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Equilibrium of Rigid Bodies
FORCES ARE VECTORS
THEREFORE WE NEED
TO USE THE
TECHNIQUES OF
VECTOR ALGEBRA
Catagories
2 D FORCE SYSTEMS
EQUILIBRIUMEQUATIONS
CONDITIONSOFEQUILIBRIUM
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
Only ONE unknownOnly ONE unknown (Force component)(Force component) can be foundcan be found
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
Example:Example:
P
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
Determine the value of the force P so as toDetermine the value of the force P so as to
satisfy the equilibrium?satisfy the equilibrium?
F 0 -350+250-80+P=0 P=180 kN
x
+
→ = →∑
Example
Consider the particle subjected to two forces
 Assume unknown force F acts to the right for
equilibrium
∑Fx = 0 ; + F + 10N = 0
F = -10N
 Force F acts towards the left for equilibrium
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
F1
Example:
If the stepped bar is in equilibrium find the force F1.
Resultant of Collinear Forces
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
EQUILIBRIUMEQUATIONS
CONDITIONSOFEQUILIBRIUM
STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
A particle when is subjected to coconcurrentncurrent forcesforces in the x-y
plane its equilibrium condition equation can be written as

 ΣFx i + ΣFy j = 0
Both of these vector equations above to be valid, implies that
both the x and the y components should be equal to zero. Hence,
 +→ ΣFx = 0 F1x + F2x + ….. = 0
 +↑ ΣFy = 0 F1y + F2y + ….. = 0
Both algebraic sums equal to zero.
∑ = 0F

Only TWO unknowns can be foundOnly TWO unknowns can be found
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
‘‘CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces’’
∑ = 0F

0=∑ xF 0=∑ yF
0=+ ∑∑ jFiF yx

andand
TwoTwo Force componentForce component unknownunknownss cancan
be foundbe found
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
• Resolve the given forces into i and j
components and apply the equilibrium
+→ ∑F∑Fxx = 0= 0
+↑ ∑F∑Fyy = 0= 0
• Scalar equations of equilibrium
require that the algebraic sum
of the x and y components to
equal to zero.
Only TWO unknowns can be foundOnly TWO unknowns can be found!!
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Determine the magnitudes of F1 and F2 for
equilibrium. Set θ=60°.
Example:
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
FF11=1.827 kN F=1.827 kN F22=9.596 kN=9.596 kN
Only TWO unknowns can be foundOnly TWO unknowns can be found
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Example: (T)
53°
24°
Example
Determine the tension in
cables AB and AD for
equilibrium of the 250kg
engine.
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
SINCE the mass of the
engine is given i.e. unit is
‘kg’ (scalar) and not the
weight (FORCE)
the calculations should be
corrected to a vector having
a unit of Newton.
(mass * gravity )
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Procedure for Analysis
1. Free-Body Diagram
- Establish the x, y axes in any suitable
orientation
- Label all the unknown and known forces
magnitudes and directions
- Sense of the unknown force can be
assummed
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Procedure for Analysis
2. Equations of Equilibrium
- Apply the equations of equilibrium
+→ ∑Fx = 0 +↑ ∑Fy = 0
- Components are positive if they are
directed along the positive axis and
negative, if directed along the negative
axis
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Solution
FBD at Point A
- Initially, two forces acting, forces
of cables AB and AD
- Engine Weight [W=m.g]
= (250kg)(9.81m/s2
)
= 2.452 kN supported by cable CA
- Finally, three forces acting, forces
TB and TD and engine weight
on cable CA
FBD of the ring AFBD of the ring A
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Solution
+→ ∑Fx = 0; TB cos30° - TD = 0
+↑ ∑Fy = 0; TB sin30° - 2.452 = 0
Solving,
TB = 4.904 kN
TD = 4.247 kN
*Note: Neglect the weights of the cables since they
are small compared to the weight of the engine
FBD of the ring AFBD of the ring A
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Example
If the sack at A has a weight
of 20 N , determine
the weight of the sack at B
and the force in each cord
needed to hold the system in
the equilibrium position shown.
