ch. 2: force systems
TRANSCRIPT
Ch. 2: Force Systems
2.0 Outline 27 Overview of Forces 282-D Force Systems Rectangular Coordinate Systems 33 Force, Moment, and Couple 45 Resultants 613-D Force Systems Rectangular Coordinate Systems 85 Force, Moment, and Couple 95 Resultants 109
2.0 Outline
27
Ch. 2: Force Systems
2.1 Overview of Forces
Force The measure of the attempt to move a body. It isa fixed vector. For the rigid body problems or only the external effects of the external force onto the objects are of interested, force can be treated as a sliding vector. Hence, the complete description must include magnitude, direction, and line of action. And the problem makes use of the principle of transmissibility.
Contact vs. Body Force
Concentrated vs. Distributed Force
2.1 Overview of Forces
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Ch. 2: Force Systems
2.1 Overview of Forces
Contact vs. Body Force
Concentrated vs. Distributed Force
2.1 Overview of Forces
A
P P
B
29
Ch. 2: Force Systems
2.1 Overview of Forces
Force Measurement by comparison or by deformation of an elastic element (force sensor)
Action vs. Reaction Force Isolate the object and the force exerted on that body is represented FBD
Combining Force by parallelogram law and principle of moment
Force Components along the specified coordinate system to satisfy the parallelogram law (reverse step)
2.1 Overview of Forces
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Ch. 2: Force Systems
2.1 Overview of Forces
Orthogonal Projection along the specified direction. The components of a vector are not the same as the orthogonal projections onto the same coord. system, except the coordinate system is the orthogonal (rectangular) coordinate system.
2.1 Overview of Forces
R
F1
F2
Fa
Fb
a
b
R
Fa’
Fb’
a'
b'
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Ch. 2: Force Systems
2.1 Overview of Forces
Addition of Parallel Forces by graphic or algebraic (principle of moment) approach
2.1 Overview of Forces
F2
F1
(1)
F2
F1
F -F
R1 R2
(2)
R1
R2
R
(3)
32
Ch. 2: Force Systems2.2 2-D Simple Rectangular Coordinate Systems
2.2 2-D Rectangular Coord. Systems
y
xFx
FyF
θ
( )
x y
2 2x y
x y
y x
ˆˆF F
F = F F
F Fcos F Fsin
arctan 2 F ,F
θ θ
θ
+
+
= =
=
x yF = F F
F = i + j
Magnitude is always positiveScalar component includes sign information too!
33
Ch. 2: Force Systems
2.2 2-D Rectangular Coord. Systems
2.2 2-D Arbitrary Rectangular Coordinate Systems
by convenience, right-handed, geometry of the problem
β
x
y
F
Fx = Fsinβ
Fy = Fcosβ
βx
yF
Fx = Fsin(π−β)
Fy = -Fcos(π−β)
x
Fx = Fcos(α−β)
Fy = Fsin(α−β)
β
y F
α
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Ch. 2: Force Systems
P. 2/1 If the two equal tension T in the pulley cabletogether produce a force of 5 kN on thepulley bearing, calculate T.
5 kN
35
Ch. 2: Force Systems
P. 2/1
60゜
T
T 5 kN5 kN
By cosine law: 2 2 25 T T 2T Tcos60= + + ⋅ °
T = 2.89 kN
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Ch. 2: Force Systems
P. 2/2 While steadily pushing the machine upan incline, a person exerts a 180 N force Pas shown. Determine the components of Pwhich are parallel and perpendicular to theincline.
37
Ch. 2: Force Systems
P. 2/2
t
n
180 N10゜
15゜15゜
( )( )
t
n
P 180cos 10 15 163.1 N
P 180sin 10 15 76.1 N
= + =
= − + = −
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Ch. 2: Force Systems
P. 2/3 Determine the resultant R of the two forcesapplied to the bracket. Write R in terms ofunit vectors along the x- and y-axes shown.
