The document discusses static equilibrium, where an object experiences no net forces or torques acting upon it. It defines static and dynamic equilibrium, and explores conditions for equilibrium through examples of masses attached to levers or hanging from rods. Key concepts covered include the definitions of force, torque, and their relationships. Torque is described as a twisting force that depends on the distance from the axis of rotation and the perpendicular component of the applied force.

1. Static Equilibrium

2. Equilibrium
An object is in “Equilibrium” when:
1. There is no net force acting on the object
2. There is no net Torque (we’ll get to this later)
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.
a 
v
t
 0
 

t
 0 We’ll get to this later

3. Static Equilibrium
An object is in “Static Equilibrium” when it is
NOT MOVING.

v 
x
t
= 0
a 
v
t
 0

4. Dynamic Equilibrium
An object is in “Dynamic Equilibrium” when it is
MOVING with constant linear velocity
and/or rotating with constant angular velocity.

a 
v
t
 0
 

t
 0

5. Equilibrium
Let’s focus on condition 1: net force = 0
The x components of force cancel
The y components of force cancel
 F 0  F x0  F y0

6. Condition 1: No net Force
We have already looked at situations where the net force = zero.
Determine the magnitude of the forces acting on each of the
2 kg masses at rest below.
60°
30°
30°
30°

7. Condition 1: No net Force
ΣFx = 0 and ΣFy = 0
mg = 20 N
N = 20 N
ΣFy = 0
N - mg = 0
N = mg = 20 N

8. Condition 1: No net Force
ΣFx = 0 and ΣFy = 0
ΣFy = 0
T1 + T2 - mg = 0
T1 = T2 = T
T + T = mg
2T = 20 N
T = 10 N
T1 = 10 N
20 N
T2 = 10 N

9. Condition 1: No net Force
60°
mg =20 N
Ff = ?
N = ?
ΣFx = 0
ΣFy = 0

10. Condition 1: No net Force
ΣFx = 0 and ΣFy = 0
30°
30°
mg = 20 N
T1 = 20 N
T2 = 20 N
ΣFy = 0
T1y + T2y - mg = 0
2Ty = mg = 20 N
Ty = mg/2 = 10 N
T2 = 20 N
T2x
T2y = 10 N
30°
Ty/T = sin 30
T = Ty/sin 30
T = (10 N)/sin30
T = 20 N
ΣFx = 0
T2x - T1x = 0
T1x = T2x
Equal angles ==> T1 = T2
Note: unequal angles ==> T1 ≠ T2

11. Condition 1: No net Force
ΣFx = 0 and ΣFy = 0
Note:
The y-components cancel, so
T1y and T2y both equal 10 N
30°
30°
mg = 20 N
T1 = 20 N
T2 = 20 N

12. Condition 1: No net Force
ΣFx = 0 and ΣFy = 0
30°
20 N
T1 = 40 N
T2 = 35 N
ΣFy = 0
T1y - mg = 0
T1y = mg = 20 N
T1y/T1 = sin 30
T1 = Ty/sin 30 = 40 N
ΣFx = 0
T2 - T1x = 0
T2 = T1x= T1cos30
T2 = (40 N)cos30
T2 = 35 N Okay
Good

13. Condition 1: No net Force
ΣFx = 0 and ΣFy = 0
Note:
The x-components cancel
The y-components cancel
30°
20 N
T1 = 40 N
T2 = 35 N

14. 30°
60°
Condition 1: No net Force
A Harder Problem!
a. Which string has the greater tension?
b. What is the tension in each string?

15. a. Which string has the greater tension?
ΣFx = 0 so T1x = T2x
30°
60°
T1
T2
T1 must be greater in order to have the same x-component as T2.

16. ΣFy = 0
T1y + T2y - mg = 0
T1sin60 + T2sin30 - mg = 0
T1sin60 + T2sin30 = 20 N
Solve simultaneous equations!
ΣFx = 0
T2x-T1x = 0
T1x = T2x
T1cos60 = T2cos30
30°
60°
T1
T2
Note: unequal angles ==> T1 ≠ T2
What is the tension in each string?

17. Equilibrium
An object is in “Equilibrium” when:
1. There is no net force acting on the object
2. There is no net Torque
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.
a 
v
t
 0
 

t
 0

18. What is Torque?
Torque is like “twisting force”
The more torque you apply to a wheel the more quickly its rate of spin changes
Torque = Fr

19. Math Review:
1. Definition of angle in “radians”
2. One revolution = 360° = 2π radians
ex: π radians = 180°
ex: π/2 radians = 90°

  s /r
 
arc length
radius


s
r

20. Torque
Torque is like “twisting force”
The more torque you apply to a wheel, the more quickly its rate of spin changes

21. Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
If the force is the same in each case, which case produces
a more effective “twisting force”?
This one!

22. Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
What affects the torque?
1. The place where the force is applied: the distance “r”
2. The strength of the force
3. The angle of the force
r
ø
F

F// to r

F to r
ø
Torque = r . Fsinø

23. Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
What affects the torque?
1. The distance from the axis rotation “r” that the force is applied
2. The component of force perpendicular to the r-vector
r
ø
F

F to r
ø

24. Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
r
ø
F

F  Fsinø
ø
  (F )(r)
  (F sinø)(r)
  Fr sinø

25. Torque = (Fsinø)(r)
Two different ways of looking at torque
r
ø
F

F  Fsinø
ø
Torque = (F)(rsinø)
r ø
F

r
ø
r

F  Fsinø

r
F
Torque  (F)(r)

Torque  (F)(r )

26. Imagine a bicycle wheel that can only spin about its axle.
Torque = (F)(rsinø)
r ø
F

r
ø

r
F
r is called the "moment arm" or "moment"

27. Equilibrium
An object is in “Equilibrium” when:
1. There is no net force acting on the object
2. There is no net Torque
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.
a 
v
t
 0
 

t
 0

28. Condition 2: net torque = 0
Torque that makes a wheel want to rotate clockwise is +
Torque that makes a wheel want to rotate counterclockwise is -




Positive Torque Negative Torque

29. Condition 2: No net Torque
Weights are attached to 8 meter long levers at rest.
Determine the unknown weights below
20 N
??
20 N
??
20 N
??

30. Condition 2: No net Torque
Weights are attached to an 8 meter long lever at rest.
Determine the unknown weight below
20 N
??

31. Condition 2: No net Torque
F1 = 20 N
r1 = 4 m
r2 = 4 m
F2 =??
ΣT’s = 0
T2 - T1 = 0
T2 = T1
F2r2= F1r1
(F2)(4)= (20)(4)
F2 = 20 N … same as F1
Upward force from the fulcrum produces no torque (since r = 0)

32. Condition 2: No net Torque
20 N
20 N

33. Condition 2: No net Torque
20 N
??
Weights are attached to an 8 meter long lever at rest.
Determine the unknown weight below

34. Condition 2: No net Torque
ΣT’s = 0
T2 - T1 = 0
T2 = T1
F2r2sinø2 = F1r1sinø1
(F2)(2)(sin90) = (20)(4)(sin90)
F2 = 40 N
F2 =??
F1 = 20 N
r1 = 4 m
r2 = 2 m
(force at the fulcrum is not shown)

35. Condition 2: No net Torque
20 N
40 N

36. Condition 2: No net Torque
20 N
??
Weights are attached to an 8 meter long lever at rest.
Determine the unknown weight below

37. Condition 2: No net Torque
F2 =??
F1 = 20 N
r1 = 3 m
r2 = 2 m
ΣT’s = 0
T2 - T1 = 0
T2 = T1
F2r2sinø2 = F1r1sinø1
(F2)(2)(sin90) = (20)(3)(sin90)
F2 = 30 N
(force at the fulcrum is not shown)

38. Condition 2: No net Torque
20 N
30 N

39. In this special case where - the pivot point is in the middle of the lever,
- and ø1 = ø2 F1R1sinø1 = F2R2sinø2 F1R1= F2R2
20 N
20 N
20 N
40 N
20 N
30 N
(20)(4) = (20)(4)
(20)(4) = (40)(2)
(20)(3) = (30)(2)

40. More interesting problems (the pivot is not at the center of mass)
Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.
Determine the unknown weight below.
20 N
??
CM

41. Trick: gravity applies a torque “equivalent to” (the weight of the lever)(Rcm)
Tcm =(mg)(rcm) = (100 N)(2 m) = 200 Nm
More interesting problems (the pivot is not at the center of mass)
20 N
??
CM
Weight of lever
Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.

42. Masses are attached to an 8 meter long lever at rest.
The lever has a mass of 10 kg.
Determine the unknown weight below.
F1 = 20 N
CM
R1 = 6 m
R2 = 2 m
F2 = ??
Rcm = 2 m
Fcm = 100 N
ΣT’s = 0
T2 - T1 - