4.4 principles of moments also known as varignon’s theorem “moment of a force about a point is...
TRANSCRIPT
![Page 1: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/1.jpg)
4.4 Principles of Moments4.4 Principles of Moments
Also known as Varignon’s Theorem“Moment of a force about a point is equal to the sum of the moments of the forces’ components about the point”
For F = F1 + F2,
MO = r X F1 + r X F2
= r X (F1 + F2)
= r X F
![Page 2: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/2.jpg)
4.4 Principles of Moments4.4 Principles of Moments
The guy cable exerts a force F on the pole and creates a moment about the base at A
MA = Fd If the force is replaced
by Fx and Fy at point B where the cable acts on the pole, the sum of moment about point A yields the same resultant moment
![Page 3: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/3.jpg)
4.4 Principles of Moments4.4 Principles of Moments
Fy create zero moment about A
MA = Fxh Apply principle of
transmissibility and slide the force where line of action intersects the ground at C, Fx create zero moment about A
MA = Fyb
![Page 4: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/4.jpg)
4.4 Principles of Moments4.4 Principles of Moments
Example 4.6A 200N force acts on the bracket.
Determine the moment of the force about point A.
![Page 5: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/5.jpg)
4.4 Principles of Moments4.4 Principles of Moments
SolutionMethod 1:From trigonometry using triangle BCD,
CB = d = 100cos45° = 70.71mm = 0.07071m
Thus, MA = Fd = 200N(0.07071m)
= 14.1N.m (CCW)As a Cartesian vector,
MA = {14.1k}N.m
![Page 6: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/6.jpg)
4.4 Principles of Moments4.4 Principles of Moments
SolutionMethod 2: Resolve 200N force into x and y components Principle of Moments
MA = ∑Fd
MA = (200sin45°N)(0.20m) – (200cos45°)(0.10m) = 14.1 N.m (CCW)
Thus, MA = {14.1k}N.m
![Page 7: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/7.jpg)
4.4 Principles of Moments4.4 Principles of Moments
Example 4.7The force F acts at the end of the angle bracket. Determine the moment of the
force about point O.
![Page 8: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/8.jpg)
4.4 Principles of Moments4.4 Principles of Moments
SolutionMethod 1MO = 400sin30°N(0.2m)-400cos30°N(0.4m)
= -98.6N.m = 98.6N.m (CCW)
As a Cartesian vector, MO = {-98.6k}N.m
View Free Body Diagram
![Page 9: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/9.jpg)
4.4 Principles of Moments4.4 Principles of Moments
SolutionMethod 2: Express as Cartesian vector
r = {0.4i – 0.2j}NF = {400sin30°i – 400cos30°j}N = {200.0i – 346.4j}N
For moment,
mNk
kji
FXrMO
.6.98
04.3460.200
02.04.0
![Page 10: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/10.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
For moment of a force about a point, the moment and its axis is always perpendicular to the plane containing the force and the moment arm
A scalar or vector analysis is used to find the component of the moment along a specified axis that passes through the point
![Page 11: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/11.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Scalar AnalysisExample Consider the pipe assembly that lies in the
horizontal plane and is subjected to the vertical force of F = 20N applied at point A.