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Ch. 2: Force Systems
P. 2/3 clearly and carefully draw the picture!
10゜
20゜
15゜20゜
150 N200 N
y’ y
x
x’
( ) ( )( ) ( )
'
'
x
y
x
y
R 200cos 15 + 20 150sin 10 20 88.8 N
R 200sin 15 20 150cos 10 20 244.6 N
R 200cos15 150sin10 167.1 N
R 200sin15 150cos10 199.5 N
88.8 244.6 N 167.1 199.5 N
Non-Orthogonal Coordinate System x'-yby l
− + =
= + + + =
= − =
= + =
= +
=
R = i + j i' j'
( ) ( )
aw of sine and cosine200 N --> 174.34 55.1 N150 N --> -79.8 157.2 N
174.34-79.8 55.1+157.2 94.54 212.3 N
i' + ji' + j
R = i' + j = i' + j
200 N
110゜15゜
30゜
70゜80゜
150 N
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Ch. 2: Force Systems
P. 2/4 It is desired to remove the spike from thetimber by applying force along its horizontalaxis. An obstruction A prevents direct access,so that two forces, one 1.6 kN and the other P,are applied by cables as shown. Computethe magnitude of P necessary to ensureaxial tension T along the spike. Also find T.
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Ch. 2: Force Systems
P. 2/4
x
y
No net force in y-direction
y
x
100 150R Psin atan 1.6sin atan 0200 200
P = 2.15 kN
100 150T = R Pcos atan 1.6cos atan200 200
= 3.20 kN
= − = ∴
= +
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Ch. 2: Force Systems
P. 2/5 As it inserts the small cylindrical part intoa close fitting circular hole, the robot armexerts a 90 N force P on the part parallel tothe axis of the hole as shown. Determinethe components of the force which the partexerts on the robot along axes (a) paralleland perpendicular to the arm AB, and (b)parallel and perpendicular to the arm BC.
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Ch. 2: Force Systems
P. 2/5Quasi-EquilibriumP is the force done by the robot on the part-P is the force done by the part on the robot
may be part of strength analysisof the robot arm
15゜45゜60゜
90 Nn1
t1n2
t2 -90cos45 90sin 45 63.6 63.6 N90cos60 90sin 60 45 77.9 N
− + = − +− + = +
1 1 1 1
2 2 2 2
P = n t n tP = n t n t
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Ch. 2: Force Systems
2.3 2-D Moment and Couple
2.3 2-D Moment and Couple
Moment The measure of the attempt to rotate a body. It is induced by force. The moment vector’s direction is perpendicular to the plane established by the point and the line of action of the force. It is a fixed vector. For the rigid body problems or only the external effects of the external moment onto the objects are of interested, moment can be treated as a sliding vector. Hence, the complete description must include magnitude, direction, and line of action. And the problem makes use of the principle of transmissibility.
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Ch. 2: Force Systems
AM F
2.3 2-D Moment and Couple
α
FA r
d
MA
For 2-D (in-plane rotation) problem, moment vector alwayspoints perpendicular to the plane. So it can be indicated bythe magnitude and the sense of rotation (CCW +, CW -)about the point.
Moment of about point A
( )AM F r sin Fdα= ×
= ⋅ =AM r F
*** Sign’s consistency throughout the problem ***
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Ch. 2: Force Systems
2.3 2-D Moment and Couple
Varignon’s theorem The moment of a force about any point is equal to the sum of the moments of the components of the force about the same point
F
Ar
(MA ) FP
Q
(MA ) Q
(MA ) P ( )( )
( ) ( )
if
= ×
×
+
A F
A AP Q
M r F
= r P + Q
= M M
F = P + Q
Usage Calculate the moment of the force fromits components. Moments of some componentsmay be trivial to calculate.
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Ch. 2: Force Systems
2.3 2-D Moment and Couple
Couple The measure of the attempt to purely rotate a body. It is produced by two equal, opposite, and non-collinear force. The couple vector’s direction is perpendicular to the plane established by those two lines of action of the force. It is a free vector and so no moment center. Only the magnitude and direction are enough to describe the couple.
