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MOMENTS
Created by The North Carolina School of Science and Math. Copyright 2012. North Carolina Department of Public Instruction.
If our parrot isn’t sitting in the center, how do I calculate T1 and T2?
W
T2T1
We need another equation!
Creative commons image from: http://upload.wikimedia.org/wikipedia/commons/thumb/0/02/Amazona_albifrons_-in_tree2-3c.jpg/444px-Amazona_albifrons_-in_tree2-3c.jpg
Static Equilibrium
• If a system is in static equilibrium– Σ Fx = 0
– Σ Fy = 0
and …– Σ M = 0
• Just like with forces,moments havedirection.
Creative commons image from: http://upload.wikimedia.org/wikipedia/commons/thumb/0/02/Amazona_albifrons_-in_tree2-3c.jpg/444px-Amazona_albifrons_-in_tree2-3c.jpg
Moments• A moment of a force is a measure of
its tendency to cause a body to rotate about a point or axis.
• A moment (M) is calculated using the formula:
Moment = Force * Distance
M = F * D
Always use the perpendicular distance between the force and the point!
Moment = Force * Perpendicular Distance M = F * D
What distance would you use here?
L
F = 20 lb
L = 2 ft
Find the moment about point O:
M = F * D
M = (20 lb) * (2 ft)
M = 40 ft-lb clockwise
Moments
Moment = Force * Perpendicular Distance
M = F * D
Can you visualize the result of the force acting on the beam?
The beam has a tendency to rotate clockwise about point O.
Moments
Moment = Force * Distance
M = F * D
w
L
F = 20 lbf
L = 0.5 ft
Find the moment about point O:
M = F * D
M = (20 lb) * (0.5 ft)
M = 10 ft-lb clockwise
Moments
Moment = Force * Distance
M = F * D
Can you visualize the result of the force acting on the beam?
The beam has a tendency to rotate clockwise about point O.
Moments
Moment = Force * Perpendicular Distance M = F * D What distance would you use here?
L
Moments – Direct Method
d'
M = F * d’Use trigonometry to find d’
θ
1 2
θ d‘ = L cos(θ)
M = F * L cos(θ)
Moment = Force * Perpendicular Distance M = F * D Alternative Method …
L
Moments – Component Method
M = Fx * Dx + Fy * Dyθ
Let’s find the components
Fx = F sin(θ) Fy = F cos(θ)
Dx = 0 Dy = L
M = Fx * Dx + Fy * Dy
M = F sin(θ) * 0 + F cos(θ) * L
M = F * L cos(θ) (same as before)
Fx
Fy
Advantages of Component Method
• Use static equilibrium equations directly to find reaction forces.
• Find reaction forces for the beam in the prior slide.– Σ Fx = 0: Fpx – Fx = 0 -> Fpx = Fx = F sin(θ)
– Σ Fy = 0: Fpy + Fy = 0 -> Fpy = - F cos(θ)
– Σ M = 0: Mp + Fy L + Fx 0 = 0 -> Mp = - F L cos(θ)
L θ
Fx
Fy
𝐹𝑃𝑋
𝐹𝑃𝑌
𝑀𝑃
HOW TO SOLVE A MOMENT PROBLEM
• Draw a free body diagram that shows all forces.
• List the knowns and unknowns• Use the following three formulas to determine
unknowns (system is in static equilibrium)– Σ Fx = 0
– Σ Fy = 0
– Σ M = 0
Moment Problem, example
• Use static equilibrium equations directly to find reaction forces.
• Find reaction forces for the beam in the prior slide.– Σ Fx = 0: Fpx – Fx = 0 -> Fpx = Fx = F sin(θ)
– Σ Fy = 0: Fpy – Fy = 0 -> Fpy = - F cos(θ)
– Σ M = 0: Mp – Fy L + Fx 0 = 0 -> Mp = - F L cos(θ) (CCW)
L θ
Fx
Fy
𝐹𝑃𝑋
𝐹𝑃𝑌
𝑀𝑃
If our parrot is not sitting in the center, how do I calculate T1 and T2?
W
T2T1
Parrot on the Perch: Revisited
41
𝑾 −𝟓∗𝑻 𝟐=𝟎𝟓∗𝑻 𝟐=𝑾𝑻 𝟐=
𝑾𝟓
Σ M = 0 (about where?):
Σ Fy = 0
𝑻 𝟏+𝑻𝟐=𝑾 𝑻 𝟏=𝟒𝑾𝟓
• The sum of moments is zero around any point!• Choose any location, but the same for the whole problem.• Some places are easier than others.
W
T2T1
Where to take the moment
41
𝑻 𝟏∗𝟏−𝑻𝟐∗𝟒−𝟎∗𝑾=𝟎Σ M = 0 (this time about W)
Σ Fy = 0𝑻 𝟏+𝑻𝟐=𝑾
𝑻 𝟏−𝟒𝑻𝟐=𝟎𝑻 𝟏=𝟒𝑻𝟐
Solve:𝟒𝑻 𝟐+𝑻𝟐=𝑾 5
𝑻 𝟐=𝑾𝟓
Supports are translated into forces and moments in a free body diagrams. The following are three common supports and the forces and moments used to replace them.
Roller:
Pin Connection:
Fixed Support:
Fy
Fy
Fx
Fx
Fy
Mo
Review: Types of Supports
8kN 9kN
5kN compressed spring
ΣFX = 0, therefore RAx = 0
ΣFy = 0 = RAy + RB - 8kN + 5kN - 9 kN
RAy + RB = 12kN -> RAy = 12kN - RB
ΣMA = 0 = -8kN·2m + 5kN·4m + RB·5m - 9kN·7m
(16 - 20 + 63) kN-m = RB·5m
RB = 11.8 kN
Put into ΣFy result: RAy = 22kN-11.8 kN = 10.2 kN
RAx
RAy RB
8kN 5kN9kN
2m 2m 1m 2m
2m 2m 1m 2m
C
A
B30°
30°
3m
2m
3kN
The rod AC is hinged at point A and is suspended by a cable (BD) at Point B. A weight of 3kN hangs off the rod at Point C. Calculate the reaction forces at Point A and the tension in the cable BD.
D
C
A
B30°
30°
3m
2m
3kN
RAy
RAx
FBy
FBx
5 sin(30)=2.5m
3 sin(30)= 1.5m
3 cos(30)=2.6m
5 cos(30) = 4.3m
D
FreeBodyDiagram
3kN
RAy
RAx
FBy
FBx
5 sin(30)=2.5m
3 sin(30)= 1.5m
3 cos(30)=2.6m
5 cos(30) = 4.3m
FBy = Fb sin(30°) = 0.5 FB
FBx = Fbcos(30°) = 0.87 FB
ΣMA = 0 = -3kN·4.3m + FBy·2.6m + FBx·1.5m
12.9kN-m = 0.5 FB·2.6m + 0.87 FB·1.5m
FB = 4.95 kN
3kN
RAy
RAx
FBy
FBx
5 sin(30)=2.5m
3 sin(30)= 1.5m
3 cos(30)=2.6m
5 cos(30) = 4.3m
FB = 4.95 kN
FBy = Fb sin(30°) = 0.5 FB = 0.50*4.95 kN = 2.5 kN FBx = Fbcos(30°) = 0.87FB= 0.87*4.95 kN = 4.3 kN
ΣFx = 0 = RAx – FBx Rax = FBx = 4.3 kN ΣFy = 0 = RAy + FBy - 3kN
Ray = FBx = 3kN – Fby= 3kN – 2.5kN = 0.5 kN