For magnitude of moment, MO = (20N)(0.5m) = 10N.m
For direction of moment, apply right hand rule
![Page 12: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/12.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Scalar Analysis Moment tends to turn pipe about the Ob
axis Determine the component of MO about
the y axis, My since this component tend to unscrew the pipe from the flange at O
For magnitude of My,
My = 3/5(10N.m) = 6N.m
For direction of My, apply right hand rule
![Page 13: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/13.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Scalar Analysis If the line of action of a force F is
perpendicular to any specified axis aa, for the magnitude of the moment of F about the axis,
Ma = Fda
where da is the perpendicular or shortest distance from the force line of action to the axis
A force will not contribute a moment about a specified axis if the force line of action is parallel or passes through the axis
![Page 14: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/14.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
A horizontal force F applied to the handle of the flex-headed wrench, tends to turn the socket at A about the z-axis
Effect is caused by the moment of F about the z-axis
Maximum moment occurs when the wrench is in the horizontal plane so that full leverage from the handle is achieved
(MZ)max = Fd
![Page 15: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/15.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
For handle not in the horizontal position,
(MZ)max = Fd’where d’ is the perpendicular distance from the force line of action to the axis
Otherwise, for moment, MA = Fd
MZ = MAcosθ
![Page 16: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/16.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Vector AnalysisExampleMO = rA X F = (0.3i +0.4j) X (-20k)
= {-8i + 6j}N.mSince unit vector for this axis is ua = j,
My = MO.ua
= (-8i + 6j)·j = 6N.m
![Page 17: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/17.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Vector Analysis Consider body subjected to
force F acting at point A To determine moment, Ma,
- For moment of F about any arbitrary point O that lies on the aa’ axis
MO = r X F where r is directed from O to A- MO acts along the moment axis bb’, so projected MO on the aa’ axis is MA
![Page 18: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/18.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Vector Analysis- For magnitude of MA,
MA = MOcosθ = MO·ua
where ua is a unit vector that defines the direction of aa’ axis
MA = ua·(r X F)
- In determinant form,
zyx
zyxazayaxa
FFF
rrr
kji
kujuiuM
)(
![Page 19: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/19.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Vector Analysis- Or expressed as,
where uax, uay, uaz represent the x, y, z components of the unit vector defining the direction of aa’ axis and rx, ry, rz represent that of the position vector drawn from any point O on the aa’ axis and Fx, Fy, Fz represent that of the force vector
zyx
zyx
azayax
axa
FFF
rrr
uuu
FXruM )(
![Page 20: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/20.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Vector Analysis The sign of the scalar indicates the
direction of Ma along the aa’ axis
- If positive, Ma has the same sense as ua
- If negative, Ma act opposite to ua Express Ma as a Cartesian vector,
Ma = Maua = [ua·(r X F)] ua
For magnitude of Ma,
Ma = ∑[ua·(r X F)] = ua·∑(r X F)
![Page 21: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/21.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Wind blowing on the sign causes a resultant force F that tends to tip the sign over due to moment MA created about the aa axis
MA = r X F For magnitude of projection
of moment along the axis whose direction is defined by unit vector uA is
MA = ua·(r X F)
![Page 22: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/22.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Example 4.8 The force F = {-40i + 20j + 10k} N acts on the
point A. Determine the moments of this force about the x and a axes.
![Page 23: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/23.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
SolutionMethod 1
Negative sign indicates that sense of Mx is opposite to i
mN
FXriM
iu
mkjir
Ax
x
A
.80
102040
643
001
)(
}643{
![Page 24: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/24.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
SolutionWe can also compute Ma using rA as rA
extends from a point on the a axis to the force
mN
FXruM
jiu
AAa
A
.120
102040
643
054
53
)(
54
53
![Page 25: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/25.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
SolutionMethod 2 Only 10N and 20N forces contribute moments
about the x axis Line of action of the 40N is parallel to this axis
and thus, moment = 0 Using right hand rule
Mx = (10N)(4m) – (20N)(6m) = -80N.m
My = (10N)(3m) – (40N)(6m) = -210N.m
Mz = (40N)(4m) – (20N)(3m) = 100N.m
![Page 26: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/26.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Solution
![Page 27: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/27.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Example 4.9 The rod is supported by two brackets at A and
B. Determine the moment MAB produced by F = {-600i + 200j – 300k}N, which tends to
rotate the rod about the AB axis.
![Page 28: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/28.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Solution Vector analysis chosen as moment arm
from line of action of F to the AB axis is hard to determine
For unit vector defining direction of AB axis of the rod,
For simplicity, choose rD
ji
jirr
u
FXruM
B
BB
BAB
447.0894.0
)2.0()4.0(
2.04.0
)(
22
View Free Body Diagram
![Page 29: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/29.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Solution For force,
In determinant form,
Negative sign indicates MAB is opposite to uB
mN
FXruM
NkjiF
DBAB
.67.53
300200600
02.00
0447.0894.0
)(
}300200600{
![Page 30: 4.4 Principles of Moments Also known as Varignon’s Theorem “Moment of a force about a point is equal to the sum of the moments of the forces’ components](https://reader035.vdocument.in/reader035/viewer/2022062720/56649f115503460f94c23af9/html5/thumbnails/30.jpg)
4.5 Moment of a Force about a Specified Axis
4.5 Moment of a Force about a Specified Axis
Solution In Cartesian form,
*Note: if axis AB is defined using unit vector directed from B towards A, the above formulation –uB should be used.
MAB = MAB(-uB)
mNji
jimNuMM BABAB
.}0.240.48{
)447.0894.0)(.67.53(