α
-F
r
M
d
FrArB
( )
( ) M = F r sin Fdα
× ×
= ×
⋅ =
A BM = r F + r -Fr F
arbitrary chosen
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Ch. 2: Force Systems
2.3 2-D Moment and Couple
M
d
F
-
F
M
d-F -F
M
dF
-2FM
d/2
2F
For rigid body, several pairs of equal & opposite forcescan determine the same couple. It is unique tocalculate the couple from the given pair of forces butit is non-unique to determine the pair of forces whichproduce that value of couple.
Usage Effect of the couple can be determined fromthe equivalent pair of forces. Effect from some specificpairs of forces may be trivial to calculate.
-F
F
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Ch. 2: Force Systems
P. 2/6 Calculate the moment of the 250 N force onthe handle of the monkey wrench about thecenter of the bolt.
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Ch. 2: Force Systems
P. 2/6
x
y
Varignon’s Theorem
OM 250cos15 0.2 250sin15 0.03 46.4 Nm CW
= − × + ×=
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Ch. 2: Force Systems
P. 2/7
x
y
Vector approach
0.03 0.35 m240cos10 - 240sin10 N
84.0 Nm= ×O
r = i + jF = i jM r F = - k
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Ch. 2: Force Systems
P. 2/8 The force exerted by the plunger of cylinderAB on the door is 40 N directed along the lineAB, and this force tends to keep the doorclosed. Compute the moment of this forceabout the hinge O. What force Fc normal tothe plane of the door must the door stop at Cexert on the door so that the combined momentabout O of the two forces is zero?
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Ch. 2: Force Systems
P. 2/8
( )O
C O
= atan 100/400 0.245 radM 40cos 0.075 40sin 0.425 7.03 Nm CWF M / 0.825 8.53 N
θθ θ
=
= − × − × == =
40 NFC
MO
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Ch. 2: Force Systems
P. 2/9 While inserting a cylindrical part into thecircular hole, the robot exerts the 90 N forceon the part as shown. Determine the momentabout points A, B, and C of the force whichthe part exerts on the robot.
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Ch. 2: Force Systems
P. 2/9
( ) ( )
at about
at about
0.55cos60 + 0.45cos45 0.55sin60 - 0.45sin45 0.593 0.158 m-90sin15 90cos15 23.29 86.93 N
90 0.15 13.5 Nm CCW55.23 Nm CCW
68.7 Nm CCW
= × == ×
= + =
AC
C
-P C A AC
A C -P C A
r = i + j = i + jF = -P = i + j = - i + jMM r F =M M M
-P
MC
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Ch. 2: Force Systems
P. 2/10 As part of a test, the two aircraft engines arerevved up and the propeller pitches areadjusted so as to result in the fore and aftthrusts shown. What force F must be exertedby the ground on each of the main brakedwheels at A and B to counteract the turningeffect of the two propeller thrusts? Necglectany effects of the nose wheel C, which isturned 90°and unbraked.
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Ch. 2: Force Systems
P. 2/10
CM 2 5 F 3 = 0F = 3.33 kN
= × − ×
no rotation resultant couple = 0
F
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Ch. 2: Force Systems
P. 2/11 A lug wrench is used to tighten a square-headbolt. If 250 N forces are applied to the wrenchas shown, determine the magnitude F of theequal forces exerted on the four contact pointson the 25 mm bolt head so that their externaleffect on the bolt is equivalent to that of thetwo 250 N forces. Assume that the forces areperpendicular to the flats of the bolt head.
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Ch. 2: Force Systems
P. 2/11
( )250 0.7 2 F 0.025F = 3500 N
× = ×
Equivalent couple system at bolt head
F
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Ch. 2: Force Systems
2.4 2-D Resultants
2.4 2-D Resultants
Force – push / pull body in the direction of force–- rotate the body about any axis except the
intersection line to the line of force
Dual effects : force and couple to separate the push / pulland rotate effect while maintaining theresultant force and moment (external effect)
FA
B
-F d
FA
B F
M=FdA
B F
Force-Couple System
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Ch. 2: Force Systems
2.4 2-D Resultants
Resultant is the simplest force combination which can replace the original system of forces, moments, and couples without altering the external effect of the system on the rigid body. The resultant force determination will be used in the Newton’s 2nd law :
m=∑F a
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Ch. 2: Force Systems
2.4 2-D Resultants
Resultant Determination• Force Polygon : head to tail of force vectorsNote: only magnitude and direction are ensured
i.e., line of action may be incorrect!