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Solution
TEC
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
TEC
FBD of the ring EFBD of the ring E
FBD of the ring CFBD of the ring C
TEC
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Solution
FBD at Point E.
Three forces acting,
forces of cables EG
and EC and the weight
of the sack on cable EA
FBD of the ring EFBD of the ring E
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Solution
Use equilibrium at the ring to determine tension in
CD and weight of B with TEC known
+→ ∑Fx = 0; TEG sin30° - TECcos45° = 0
+↑ ∑Fy = 0; TEG cos30° - TECsin45° - 20 = 0
Solving,
TEC = 38.637 N
TEG = 54.641 N
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
FBD of the ring EFBD of the ring EFBD of the ring CFBD of the ring C
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Solution
FBD at Point C
Three forces acting, forces by cable CD
and EC (known) and
weight of sack B on
cable CB.
FBD of the ringFBD of the ring CC
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Solution
+→ ∑Fx = 0; 38.637cos45° - (4/5)TCD = 0
+↑ ∑Fy = 0; (3/5)TCD + 38.637sin45° – WB = 0
Solving,
TCD = 34.151 N
WB = 47.811 N
*Note: components of TCD are proportional to the slope
of the cord by the 3-4-5 triangle
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Example: (T)
The 50-kg homogenous smooth sphere rests on the 30°
incline A and bears against the smooth vertical wall B.
Calculate the contact forces at A and B?
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
FBD of the sphereFBD of the sphere
30°
A
B
30°
A
B
30°
RRAA
RRBB
CC
WW
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Example:
y A A
+
x B B
W 50x9.81 490.5
F 0 F cos30 -490.5= 0 F 566.381 N
(assumed direction correct)
F 0 566.381sin30 -F 0 F 283.191 N
+
= =
↑ = ° ⇒ =
→ = ° = ⇒ =
∑
∑
(assumed direction correct)
2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem
CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
Example:
N
STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS
STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS
EQUILIBRIUMEQUATIONS
CONDITIONSOFEQUILIBRIUM
3 unknowns3 unknowns
5 unknowns5 unknowns
3 unknowns3 unknowns
6 unknowns6 unknowns
A particle when is subjected to coconcurrentncurrent forcesforces in the x-y-z
axes, its equilibrium condition equation can be written as

 ΣFx i + ΣFy j + ΣFz k = 0
Both of these vector equations above to be valid, implies that the
x, the y and the z components be equal to zero separately.
Hence,
 +→ ΣFx = 0 F1x + F2x + ….. = 0
 +↑ ΣFy = 0 F1y + F2y + ….. = 0
 + ΣFz = 0 F1z + F2z + ….. = 0
∑ = 0F

Only TOnly THREEHREE unknowns can be foundunknowns can be found
Three-D Force Systems
Concurrent at a point
When the system of external 3 dimensional
forces acting on an object in equilibrium:
Σ F = (ΣFx) i + (ΣFy) j + (ΣFz) k = 0
so each component of this equation must
be determined separately:
ΣΣFFxx =0,=0, ΣΣFFyy =0=0,, ΣΣFFzz =0.=0.
Three-D Force Systems
Concurrent at a point
• Resolve the given forces into i, j and k
components and apply the equilibrium
+→ ∑F∑Fxx = 0= 0
+↑ ∑F∑Fyy = 0= 0
+ ∑F∑Fzz = 0= 0
• Equations of equilibrium require that the algebraic
sum of x, y and z components must be equal to
zero.
TTHREEHREE unknowns can be foundunknowns can be found!!
Three-D Force Systems
Concurrent at a point
The 100-kg cylinder is
suspended from the
ceiling by cables
attached at points B, C
and D.
What are the tensions in
cables AB, AC & AD ?
Note that:
the gravity effect is in –the gravity effect is in –veve
y direction.y direction.
Example:
Three-D Force Systems
Concurrent at a point
Solution Strategy:
•Isolate the part of the cable system near point A,
•Obtain a free-body diagram subjected to forces due to the
tensions in the cables.
•Because the sums of the external forces in the x, y, and z
directions must IN BALANCE, obtain 3 INDEPENDENT
equations for the three unknown cables that are in tension.
