Statics - Part1

8/2/2019 Statics - Part1

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1st. Chapter

Determination of coplanar resultant force acting on a particle, resolution of forces into

components and their moments.

Theorem: Statics 43 54. p.

1stProblem

Consider a particle A acted upon by two coplanar forces. Theforces considered here all pass through A, they are also said to beconcurrent. Determine the resultant force of the acting forces inthe A point by calculation and construction!Given data F1= 250 N; F2= 500 N

Solution by calculation (Statics 52.page.)In order to obtain the resultant force, the two vectors, F1 and F2, must be added together. So tosay, summarized:

Fr = FiThe force vectors:

F1 = F1i= 250i; F2 = F2j= 500jBy substituting into the resultant force:

Fr = F1i F2j = 250i 500j [N]

Result

The resultant force (Statics 49.page.):

By vector coordinates:Fr = 250i 500j [N]

By magnitude and unit-vectors:

N559)500()250( 22222

1 =+=+= FFFr

The resultant is situated in the 3rd plane section (both of its coordinates are negative), thus itsangle:

4,243500

250tgarc180tgarc180

1

2=+=+=

F

Fx .

By the use of the unit vector:

The unit vector of the resultant force: jijie 89415,044776,0sincos=+=

xx [ ]N559r eeF == rF

Solution by construction (Statics 46. pages)The Parallelogram-law will be applied after set-ting the scale: By adding the forces together (oneafter the other, without any concern of their order)the resultant vector can be obtained by connectingthe tail of the first vector with the tip of the lastvector.


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Result

According to the scale the resultant is approximately: Fr 550 N, angle of the resultant: x250, while its line of action crosses the intersection of the line of action of the two forces.

2nd

Problem

Determine the resultant ofF1 and F2!Given data F1= 500 N; F2= 400 N; 1= 15; 2 = 75

Solution by calculation (Statics 52. pages.)The resultant of the forces can be determined by adding F1 and F2 to-gether, so to say summarizing the given vectors.

Fr = FiThe Fr resultant vector components can be derived by the deconvolution ofF1 and F2 scalarcomponents:

Frx = Fix; Fry = FiyThe components:

F1x= F1cos1= 500cos15 = 483 NF1y= F1sin1= 500sin15 = 129,4 NF2x= F2cos2= 400cos75 = 103,5 NF2y= F2sin2 = 400sin75 = 386,4 N

The components of the resultant force:Frx= F1x+ F2x= 483 + 103,5 = 586,5 N

Fry= F1y+ F2y= 129,4 + 386,4 = 515,8 N

Result

The resultant: By vector coordinates:

Fr = Frxi + Fryj = 586,5i + 515,8j [N] By magnitude and unit vector:

N04,7818,5155,586 2222 =+=+= eyexe FFF

The resultant is situated in the 1st plane section (both of its coordinates are positive), thus itsangle:

=== 33,415,5868,515tgarctgarc

ex

eyx

FF

Solution by construction (Statics 46. pages)By setting the scale, the forces are added together by appropriate scaling, direction and senseone after the other. The resultant vector can be obtained by connecting the tail of the first vec-tor with the tip of the last vector.

According to the scale the resultant is approximately: Fr 790 N, angle of the resultant:

x 40, while its line of action crosses the intersection of the line of action of the two forces.


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3rd

Problem

Consider a particle denoted by O acted upon a certain system offorces. Determine the resultant of the given system of force by cal-culation and construction!

Given data F1= 2000 N; F2= 3000 N; F3= 7 kN; F4= 4 kN;1= 30; 2= 80; 3= 100; 4= 40

Solution by calculation (Statics 52. pages.)The resultant of the forces can be determined by adding the forcestogether, so to say summarizing the given vectors.