For the specified rectangular coordinate system,
( ) ( )( )
22
x x y y x y
y x
R F R F R= F F
arctan 2 R ,Rθ
+ + =
= = +
=
∑∑ ∑ ∑ ∑1 2 3R = F F F F
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Ch. 2: Force Systems
2.4 2-D Resultants
Resultant Determination• Prin. Transmissibility & Parallelogram Law :
graphical method quick and easy visualizable butlow accuracy
Note: magnitude, direction, and line of action are correct
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Ch. 2: Force Systems
2.4 2-D Resultants
Resultant Determination• Force-Couple Equivalent Method:
algebraic method high accuracyNote: magnitude, direction, and line of action are correct
1. Specify a convenient reference point2. Move all forces so the new lines of action
pass through point O Force-Couple Equivalence
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Ch. 2: Force Systems
2.4 2-D Resultants
Resultant Determination• Force-Couple Equivalent Method:
algebraic method high accuracyNote: magnitude, direction, and line of action are correct
3. Sum forces and couples to 4. Locate the correct line of action of Force-Couple Equivalence
and OR MR
( )O i i ii i
M M Fd Rd Principle of Moment= = =∑∑ ∑
R = F
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Ch. 2: Force Systems
P. 2/12 In the design of the lifting hook the action ofthe applied force F at the critical section ofthe hook is a direct pull at B and a couple.If the magnitude of the couple is 4000 Nm,determine the magnitude of F.
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Ch. 2: Force Systems
P. 2/12Equivalent force-couple systemat the critical section
F 0.1 = 4000 F=40 kN× ∴
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Ch. 2: Force Systems
P. 2/13 Calculate the moment of the 1200 N forceabout pin A of the bracket. Begin by replacingthe 1200 N force by a force couple systemat point C. Calculate the moment of the1200 N force about the pin at B.
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Ch. 2: Force Systems
P. 2/13
C
A C
B A
M 1200 0.2 240 Nm CCW1M M 1200 0.6 562 Nm CCW5
2M M 1200 0.5 1099 Nm CCW5
= × =
= + × × =
= + × × =
Force-Couple equivalent systemmakes the moment calculation intuitive
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Ch. 2: Force Systems
P. 2/14 The combined drive wheels of a front-wheel-drive automobile are acted on by a 7000 Nnormal reaction force and a friction force F,both of which are exerted by the road surface.If it is known that the resultant of these twoforces makes a 15°angle with the vertical,determine the equivalent force-couple systemat the car mass center G. Treat this as a2D problem.
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Ch. 2: Force Systems
P. 2/14
G
Rcos15 = 7000 R = 7246.9 NM 7000 1 7246.9sin15 0.5 7937.8 Nm CW
∴= × + × =
R
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Ch. 2: Force Systems
P. 2/15 Determine and locate the resultant R of thetwo forces and one couple acting on the I-beam.
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Ch. 2: Force Systems
O
R = 8-5 = 3 kN downwardM 25 5 2 8 2 1 kNm CW= − × − × =
P. 2/15 typical step in strength analysis
First, find the equivalent force-couple at point O
Then, locate the correct line of action by prin. of moment13d = 1 d = 1/3 m & x = 4 m3
∴
O
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Ch. 2: Force Systems
P. 2/16 If the resultant of the two forces and couple Mpasses through point O, determine M.
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Ch. 2: Force Systems
P. 2/16 Resultant passes through point Omeans there is no moment at point O
OM M - 400 0.15cos30 -320 0.3 = 0M = 148 Nm CCW
= × ×
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Ch. 2: Force Systems
P. 2/17 The directions of the two thrust vectors of anexperimental aircraft can be independentlychanged from the conventional forward directionwithin limits. For the thrust configuration shown,determine the equivalent force-couple systemat point O. Then replace this force-couplesystem by a single force and specify the pointon the x-axis through which the line of actionof this resultant passes.
