3-4 Three-Dimensional Force Systems PDF

3.
4 Three-Dimensional
Force Systems
For particle equilibrium
∑F = 0
Resolving into i, j, k components
∑Fxi + ∑Fyj + ∑Fzk = 0
Three scalar equations representing algebraic
sums of the x, y, z forces
∑Fxi = 0
∑Fyj = 0
∑Fzk = 0
3.4 Three-Dimensional
Force Systems
Make use of the three scalar equations
to solve for unknowns such as angles
or magnitudes of forces
3.4 Three-Dimensional Force
Systems
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
Force Systems
Procedure for Analysis
Free-body Diagram
- Establish the z, y, z axes in any suitable
orientation
- Label all known and unknown force
magnitudes and directions
- Sense of a force with unknown
magnitude can be assumed
Force Systems
Procedure for Analysis
Equations of Equilibrium
- Apply ∑Fx = 0, ∑Fy = 0 and ∑Fz = 0 when forces
can 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.
Force Systems
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.
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.
Force Systems
Solution
FBD at Point A
- Point A chosen as the forces are
concurrent at this point
Force Systems
Solution
Equations of Equilibrium,
∑Fx = 0; FDsin30° - (4/5)FC = 0
∑Fy = 0; -FDcos30° + FB = 0
∑Fz = 0; (3/5)FC – 90N = 0
Solving,
FC = 150N
FD = 240N
FB = 208N
Force Systems
Solution
For the stretch of the spring,
FB = ksAB
208N = 500N/m(sAB)
sAB = 0.416m
Force Systems
Example 3.6
Determine the magnitude
and coordinate direction
angles of force F that are
required for equilibrium of
the particle O.
Force Systems
Solution
FBD at Point O
- Four forces acting on
particle O
Force Systems
Solution
Expressing each forces in Cartesian vectors,
F1 = {400j} N
F2 = {-800k} N
F3 = F3(rB / rB)
= {-200i – 300j + 600k } N
F = Fxi + Fyj + Fzk
Force Systems
Solution
For equilibrium,
∑F = 0; F1 + F2 + F3 + F = 0
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
3.4 Three-Dimensional Force
Systems
Solution
r r r r
F = {200i − 100 j − 200k }N
r
F = (200 )2 + (− 100 )2 + (200 )2 = 300 N
r
r F 200 r 100 r 200 r
uF = r = i− j− k
F 300 300 300
−1 200 
α = cos   = 48 . 2o
 300 
− 100 
β = cos −1  = 109 o
 300 
γ = cos −1
200  o
 = 48.2
 300 
Force Systems
Example 3.7
Determine the force
developed in each cable
used to support the 40kN
(≈ 4 tonne) crate.
Force Systems
Solution
FBD at Point A
- To expose all three
unknown forces in the
cables
Force Systems
Solution
Expressing each forces in Cartesian vectors,
FB = FB(rB / rB)
= -0.318FBi – 0.424FBj + 0.848FBk
FC = FC (rC / rC)
= -0.318FCi – 0.424FCj + 0.848FCk
FD = FDi
W = -40k
Force Systems
Solution
For equilibrium,
∑F = 0; FB + FC + FD + W = 0
-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
Force Systems
Solution
Solving,
FB = FC = 23.6kN
FD = 15.0kN
Force Systems
Example 3.6
The 100kg crate is
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.
Force Systems
Solution
FBD at Point A
- Weight of the crate = 100 (9.81) = 981
N
- To expose all three
unknown forces in the
cables
Force Systems
Solution
Expressing each forces in Cartesian
vectors,
FB = FBi
FC = FCcos120°i + FCcos135°j –
FCcos60°k
FD = -0.333FDi + 0.667FDj + 0.667FDk
W = -981k
Force Systems
Solution
For equilibrium,
∑F = 0; FB + FC + FD + W = 0
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
Force Systems
Solution
Solving,
FB = 693.7N
FC = 813N
FD = 693.7N
For the stretch of the spring,
FB = ks
693.7N = 1500s
s = 0.462m

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