Fe = FiThe Fr resultant vector components can be derived by the deconvolution of scalar componentsof the system of forces:

Fex = Fix; Fey = FiyCalculation of the components:

Fix= Ficosi Fiy= Fisini (1.1)For example the components of the 2nd force:

F2x= F2cos2= 3000cos80 = 520,94 NF2y= F2sin2= 3000sin80 = 2954,42 N

In order to make the calculation more transparent, the components ofthe forces can be summarized in a table. Naturally, all the components must be calculated as itis shown in the example.

i Fix [N] Fiy [N]

1 1732,05 1000

2 520,94 2954,42

3 1215,53 6893,65

4 3064,17 2571,15

4101,63 5510,38

Result

The resultant: By vector coordinates::

Fr = 4101,63i 5510,38j [N] By magnitude and unit vector:

N32,686938,551063,4101 2222 =+=+= ryrxr FFF The resultant is situated in the 4th plane section thus its angle:

=== 66,30663,4101

38,5510tgarc360tgarc360

ex

ey

xF

F

Solution by construction (Statics 46. pages)

By setting the scale, the forces are added together by appropri-

ate scaling, direction and sense one after the other. The resultantvector can be obtained by connecting the tail of the first vector

with the tip of the last vector. Since the constructed forces do

not depend upon the order in whichF

1,F

2,F

4,F

3 are selected, itcan be concluded that the addition of vectors is commutative.


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In this case the order is: F1, F2, F4, F3.

Result

According to the scale the resultant is approximately: Fr 7000 N, angle of resultant: x 55. The line of action of resultant crosses the intersection ofF1, F2, F4, F3.

4thProblem

Break up the F force vector into u and v components. Apply calculation

and construction!

Data F= 500 N; 1 = 30; 2= 60

Solution

By setting the scale, the F force vector is plotted by appropriate scal-

ing, direction and sense. Then drawing parallel lines with the u and vfrom the tip of the vector, the lines will intersect the magnitude of the

u and v. This is the inverse Parallelogram Law. (Statics 46.p.).

Results

According to the scaling, the magnitude of the u and v components are: Fu 280 N; Fv 280N. Their senses can be determined from the vector plot.

5thProblemt

Break up the F force vector intox andy components with calculation.

Data F= 500 N; = 60

SolutionIn a Cartesian x,y system, the perpendicular components can be derived by

deconvoluting the components of the F vector (Statics 50th. p). According to

the OAB triangle:

N25060cos500cos === FFx

N01,43360sin500sin === FFy

Result

Thex andy components of the F force vector: Fx = 250 N, Fy = 433,01 N(By the use of these, the F vector can be described as follows:F = Fxi + Fyj = 250i + 433,01j [N] )


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6thProblem

Determine the moment ofF related to the O point! Data F= 100 N; = 60;xF= 2,5 m;yF= 1,4 m

Solution

a., According to the secondary school studies, the magnitudeof the moment can be obtained by: Mo = rF, where r is theperpendicular distance between the O point and the line ofaction of the force.It is not easy to define this distance. According to the figure:

m86,24,15,2 2222 =+=+=FFA

xyr

25,295,2

4,1tgarctgarc ===

F

F

x

y

( ) ==== 75,3025,29609090 m46,275,30cos86,2cos === Arr

The magnitude of the moment taking into consideration the sense is the following:Nm24610046,2 === rFM

b., Let us find the solution of the earlier problem by the useof the Superposition law: Each components of the force vec-

tor has an effect as a moment around the O point, andthese effects can be summed. The force vector is situated inthe x-y plan, thus the moment calculated on the O pointwhich is also situated in thex-y plan is in thez direction.The components of the F force vector are:

Fx= Fsin = 100sin60 = 86,6 NFy= Fcos = 100cos60 = 50 N

According to this result, the moment taking into consid-eration the sense of rotation is calculated in the following way:

M0= yFFxxFFy = 1,486,6 2,550 = 246,24 NmThe difference between this result and the result given in the a.), is caused by rounding error.

Result

The moment vectorr: Mo = 246,24k [Nm], since the F force vector rotates clockwise,the moment vector points to the negative direction ofz.

Statics - Part1

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