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Ch. 2: Force Systems
( ) ( )O
T + Tcos15 Tsin15 1.966T 0.259T NM Tcos15 3- T 3- Tsin15 10 = 2.69T Nm CW= × × ×
R = i + j = i + j P. 2/17
0.259T x = -2.69T x = -10.4 m× ∴
R
MO
Rx
Ry
Since Rx of the new force systemdoes not contribute moment about O,only Ry can be used in calculation.
MO
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Ch. 2: Force Systems
P. 2/18 Two integral pulleys are subjected to the belttensions shown. If the resultant R of theseforces passes through the center O, determineT and the magnitude of R and the CCW angleθit makes with the x-axis.
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Ch. 2: Force Systems
P. 2/18 Resultant force passes O MO = 0
( ) ( )( ) ( )
160 T 100 150 200 200 0 T = 60 N
200 +150 -160cos30 - 60cos30 160sin30 + 60sin30159.5 110 N
R = 193.7 N = 34.6θ °
− × + − × = ∴
R = i + j = i + j
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Ch. 2: Force Systems
P. 2/19 A rear-wheel-drive car is stuck in the snow between otherparked cars as shown. In an attempt to free the car, threestudents exert forces on the car at points A, B, and Cwhile the driver’s actions result in a forward thrust of 200 Nacting parallel to the plane of rotation of each rear wheel.Treating the problem as 2D, determine the equivalentforce-couple system at the car center of mass G and locatethe position x of the point on the car centerline throughwhich the resultant passes. Neglect all forces not shown.
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Ch. 2: Force Systems
( ) ( )
G
200 + 400 + 200 + 250sin30 250cos30 350925 + 566.5 N
M 350 1.65 250sin 30 0.9 690 Nm CCW
+
= × + × =
R = i + j = i j
P. 2/19 • 400 N and y-direction of 250 Ncause no moment about O.• Moments by thrust force 200 Ncancel each other.
566.5 x = 690 x = 1.218 m× ∴
Rx
Ry
Since Rx of the new force systemdoes not contribute moment about G,only Ry can be used in calculation.
MO
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Ch. 2: Force Systems
P. 2/20 An exhaust system for a pickup truck is shown in the figure.The weights Wh, Wm, and Wt of the headpipe, muffler, andtailpipe are 10, 100, and 50 N, respectively, and act at theindicated points. If the exhaust pipe hanger at point A isadjusted so that its tension FA is 50 N, determine the requiredforces in the hangers at points B, C, and D so that theforce-couple system at point O is zero. Why is a zeroforce-couple system at O desirable.
83
Ch. 2: Force Systems
P. 2/20
So the pipe is in equilibrium w/ no external reaction force at support O.Therefore stress at O is zero no breakage
Force-couple at point O is zero force-couple at any point is zero too!At point E,
E
( ) ( ) ( )h m t A B
B
W 0.2 1.3 0.9 W 0.65 0.9 W 0.4 F 1.3 0.9 F 0.9 0F 98.9 N
× + + + × + + × − × + − × =
=
A B C D h m t
D C
C D
F F F cos30 F cos30 W W W 0F sin 30 F sin 30 0F F 6.415 N
+ + + − − − =− =
= =
Force components in horizontal and vertical direction = 0
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Ch. 2: Force Systems
2 2 2x y z x y z
x x y y z z
F x y z
F
F F F F = F F F
F Fcos F Fcos F Fcos
directional unit vector cos cos cos
F
θ θ θ
θ θ θ
+ +
= = =
=
F = i + j+ k
n i + j+ kF = n
2.5 3-D Rectangular Coord. Systems
2.5 3-D Rectangular Coordinate Systems
Force Vector description : directional unit vectorand magnitude
xz y xy z yz xF Fsin F Fsin F Fsinθ θ θ= = =
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Ch. 2: Force Systems
2.5 3-D Rectangular Coord. Systems
Direction of Force Vector by Two Points
( ) ( ) ( )( ) ( ) ( )
2 1 2 1 2 1F 2 2 2
2 1 2 1 2 1
FF
x x y y z zABAB x x y y z z
− − −= =
− + − + −
i + j+ kn
F = n
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Ch. 2: Force Systems
2.5 3-D Rectangular Coord. Systems
Direction of Force Vector by Two Angles
( ) ( ) ( )F
F
F
cos cos cos sin sin
1F
φ θ φ θ φ=
=
n i + j+ k
nF = n
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Ch. 2: Force Systems
F n
2.5 3-D Rectangular Coord. Systems
Orthogonal Projection of the Force Vector may not be equal to its component. They are equal when the rectangular coordinate system is used.
