Tutorial 1
MECH 101
Liang Tengfei
tfliang@ust.hk
Office phone : 2358-8811
Mobile : 6497-0191
Office hour : 14:00-15:00 Fri
1
Outline
 Cartesian vector calculation(+ • ×)
 2D forces equilibrium problem
 Moment calculation
 Reduction of force system
2
Express a Force as a Cartesian Vector
 Magnitude + direction (daily life)
a Cartesian Vector (statics analysis)
Fy  F cos(60)  200  0.5  100( N )
Fz  F cos(45)  200  0.707  141.4( N )
Fx  F cos( )  ?
z
Fz
F=200N
F 2  Fx 2  Fy 2  Fz 2
45۫

Fx
60۫
 [ F cos(60)]2  [ F cos(45)]2  [ F cos( )]2
Fy
 F 2 [cos(60)2  cos(45)2  cos( )2 ]
y
x
 cos(60) 2  cos(45)2  cos( )2  1
   600
3
Exercise 1
 + : resultant force
F1=3 i-5 j+7 k , F2=-9 i+4 j-3 k F1+F2=?-6 i-1 j+4 k
 • : Projection
F1=3 i-5 j+7 k, e =0.6 j-0.8 k F1 • e=?-8.6
 ×: Moment
F1=3 i-5 j+7 k, r =-9 i+4 j-3 k r×F1=?
i
j
k
C  ( Ax i  Ay j  Az k )  ( Bxi  By j  Bz k )  Ax
Ay
Az
Bx
By
Bz
Pay attention to the order of the vector.
i
j
k
r  F  9
4
3  13i  54 j  33k
3
5
7
4
2D forces equilibrium problem
5
Solution
Step 1: Draw the Free-Body Diagram
TBA
TEA
Isolating part of the cords system
near point A , add the force.
W=mg=(30kg) (10.0m/s2 ) =300N,
TBA,TEA are unknowns.
W
6
Solution
Step2
Select a coordinate system, and resolve the cord tensions into
x and y components.
y
TBAx  TBA cos60o
TBAy  TBA sin 60o
TBA
TEAx  TEA
TEA
x
TEAy  0
Wx  0
W
Wy  300N
7
Solution
Step3
Apply the Equilibrium Equation
y
F  T
F  T
x
TBA
TEA
x
W
y
EA
 TBA cos60  0
o
o
sin
60
 300  0
BA
Step4
Solving these two equations, we find:
TBA  200 3N  346.4N
TEA  100 3N  173.2N
8
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?
9
practice
y
F
x
 TCB  TDB
 Fy  TDB
4
 TAB cos 600  0
5
3
 TAB sin 60o  0
4
x
 TDB  500N , TCB  573.2N
TBA  200 3N  346.4N
10
Review the Steps
1.
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.
11
Exercise 2
N
dy
dx
mB  3.58kg, N  19.7 N
12
Exercise 3
How many unkowns?
Find out the third equation:
AB cos( )  x  5  10
FBC
x 
k
  40.2
0
13
Objects in Equilibrium
An 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
14
Moment
 The 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
15
Moment about a point
M  r F
o
r
(general method) & (good for 3D)
(don’t exchange r and F)
F
F
o
d
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.
  

C  A B  A  B  sin ec
If F passes o, the moment of F about O is zero.
16
Example
A 200-N force acts on the bracket shown in fig (a).Determine the
moment of the force about point A
17
18
19
Which method is the best?
20
Moment about an axis
What kind of force
can roll the handle?
Ml   r  F   l  l
If the force is in the same plane
with the axis, it doesn’t cause
moment about the axis
21
Force system reduction
3 forces =>1 force
to satisfy the mechanical effect is the same.
1. Resultant force should be the same
2. The resultant moment about an arbitrary point should be the same
22
Resultant force
23
Resultant moment
24
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?
25
Reduce the distributed force
Where is the centroid of the How to calculate the equivalent force?
area of the load diagram?
How to calculate the equivalent force? And where dose this
concentrated force locate?
26
Example
L
L
F 

f ( x)d x
F .d   f ( x) xdx
0
0
L


0
q
xd x
L
q 1 2

x
L 2
1
 qL
2
L
0
L

0
q
xxdx
L
q1 3L

x
L3 0
1
 qL2
3
1
qL2
2
d  3
 L
1
3
qL
2
27
Thank You!
28