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Tutorial 1 MECH 101
Liang [email protected]
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Office hour : 14:00-15:00 Fri
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Outline
Cartesian vector calculation(+ • ×)
2D forces equilibrium problemMoment calculationReduction of force system
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Express a Force as a Cartesian VectorMagnitude + direction (daily life) a Cartesian Vector (statics analysis)
cos(60) 200 0.5 100( )
cos(45) 200 0.707 141.4( )
cos( ) ?
Fy F N
Fz F N
Fx F
45۫
60۫
F=200N
z
y
x
Fz
FyFx
2 2 2 2
2 2 2
2 2 2 2
2 2 2
0
[ cos(60)] [ cos(45)] [ cos( )]
[cos(60) cos(45) cos( ) ]
cos(60) cos(45) cos( ) 1
60
F Fx Fy Fz
F F F
F
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Exercise 1+ : resultant force F1=3 i-5 j+7 k , F2=-9 i+4 j-3 k F1+F2=? • : Projection F1=3 i-5 j+7 k, e =0.6 j-0.8 k F1 • e=? ×: Moment F1=3 i-5 j+7 k, r =-9 i+4 j-3 k r×F1=?
-6 i-1 j+4 k
-8.6
( ) ( )
Pay attention to the order of the vector.
9 4 3 13 54 33
3 5 7
x y z x y z x y z
x y z
i j k
C A i A j A k B i B j B k A A A
B B B
i j k
r F i j k
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2D forces equilibrium problem
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W
EAT
BAT
SolutionStep 1: Draw the Free-Body Diagram
Isolating part of the cords system near point A , add the force.
W=mg=(30kg) (10.0m/s2 ) =300N, TBA,TEA are unknowns.
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Step2Select a coordinate system, and resolve the cord tensions into
x and y components. cos 60oBAx BAT T
sin 60oBAy BAT T
0EAyT EAx EAT T
0xW
300yW N
SolutionSolution
W
EAT
BAT
x
y
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Step3Apply the Equilibrium Equation
cos 60 0ox EA BAF T T
sin 60 300 0oy BAF T
200 3 346.4BAT N N 100 3 173.2EAT N N
Step4Step4
Solving these two equations, we find:Solving these two equations, we find:
SolutionSolution
x
W
EAT
BAT
y
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Objects in Equilibrium
An particle is in equilibrium
ΣF = 0 2D Rx=0,Ry=0 2 Equations, 2 Unknowns
Why I don’t choose point B to analyze first?
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practice
x
y
04cos 60 0
5x CB DB ABF T T T
3sin 60 0
4o
y DB ABF T T
500 , 573.2DB CBT N T N
200 3 346.4BAT N N
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Review the Steps1. Draw the Free-Body Diagram
2. Select a coordinate system and find the x and y
components of every force
3. Apply the Equilibrium Equations
4. Solve the equations.
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Exercise 2
dx
dyN
3.58 , 19.7Bm kg N N
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Exercise 3
cos( ) 5 10
BC
AB x
Fx
k
How many unkowns?
040.2 13
Find out the third equation:
Objects in EquilibriumAn object is in equilibrium ΣF = 0 All the forces pass the same point. (particle)
ΣF = 0 An object is in equilibrium
If the forces don’t pass the same point, we need: ΣF = 0 & ΣM = 0 An object is in equilibrium
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MomentThe moment of a force about a point or an
axis: a measure of the tendency of the force to rotate a body about that point or the axis.
moment about a point ----A ball on the sea & The earth
Moment about an axis ----The door & the handle
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Moment about a point M r F (general method) & (good for 3D)
(don’t exchange r and F)
oF
d
r
If you can find the Moment arm d, it’s easier to use M=F*d. Especially in 2D problem, the direction of M is obvious.
o
F
If F passes o, the moment of F about O is zero.
ceBABAC sin
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Example A 200-N force acts on the bracket shown in fig (a).Determine the moment of the force about point A
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Which method is the best?
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Moment about an axis
l l l M r F
What kind of force can roll the handle?
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If the force is in the same plane with the axis, it doesn’t cause moment about the axis
Force system reduction3 forces =>1 force to satisfy the mechanical effect is the same.
1. Resultant force should be the same2. The resultant moment about an arbitrary point should be the same
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Resultant force
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Resultant moment
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Think about:Force is an vector. Can this vector move
arbitrarily?Can a couple move around?What’s the mechanical effect of a force?What’s the mechanical effect of a couple?
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Reduce the distributed force
Where is the centroid of the area of the load diagram?
How to calculate the equivalent force?
How to calculate the equivalent force? And where dose this concentrated force locate?
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Example
qL
xL
q
xdxL
q
dxxfF
L
L
L
2
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1
)(
0
2
0
0
2
0
3
0
0
3
13
1
)(.
qL
xL
q
xxdxL
q
xdxxfdF
L
L
L
LqL
qLd
3
2
2131 2
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Thank You!
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