Orthogonal Projection of in the - direction
θ
Fn
( )
F
n F
n n F
F
F F cosF
F F
θ
=
= =
=
F nF nF n = n n
F n = n n n
magnitude = dot product of with F n
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Ch. 2: Force Systems
P. 2/21 In opening a door which is equipped witha heavy duty return mechanism, a personexerts a force P of magnitude 32 N as shown.Force P and the normal n to the face ofthe door lie in a vertical plane. Express P asa vector and determine the angles θx θy θzwhich the line of action P makes with thepositive x-, y-, and z-axes.
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Ch. 2: Force Systems
P. 2/21
x
y
z
Pcos30cos20 Pcos30sin20 Psin3026.0 + 9.48 +16 N
acos 35.5P
acos 72.8P
acos 60P
θ
θ
θ
°
°
°
= =
= =
= =
P = i + j+ k = i j k
P i
P j
P k
top view
20゜
Pxy=Pcos30
angle description
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Ch. 2: Force Systems
P. 2/22 The rectangular plate is supported by hingesalong its side BC and by the cable AE. If thecable tension is 300 N, determine the projectiononto line BC of the force exerted on the plateby the cable. Note that E is the midpoint of thehorizontal upper edge of the structural support.
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Ch. 2: Force Systems
P. 2/22 ( )( )( )( )( )
BC
BC BC
0.4,0,1.2sin 25
0,0,1.2sin 25
0,1.2cos 25,0
0.4,1.2cos 25,0
0,0.6cos 25,0
T 142.1 193.2 180.2
0.9063 0.4226
T 251.2 N
A
B
C
D
E
AEAE
BCBC
= −
=
=
= −
=
= + −
= = −
= =
T = i j k
n j k
T n
Orthogonal projection in a direction: magnitude = dot product
2-point description
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Ch. 2: Force Systems
P. 2/23 The power line is strung from the power-polearm at A to point B on the same horizontalplane. Because of the sag of the cable in thevertical plane, the cable makes an angle of 15°with the horizontal where it attaches to A.If the cable tension at A is 800 N, write T asa vector and determine the magnitude of itsprojection onto the x-z plane.
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Ch. 2: Force Systems
2 2xz x z
y
xz y
1.5atan 8.5310
Tcos15cos + Tcos15sin Tsin15764.2 +114.6 207
T T T 792 N
or acos 81.76T
T Tsin 792 N
θ
θ θ
θ
θ
°
°
=
−−
= + =
= =
= =
=
T = i j k = i j k
T j
θ
P. 2/23
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Ch. 2: Force Systems
2.6 3-D Moment and Couple
2.6 3-D Moment and Couple
Scalar approach in 3-D is more difficult than vector approach
Moment of about point
and normal to the plane and through
= ×⊥ ⊥
O
O
O O
O
M F OM r FM r M FM O
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Ch. 2: Force Systems
2.6 3-D Moment and Couple
Vector Cross Product
( ) ( ) ( )y z z y z x x z x y y x
x y z
x y z
r F r F r F r F r F r F
r r r for remembranceF F F
× − − −r F = i + j+ k
i j k
Proof check & Visualizationby Prin. of Moment
x y z z y
y z x x z
z x y y x
M r F r F
M r F r F
M r F r F
= −
= −
= −
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Ch. 2: Force Systems
2.6 3-D Moment and Couple
Moment of about axis through point F λ
OM
λM
1. Find moment of about point F O= ×OM r F
OMλ
2. Orthogonally project in the-direction along axis n
( ) ( )λ = ×OM M n n = r F n n
x y z
x y z
x y z
r r rM F F F
n n nλ =
* Point O can be any point on axis λ
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Ch. 2: Force Systems
2.6 3-D Moment and Couple
3-D Couple
Couple as free vector
98
Ch. 2: Force Systems
2.6 3-D Moment and Couple
3-D Equivalent Force-Couple System
99
Ch. 2: Force Systems
P. 2/24 The helicopter is drawn here with certain3-D geometry given. During a ground test,a 400 N aerodynamic force is applied tothe tail rotor at P as shown. Determine themoment of this force about point O ofthe airframe.
