT DB
y
DBY 255385770
510
NTd
dT DB
zDBZ 160385
770
320
NkjikTjTiTT DBZDBYDBXDBˆ160ˆ255ˆ240ˆˆˆ
Solution:
Position vector of BH = 0.6 m i + 1.2 m j - 1.2 m k
Magnitude, BH = 0 6 1 2 1 2 1 8 2 2 2 . . . . m
BH
BH BH BH BH
BH
x y z
BH
BH m i m j m k
T T T BH
BH
N
m m i m j m k
T N i N j N k
F N F N F N
| | . ( . . . )
| | . | | | | .
. . .
( ) (500 ) (500 )
1
1 8 0 6 1 2 1 2
750
1 8 0 6 1 2 1 2
250
250 500 500