•To do so, express the forces exerted by the tensions in
terms of their components.
Three-D Force Systems
Concurrent at a point
Drawing the Free-Body Diagram and Applying the Equations
Three-D Force Systems
Concurrent at a point
• Isolating the part of the cable system near point A and
show the forces exerted by the tensions in the cables.
The sum of the forces must equal zero:
Σ F = TAB + TAC + TAD − (981 N)j = 0
• Writing the Forces in Terms of their Components
• Obtain a unit vector that has the same direction as the
force TAB by dividing the position vector rAB from point
A to point B by its magnitude.
rAB = (xB − xA)i + (yB − yA)j + (zB − zA)k
= 4i + 4j +2k (m)
Three-D Force Systems
Concurrent at a point
Solution
kji
r
r
e 333.0667.0667.0 ++==
AB
AB
ABλ AB
Three-D Force Systems
Concurrent at a point
Expressing the force TAB in terms of its components by
writing it as the product of the tension TAB in cable AB
and the unit vector eAB...
TAB = TABeAB == TAB (0.667 i + 0. 667 j + 0.333 k)
Express the forces TAC and TAD in terms of their
components using the same procedure.
TAC = TAC (−0.408 i + 0.816 j − 0.408 k)
TAD = TAD (−0.514 i + 0.686 j + 0.514k )
λAB =
λAB
Three-D Force Systems
Concurrent at a point
Substituting these expressions into the equilibrium equation
TAB + TAC + TAD − (981 N)j = 0
Because the i, j, and k components must each equal to
zero, this results in three equations of:
i-component: 0.667TAB − 0.408TAC − 0.514TAD = 0
j-component: 0.667TAB + 0.816TAC + 0.686TAD = 981
k-component: 0.333TAB − 0.408TAC + 0.514TAD = 0
Solving these 3 equations successively, the tensions are:
TAB = 519 N
TAC = 636 N
Three-D Force Systems
Concurrent at a point
Three-D Force Systems
Concurrent at a point
Example:

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Lecture 5 (46)

  • 2. FORCES ARE VECTORS THEREFORE WE NEED TO USE THE TECHNIQUES OF VECTOR ALGEBRA
  • 5. Only ONE unknownOnly ONE unknown (Force component)(Force component) can be foundcan be found STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
  • 6. Example:Example: P STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS Determine the value of the force P so as toDetermine the value of the force P so as to satisfy the equilibrium?satisfy the equilibrium? F 0 -350+250-80+P=0 P=180 kN x + → = →∑
  • 7. Example Consider the particle subjected to two forces  Assume unknown force F acts to the right for equilibrium ∑Fx = 0 ; + F + 10N = 0 F = -10N  Force F acts towards the left for equilibrium STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
  • 8. F1 Example: If the stepped bar is in equilibrium find the force F1. Resultant of Collinear Forces STUDY THE EQUILIBRIUM OF 2 D FORCE SYSTEMS
  • 10. A particle when is subjected to coconcurrentncurrent forcesforces in the x-y plane its equilibrium condition equation can be written as   ΣFx i + ΣFy j = 0 Both of these vector equations above to be valid, implies that both the x and the y components should be equal to zero. Hence,  +→ ΣFx = 0 F1x + F2x + ….. = 0  +↑ ΣFy = 0 F1y + F2y + ….. = 0 Both algebraic sums equal to zero. ∑ = 0F  Only TWO unknowns can be foundOnly TWO unknowns can be found 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem ‘‘CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces’’
  • 11. ∑ = 0F  0=∑ xF 0=∑ yF 0=+ ∑∑ jFiF yx  andand TwoTwo Force componentForce component unknownunknownss cancan be foundbe found 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 12. • Resolve the given forces into i and j components and apply the equilibrium +→ ∑F∑Fxx = 0= 0 +↑ ∑F∑Fyy = 0= 0 • Scalar equations of equilibrium require that the algebraic sum of the x and y components to equal to zero. Only TWO unknowns can be foundOnly TWO unknowns can be found!! 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 13. Determine the magnitudes of F1 and F2 for equilibrium. Set θ=60°. Example: 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 14. FF11=1.827 kN F=1.827 kN F22=9.596 kN=9.596 kN Only TWO unknowns can be foundOnly TWO unknowns can be found 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 15. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces Example: (T) 53° 24°
