Malaysian Institute of Aviation Technology
APPLIED SOLID MECHANICS
STATICS
Chapter 2 – Static of Particles
2D – Equilibrium
Mulia Minhat – UniKL MIAT 09
1
Malaysian Institute of Aviation Technology
OBJECTIVE
• Objectives:
– Solve the static of 2-D particle equilibrium condition.
2
Malaysian Institute of Aviation Technology
CASE STUDY 1
In a ship-unloading operation, a
3500-lb automobile is supported
by a cable. A rope is tied to the
cable and pulled to center the
automobile over its intended
position. What is the tension in
the rope and cable AB?
3
Malaysian Institute of Aviation Technology
APPROACH
- Construct a free-body diagram for the particle at the junction of the
rope and cable.
- Resolve all forces in the FBD into x and y component as per chosen
coordinate system.
- Express the condition for particle equilibrium by summing all forces
in x & y directions equals to zero.
- Solve the equilibrium equations which contains the unknowns.
4
Malaysian Institute of Aviation Technology
THEORY
- Free-body diagram (FBD): a diagram which focus on specific body
which can be idealized as particle or rigid body and is in equilibrium
condition. In FBD, all relevant forces are identified to maintain its
equilibrium conditions where all summation of forces with respect to
the chosen coordinate system is equal to zero.
B
Free-body diagram – particle A
FAB
A
C
A
y
D
10 Ib
FAC
W = 10 Ib
x
5
Malaysian Institute of Aviation Technology
THEORY
- Typical physical connections in the FBD of particle system:
Cable and pulley / Rope
Spring
θ
lo = 0.4 m
s = - 0.15 m
T
T
F = k(xf-xi)
s = 0.3 m
6
Malaysian Institute of Aviation Technology
THEORY
- Equilibrium - summation of forces equal to zero
- For 2D particle equilibrium:
F 0
where
F
x
 0 and
F
y
0
7
Malaysian Institute of Aviation Technology
SOLUTION
- Draw FBD and resolve all forces into x & y components
FAB
y
FAB y
x
3O
FAB x  FAB sin 30  0.0523FAB
FAB y  FAB cos 30  0.9986 FAB
FAB x
A
30O
FAC x
FAC y
FAC
FAC x  FAC cos 300  0.866 FAC
FAC y  FAC sin 300  0.5 FAC
3500 Ib
8
Malaysian Institute of Aviation Technology
SOLUTION
- Apply equilibrium equations for the FBD
FAB
F
FAB y
x
 0   ve
FAC x  FAB x  0
3O
0.866 FAC  0.0523 FAB  0  (1)
FAB x
A
30O
FAC x
FAC y
y
FAC
3500 Ib
F
 0   ve
FAB y  FAC y  3500  0
0.9986 FAB  0.5 FAC  3500  0  (2)
9
Malaysian Institute of Aviation Technology
SOLUTION
3624.2 Ib
- Solve two simultaneous equations.
A
From eq. (1) :
FAB 
0.866 FAC
0.0523
218.27 Ib
3500 Ib
Substitude in eq. (2)
 0.866 FAC 
0.9986
  0.5 FAC  3500  0
0
.
0523


FAC  218.27 Ib
Then substitude FAC into eq. (1) again
0.866 218.27 
0.0523
 3624.2 Ib
FAB 
FAB
10
Malaysian Institute of Aviation Technology
CASE STUDY 2
It is desired to determine the drag force at a given speed on a prototype
sailboat hull. A model is placed in a test channel and three cables are
used to align its bow on the channel centerline. For a given speed, the
tension is 40 lb in cable AB and 60 lb in cable AE.
Determine the drag force exerted on the hull and the tension in cable
AC.
11