Presentation of Fraction II

University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)







Fraction

Undergraduate Level, 1st Stage



Mr. Waleed Ali Tameemi
College of Engineering/ Babylon University
M.Sc. Civil Engineering/ the University of Kansas/ USA



2016-2017

Course Outline
Introduction
Parallelogram Law
Forces Resolution
Resultant of a Cunccurent, Coplanar Force System
Moment
Couples
Resultant of a Non-concurrent, Coplanar Force System
Resultant of a Concurrent Non-coplanar Force System
Equilibrium
Fraction
Truss
Method of Joints
Method of Sections







Fraction
Fraction or fractional force can be define as the force that tend to oppose the movement or the tendency of movement of a rigid body on a non-smooth surface. When the external force on a body is larger than the fractional force, the body will move (Limiting Fraction). Whereas, when the applied force is less than the fractional force, the body remains at rest (Static Fraction)

Where:
W: the weight of the rigid body,
P: the external force,
F: the fraction Force,
N: the surface reaction.

Law of Fraction
Experiments show that there is a relationship between the fractional force (F) and the reaction force (N) and as follows:
Coefficient of fraction ?=F/N
The direction of the fractional force is always opposite to the direction of movement.
The magnitude of the fractional force depends on the roughness/smoothness of the surface. As the roughness of the surface increase the fraction increase and as the smoothness of the surface increase the fraction decrease.

Example 1
If the body of weight (W= 200 N) shown below is resting on a rough surface with coefficient of fraction (?=0.25), calculate:
The normal reaction (N)
The fractional force (F)
The Maximum pulling force (P) keeps the body in equilibrium.



Solution:
1. The normal reaction (N)
??F_y =0 ? (+ve.)
N-W=0
N=W=200 N ? (1)

2. The fractional force (F)
By using the fraction Law:
Coefficient of fraction ?=F/N ?0.25=F/200
F=200×0.25=50 N ?
?
3. The Maximum pulling force (P) keeps the body in equilibrium
??F_x =0 ? (+ve.)
P-F=0
P=50 N

Example 2
If the body (weight W= 500 N) shown below is resting on an inclined (with angle ?=30°) rough surface with coefficient of fraction (?=0.25), calculate:
The reaction (N) that is normal to the inclined surface.
The fractional force (F).
The Maximum pulling force (P) keeps the body in equilibrium.





Solution:
Draw free-body diagram for the force system:





Since the reaction force (N) is normal (perpendicular) to the inclined surface and the vertical axis (y-axis) is perpendicular to the horizontal axis (x-axis), then the angle of the inclined surface with the horizontal axis (30°) is equal to the angle of the reaction force (N) with the vertical axis (30°).
In other words,




By using the fraction Law:
Coefficient of fraction ?=F/N ?0.25=F/N
F=0.25N (1)
??F_y =0 ? (+ve.)
N cos?30+P sin??30-? W-F sin?30=0 (2)
Substitute W and F values into equation # 2
N cos?30+P sin??30-? 500-0.25N×sin?30=0
N cos?30+P sin??30-? 500-0.25N×sin?30=0
P = (500-0.74N )/sin30=1000-1.48N (3)

??F_x =0 ? (+ve.)
N sin?30-P cos?30+F cos?30=0 (4)
Substitute P and F values into equation # 4 and solve for N
N sin?30-(1000-1.48N)×cos?30+(0.25N)×cos?30=0
0.5N-866+1.28N+0.22N=0
0.5N+1.28N+0.22N=866
2N=866
N=433 Newton [The reaction (N) that is normal to the inclined surface]
Substitute N value into equation #1 and solve for F:
F=0.25×433=108.25 Newton [The fractional force (F)]
Substitute N value into equation #3 and solve for P:
P = 1000-1.48×433=359 Newton [The Maximum pulling force (P) keeps the body in equilibrium].

Example 3
Body A which is tied to a rigid wall using a steel cable (weight 500N) is located on body B (weight 1000N) that rest on rough surface (coefficient of fraction is 1/4). If the fraction coefficient between body A and B is equal to 1/3, calculate horizontal force P.


?
Solution
Body A
Draw free-body diagram for body A:



??F_y =0 ? (+ve.)
R_A-W_A=0
R_A=W_A=500 N ?
By using the fraction Law:
Coefficient of fraction ?=F_A/R_A ?1/3=F_A/500
F_A=500/3=166.7 N ?
??F_x =0 ? (+ve.)
T-F_A=0
T=F_A=166.7 N ?
?
Body B
Draw free-body diagram for body B:




??F_y =0 ? (+ve.)
R_B-?R_A-W?_B=0
R_B=W_B+R_A=1000+500=1500 N ?
By using the fraction Law:
Coefficient of fraction ?=F_A/R_A ?1/4=F_B/1500
F_A=1500/4=750 N ?
??F_x =0 ? (+ve.)
F_A+F_B-P=0
P=F_A+F_B=166.7+750=916.7 N ?