100
Ch. 2: Force Systems
P. 2/24 Force P does not cause moment in y-direction
( ) ( )O 400 1.2 400 6 480 + 2400 N= × ×M i + k = i k
For this simple force P, we can determinethe moment component-wise
101
Ch. 2: Force Systems
P. 2/25 In picking up a load from position A, a cabletension T of magnitude 21 kN is developed.Calculate the moment that T produces aboutthe base O of the construction crane.
102
Ch. 2: Force Systems
P. 2/25
Vectorial approach
( ) ( )0,18,30 6,13,0
T 4.06 3.39 20.32 kN
18 30 m-264.2 +121.9 - 73.2 kNm
A B =
ABAB
OA
=
= − −
== ×O
T = i j k
r = j+ k M r T = i j k
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Ch. 2: Force Systems
P. 2/25Algebraic approach
( ) ( )
( )
( )
( )
x
y
z
x
T
y
T
z
T
0,18,30 6,13,0
T 4.06 3.39 20.32 kN
translate force to , moment at O by T4.06 13
moment at O by T
3.39 6
moment at O by T20.32 13 20.32 6
264.16 121.92 7
A B =
ABAB
B
=
= − −
= − ×
= − ×
= − × + ×
∴ = − + −
O
O
O
O
T = i j k
M k
M k
M i j
M i j
3.12 kNmk
104
Ch. 2: Force Systems
P. 2/26 The special-purpose milling cutter is subjectedto the force of 1200 N and a couple of 240 Nmas shown. Determine the moment of thissystem about point O.
105
Ch. 2: Force Systems
P. 2/26 MO = moment induced by force + free vector couple
1200cos30 -1200sin30 1039 - 600 N0.2 + 0.25 m
240cos30 - 240sin30 -259.8 + 327.8 +87.8 Nm= ×O
R = j k = j k r = i k M r R + j k = i j k
106
Ch. 2: Force Systems
P. 2/27 A 5 N vertical force is applied to the knobof the window-opener mechanism when thecrank BC is horizontal. Determine the momentof the force about point A and about line AB.
107
Ch. 2: Force Systems
P. 2/27
( )
( )
75cos30 + 75 + 75sin30 mm-5 375 325 Nmm
cos30 sin 30281 162.4 Nmm
= × = − +
= +
= = − −
A
AB
AB A AB AB
r = i j kM r k i jn i kM M n n i k
108
Ch. 2: Force Systems
and R MR
2.7 3-D Resultants
2.7 3-D Resultants
Resultant is the simplest force combination which can replace the original system of forces, moments, and couples without altering the external effect of the system on the rigid body.