  • 16. Example Determine the tension in cables AB and AD for equilibrium of the 250kg engine. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 17. SINCE the mass of the engine is given i.e. unit is ‘kg’ (scalar) and not the weight (FORCE) the calculations should be corrected to a vector having a unit of Newton. (mass * gravity ) 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 18. Procedure for Analysis 1. Free-Body Diagram - Establish the x, y axes in any suitable orientation - Label all the unknown and known forces magnitudes and directions - Sense of the unknown force can be assummed 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 19. Procedure for Analysis 2. Equations of Equilibrium - Apply the equations of equilibrium +→ ∑Fx = 0 +↑ ∑Fy = 0 - Components are positive if they are directed along the positive axis and negative, if directed along the negative axis 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 20. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 21. Solution FBD at Point A - Initially, two forces acting, forces of cables AB and AD - Engine Weight [W=m.g] = (250kg)(9.81m/s2 ) = 2.452 kN supported by cable CA - Finally, three forces acting, forces TB and TD and engine weight on cable CA FBD of the ring AFBD of the ring A 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 22. Solution +→ ∑Fx = 0; TB cos30° - TD = 0 +↑ ∑Fy = 0; TB sin30° - 2.452 = 0 Solving, TB = 4.904 kN TD = 4.247 kN *Note: Neglect the weights of the cables since they are small compared to the weight of the engine FBD of the ring AFBD of the ring A 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 23. Example If the sack at A has a weight of 20 N , determine the weight of the sack at B and the force in each cord needed to hold the system in the equilibrium position shown. 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 24. Solution TEC 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces TEC
  • 25. FBD of the ring EFBD of the ring E FBD of the ring CFBD of the ring C TEC 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 26. Solution FBD at Point E. Three forces acting, forces of cables EG and EC and the weight of the sack on cable EA FBD of the ring EFBD of the ring E 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 27. Solution Use equilibrium at the ring to determine tension in CD and weight of B with TEC known +→ ∑Fx = 0; TEG sin30° - TECcos45° = 0 +↑ ∑Fy = 0; TEG cos30° - TECsin45° - 20 = 0 Solving, TEC = 38.637 N TEG = 54.641 N 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 28. FBD of the ring EFBD of the ring EFBD of the ring CFBD of the ring C 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 29. Solution FBD at Point C Three forces acting, forces by cable CD and EC (known) and weight of sack B on cable CB. FBD of the ringFBD of the ring CC 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 30. Solution +→ ∑Fx = 0; 38.637cos45° - (4/5)TCD = 0 +↑ ∑Fy = 0; (3/5)TCD + 38.637sin45° – WB = 0 Solving, TCD = 34.151 N WB = 47.811 N *Note: components of TCD are proportional to the slope of the cord by the 3-4-5 triangle 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 31. Example: (T) The 50-kg homogenous smooth sphere rests on the 30° incline A and bears against the smooth vertical wall B. Calculate the contact forces at A and B? 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces
  • 32. FBD of the sphereFBD of the sphere 30° A B 30° A B 30° RRAA RRBB CC WW 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces Example:
  • 33. y A A + x B B W 50x9.81 490.5 F 0 F cos30 -490.5= 0 F 566.381 N (assumed direction correct) F 0 566.381sin30 -F 0 F 283.191 N + = = ↑ = ° ⇒ = → = ° = ⇒ = ∑ ∑ (assumed direction correct) 2D2D CConcurrentoncurrent at a point Fat a point Forceorce SSystemystem CoplanarCoplanar CConcurrentoncurrent at a point Fat a point Forcesorces Example: N
