Presentation of Parallelogram Law II

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







Parallelogram Law

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 (Moment of a Force)
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








Parallelogram Law
If F_1 and F_2 represent two forces and ? represents the angle between them, their effects can be replaced by the effect of on force, which called the resultant of forces (R) and as follows:









Then
R=?(?F_1?^2+?F_2?^2+2F_1 F_2 cos?? )
Apply triangle law:
R/sin?(180-?) =F_1/sin?? =F_2/sin??

Special Cases
Case 1, the angle ? between F_1 and F_2 equal to zero, then:




R=?(?F_1?^2+?F_2?^2+2F_1 F_2 cos?0 )=F_1 ?+F?_2
Case 2, the angle ? between F_1 and F_2 equal to 90o, then:




R=?(?F_1?^2+?F_2?^2+2F_1 F_2 cos?90 )=?(?F_1?^2+?F_2?^2 )
?=tan^(-1)??(F_1/F_2 )? the angle between R and F_2

Case 3, the angle ? between F_1 and F_2 equal to 180o, then:




R=?(?F_1?^2+?F_2?^2+2F_1 F_2 cos?180 )=?(?F_1?^2+?F_2?^2-2F_1 F_2 )=?((F_1-F_2 )^2 )=F_1-F_2


Applications

Problem1:
A cart is pulled uniformly along two cables using two horses as shown below, the tension forces along the two cables are P = 900 N and Q = 1100 N and the angle between them are ? = 60?.
Determine the magnitude of the resultant force (R)and the angles between the resultant and the two tension forces.



Solution:
Using the Parallelogram Law:



R=?(P^2+Q^2+2PQ cos?? )=?(?900?^2+?1100?^2+2×900×1100×cos?60 )=1734.9N=1.7349KN
Apply triangle law:
R/sin?(180-?) =Q/sin?? =P/sin??
1734.9/sin?(180-60) =1100/sin??
?=sin^(-1)?( 1100/(1734.9/sin?(180-60) ))=?33.3?^o

1734.9/sin?(180-60) =900/sin??
?=sin^(-1)?( 900/(1734.9/sin?(180-60) ))=?26.7?^o


?
Problem2:
A steel pulley is used in luggage’s transfer (from ground level to other roofs). If a box that weight W= 0.5 kN is being lifted by a worker that weight F = 0.7 kN using a fabric rope and the pulley.
Determine the force that which the man’s feet transfer to the ground.






Solution:
Using the Parallelogram Law:
The angle ? between W and F is equal to 180o, then:
R=?(W^2+F^2+2WF cos?180 )=?(W^2+F-2WF)=?((W-F)^2 )=W-F=0.7-0.5=0.2 kN ?(Downward)