Vectorial approach is more suitable in 3-D problems.1. Define the suitable rectangular coord. System and
specify a convenient point O2. Move all forces so the new lines of action pass
through point O force-couple equivalence3. Sum forces and couples to 4. Locate the correct line of action of force-couple equivalence solving piercing point
(2 unknowns: ) rank-2 degenerated×r R = M
109
Ch. 2: Force Systems
( )go together to determine the resultant
Principle of Moment
×
∑∑
R = F
M = r F
M
2.7 3-D Resultants
The selected point O specifies the couple
Dynamics: calculate the resultantsat C.M.G
m
I θ
=
=
∑∑
G
G
F x
M
Statics: calculate the resultants at any point0
0
=
=∑∑
F
M
110
Ch. 2: Force Systems
( ) 0× =∑ ∑R = F M = r F
= F = = ×∑ ∑OR M M r R
R
2.7 3-D Resultants
Resultants of Special Force Systems
Concurrent Forces No moment about the pointof concurrency
Parallel Forces Magnitude of = magnitude ofalgebraic sum of the given forces
Wrench Resultant as the resultantof screwdriver
R M
111
Ch. 2: Force Systems
2.7 3-D Resultants
Wrench Resultant – Force-Couple Equivalence
a) Determine the force-couple resultantat convenient point O
b) Orthogonally project along and perp. to
c) Transform couple into equiv. pair of with applied at O to cancel
d) Resultant with correct line of action andremains wrench resultant
and R M
M Rn
R
and -R R
( ) = = =R 1 R R 2 1Rn M M n n M M - MR
2M-R R
1M R
Wrench resultant is the simplest form to visualizethe effect of general force system on to the object :translate and rotate about the unique axis – screw axis
112
Ch. 2: Force Systems
2.7 3-D Resultants
axis of the wrench, which is , lies in a planethrough O and plane defined by⊥
R and R M
113
Ch. 2: Force Systems
P. 2/28 The pulley and gear are subjected to the loadsshown. For these forces, determine theequivalent force-couple system at point O.
114
Ch. 2: Force Systems
P. 2/28
( )800 + 200 -1200sin10 1200cos10792 +1182 N
800 N : 800 0.55 -800 0.1200 N : 200 0.55 200 0.11200 N : 1200sin10 0.22 1200cos10 0.075 1200cos10 0.22
260 504 28.6 Nm
+
= − × ×= − × ×= × × ×
= + = − +
1
2
3
O 1 2 3
R = i j = i j
M j kM j+ kM j+ k + i
M M + M M i j k
typical problem in shaft analysis
115
Ch. 2: Force Systems
P. 2/29 Two upward loads are exerted on the small 3Dtruss. Reduce these two loads to a singleforce-couple system at point O. Show that Ris perpendicular to Mo. Then determine thepoint in the x-z plane through which theresultant passes.
116
Ch. 2: Force Systems
P. 2/29
2400 N800 2.4 1600 2.4 1600 0.91440 + 5760 Nm
determine line of action of must be x m far from yz plane to produce 5760 Nm
5760 = 2400 x x = 2.4 m must be - z m far from xy plane to
= × × ×
× ∴
O
R = jM k + k + i = i k
RR k
R( )
produce 1440 Nm1440 = 2400 -z z = -0.6 m× ∴
i
117
Ch. 2: Force Systems
P. 2/30 Replace the two forces acting on the blockby a wrench. Write the moment M associatedwith the wrench as a vector and specify thecoordinates of the point P in the x-y planethrough which the line of action of the wrenchpasses.
118
Ch. 2: Force Systems
( )F F
F a + c Fb=O
R = i - kM j- k
P. 2/30 a) Determine force-couple resultant at O
b) Project MO || and ┴ nR
( )
( )
1 12 2
Fb Fb2 2
Fb FbF a + c2 2⊥
=
= = −
= − = − −
R
O R R
O
n i - k
M M n n i k
M M M i + j k
119
Ch. 2: Force Systems
x y
bx=a+c, y=2
⊥ = ×r = i + jM r R
P. 2/30c) Transform couple M┴ into pair of force R and –R
If r points to the piercing point of the xy plane,
d) Wrench consisting of and actsb through xy plane at x = a+c, y = 2
R M
120
Ch. 2: Force Systems
P. 2/31 The resultant of the two forces and couplemay be represented by a wrench. Determinethe vector expression for the moment M ofthe wrench and find the coordinates of the pointP in the x-z plane through which the resultantforce of the wrench passes.
121
Ch. 2: Force Systems
P. 2/31
( )( ) ( )
( ) ( )
100 100 NLet point P in xz plane, where the wrench passes, has the coordinate x, 0, z .
Moment about point P = 100 z 100 0.4 x 100 0.4 z 100 0.3 20
100z 20-100z 10-100x NmThis moment at po
× × − × − ×
=P
R = i + j
i + k + j- k - j
M i + j+ k int P must equal to the couple of the wrench passing through point P.
And since it is the wrench, .x = 0.1 m, z = 0.1 m
10 10 Nm∴
= +
P
P
M R
M i j
122