6161103 3.4 three dimensional force systems

3.4 Three-Dimensional Force Systems


� For particle equilibrium

∑F = 0

� Resolving into i, j, k components

∑Fxi + ∑Fyj + ∑Fzk = 0∑Fxi + ∑Fyj + ∑Fzk = 0

� Three scalar equations representing algebraic sums of the x, y, z forces

∑Fxi = 0

∑Fyj = 0

∑Fzk = 0



� Make use of the three scalar equations to solve for unknowns such as angles or magnitudes of forces



� Ring at A subjected to force from hook and forces from each of the three chains

� Hook force = weight of the electromagnet and the load, denoted as W

� Three scalars equations applied to FBD to determine FB, FC and FD



� Procedure for AnalysisFree-body Diagram

- Establish the z, y, z axes in any suitable orientationorientation

- Label all known and unknown force magnitudes and directions

- Sense of a force with unknown magnitude can be assumed



� Procedure for Analysis

Equations of Equilibrium

- Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0 when forces can be easily resolved into x, y, z componentscan be easily resolved into x, y, z components

- When geometry appears difficult, express each force as a Cartesian vector. Substitute vectors into ∑F = 0 and set i, j, k components = 0

- Negative results indicate that the sense of the force is opposite to that shown in the FBD.



Example 3.5

A 90N load is suspended from the hook. The

load is supported by two cables and a spring

having a stiffness k = 500N/m. having a stiffness k = 500N/m.

Determine the force in the

cables and the stretch of the

spring for equilibrium. Cable

AD lies in the x-y plane and

cable AC lies in the x-z plane.



Solution

FBD at Point A

- Point A chosen as the forces are - Point A chosen as the forces are concurrent at this point



Solution

Equations of Equilibrium,

∑Fx = 0; FDsin30° - (4/5)FC = 0

∑Fy = 0; -FDcos30° + FB = 0y D B

∑Fz = 0; (3/5)FC – 90N = 0

Solving,

FC = 150N

FD = 240N

FB = 208N



Solution

For the stretch of the spring,

FB = ksAB

208N = 500N/m(sAB)208N = 500N/m(sAB)

sAB = 0.416m



Example 3.6

Determine the magnitude

and coordinate direction and coordinate direction

angles of force F that are

required for equilibrium of

the particle O.



Solution

FBD at Point O

- Four forces acting on- Four forces acting on

particle O



Solution


Expressing each forces in Cartesian vectors,

F = {400j} NF1 = {400j} N

F2 = {-800k} N

F3 = F3(rB / rB)

= {-200i – 300j + 600k } N

F = Fxi + Fyj + Fzk



Solution

For equilibrium,

∑F = 0; F1 + F2 + F3 + F = 0

400j - 800k - 200i – 300j + 600k 400j - 800k - 200i – 300j + 600k

+ Fxi + Fyj + Fzk = 0

∑Fx = 0; - 200 + Fx = 0 Fx = 200N

∑Fy = 0; 400 – 300 + Fy = 0 Fy = -100N

∑Fz = 0; - 800 + 600 + Fz = 0 Fz = 200N



Solution

( ) ( ) ( )rrr

rr

r

rrrr

200100200

300200100200

}200100200{

222 =+−+=

−−=

F

NF

NkjiF

o

o

o

rrrr

r

2.48300200

cos

109300100

cos

2.48300200

cos

300200

300100

300200

1

1

1

=

=

=

−=

=

=

−−==

−

−

−

γ

β

α

kjiF

FuF



Example 3.7

Determine the force

developed in each cable developed in each cable

used to support the 40kN

(≈ 4 tonne) crate.



Solution

FBD at Point A

- To expose all three - To expose all three unknown forces in the cables



Solution


Expressing each forces in Cartesian vectors,

FB = FB(rB / rB) FB = FB(rB / rB)

= -0.318FBi – 0.424FBj + 0.848FBk FC = FC (rC / rC)

= -0.318FCi – 0.424FCj + 0.848FCk FD = FDi

W = -40k



Solution

For equilibrium,

∑F = 0; FB + FC + FD + W = 0

-0.318F i – 0.424F j + 0.848F k - 0.318F i -0.318FBi – 0.424FBj + 0.848FBk - 0.318FCi

– 0.424FCj + 0.848FCk + FDi - 40k= 0

∑Fx = 0; -0.318FB - 0.318FC + FD = 0

∑Fy = 0; – 0.424FB – 0.424FC = 0

∑Fz = 0; 0.848FB + 0.848FC - 40 = 0



Solution

Solving,

FB = FC = 23.6kN FB = FC = 23.6kN

FD = 15.0kN



Example 3.6

The 100kg crate is

supported by three cords, supported by three cords,

one of which is connected

to a spring. Determine the

tension in cords AC and

AD and stretch of the

spring.



Solution

FBD at Point A

- Weight of the crate = 100 (9.81) = 981 - Weight of the crate = 100 (9.81) = 981 N

- To expose all three

unknown forces in the

cables



Solution


Expressing each forces in Cartesian Expressing each forces in Cartesian vectors,

FB = FBi

FC = FCcos120°i + FCcos135°j –FCcos60°k

FD = -0.333FDi + 0.667FDj + 0.667FDk

W = -981k



Solution

For equilibrium,

∑F = 0; FB + FC + FD + W = 0

F i + F cos120°i + F cos135°j – F cos60°k FBi + FCcos120°i + FCcos135°j – FCcos60°k

-0.333FDi + 0.667FDj + 0.667FDk - 981k= 0

∑Fx = 0; FB + FCcos120° - 0.333FD = 0

∑Fy = 0; FCcos135° + 0.667FD = 0

∑Fz = 0; FCcos60° + 0.667FD - 981 = 0



Solution

Solving,

FB = 693.7N

FC = 813N FC = 813N

FD = 693.7N

For the stretch of the spring,

FB = ks

693.7N = 1500s

s = 0.462m

6161103 3.4 three dimensional force systems

Education

thethree chainshook

force systemsmake use

fbd todetermine fb

point o

threeunknown forces

fc rc rc

force systemssolution

n load