  • 34. STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS
  • 35. STUDY THE EQUILIBRIUM OF 3-FORCE SYSTEMS EQUILIBRIUMEQUATIONS CONDITIONSOFEQUILIBRIUM 3 unknowns3 unknowns 5 unknowns5 unknowns 3 unknowns3 unknowns 6 unknowns6 unknowns
  • 36. A particle when is subjected to coconcurrentncurrent forcesforces in the x-y-z axes, its equilibrium condition equation can be written as   ΣFx i + ΣFy j + ΣFz k = 0 Both of these vector equations above to be valid, implies that the x, the y and the z components be equal to zero separately. Hence,  +→ ΣFx = 0 F1x + F2x + ….. = 0  +↑ ΣFy = 0 F1y + F2y + ….. = 0  + ΣFz = 0 F1z + F2z + ….. = 0 ∑ = 0F  Only TOnly THREEHREE unknowns can be foundunknowns can be found Three-D Force Systems Concurrent at a point
  • 37. When the system of external 3 dimensional forces acting on an object in equilibrium: Σ F = (ΣFx) i + (ΣFy) j + (ΣFz) k = 0 so each component of this equation must be determined separately: ΣΣFFxx =0,=0, ΣΣFFyy =0=0,, ΣΣFFzz =0.=0. Three-D Force Systems Concurrent at a point
  • 38. • Resolve the given forces into i, j and k components and apply the equilibrium +→ ∑F∑Fxx = 0= 0 +↑ ∑F∑Fyy = 0= 0 + ∑F∑Fzz = 0= 0 • Equations of equilibrium require that the algebraic sum of x, y and z components must be equal to zero. TTHREEHREE unknowns can be foundunknowns can be found!! Three-D Force Systems Concurrent at a point
  • 39. The 100-kg cylinder is suspended from the ceiling by cables attached at points B, C and D. What are the tensions in cables AB, AC & AD ? Note that: the gravity effect is in –the gravity effect is in –veve y direction.y direction. Example: Three-D Force Systems Concurrent at a point
  • 40. Solution Strategy: •Isolate the part of the cable system near point A, •Obtain a free-body diagram subjected to forces due to the tensions in the cables. •Because the sums of the external forces in the x, y, and z directions must IN BALANCE, obtain 3 INDEPENDENT equations for the three unknown cables that are in tension. •To do so, express the forces exerted by the tensions in terms of their components. Three-D Force Systems Concurrent at a point
  • 41. Drawing the Free-Body Diagram and Applying the Equations Three-D Force Systems Concurrent at a point
  • 42. • Isolating the part of the cable system near point A and show the forces exerted by the tensions in the cables. The sum of the forces must equal zero: Σ F = TAB + TAC + TAD − (981 N)j = 0 • Writing the Forces in Terms of their Components • Obtain a unit vector that has the same direction as the force TAB by dividing the position vector rAB from point A to point B by its magnitude. rAB = (xB − xA)i + (yB − yA)j + (zB − zA)k = 4i + 4j +2k (m) Three-D Force Systems Concurrent at a point
  • 43. Solution kji r r e 333.0667.0667.0 ++== AB AB ABλ AB Three-D Force Systems Concurrent at a point
  • 44. Expressing the force TAB in terms of its components by writing it as the product of the tension TAB in cable AB and the unit vector eAB... TAB = TABeAB == TAB (0.667 i + 0. 667 j + 0.333 k) Express the forces TAC and TAD in terms of their components using the same procedure. TAC = TAC (−0.408 i + 0.816 j − 0.408 k) TAD = TAD (−0.514 i + 0.686 j + 0.514k ) λAB = λAB Three-D Force Systems Concurrent at a point
  • 45. Substituting these expressions into the equilibrium equation TAB + TAC + TAD − (981 N)j = 0 Because the i, j, and k components must each equal to zero, this results in three equations of: i-component: 0.667TAB − 0.408TAC − 0.514TAD = 0 j-component: 0.667TAB + 0.816TAC + 0.686TAD = 981 k-component: 0.333TAB − 0.408TAC + 0.514TAD = 0 Solving these 3 equations successively, the tensions are: TAB = 519 N TAC = 636 N Three-D Force Systems Concurrent at a point
  • 46. Three-D Force Systems Concurrent at a point Example: