Malaysian Institute of Aviation Technology
APPLIED SOLID MECHANICS STATICS Chapter 2 – Static of Particles 2D – Equilibrium Mulia Minhat – UniKL MIAT 09
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Malaysian Institute of Aviation Technology
OBJECTIVE • Objectives: – Solve the static of 2-D particle equilibrium condition.
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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?
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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.
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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
D 10 Ib
FAC y
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
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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
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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 y = FAB cos 30 = 0.9986 FAB
A FAC y
FAB x = FAB sin 30 = 0.0523FAB
30O
FAC x 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
FAC x − FAB x = 0
3O
FAB x
0.866 FAC − 0.0523 FAB = 0 K (1)
A FAC y
30O
FAC x FAC
3500 Ib
= 0 → + ve
∑F
y
= 0 ↑ + ve
FAB y − FAC y − 3500 = 0 0.9986 FAB − 0.5 FAC − 3500 = 0 K (2) 9
Malaysian Institute of Aviation Technology
SOLUTION 3624.2 Ib
- Solve two simultaneous equations. A
From eq. (1) : 0.866 FAC FAB = 0.0523
218.27 Ib 3500 Ib
Substitude in eq. (2) ⎛ 0.866 FAC 0.9986⎜ ⎝ 0.0523 FAC = 218.27 Ib
⎞ ⎟ − 0.5 FAC − 3500 = 0 ⎠
Then substitude FAC into eq. (1) again 0.866 (218.27 ) 0.0523 = 3624.2 Ib
FAB = FAB
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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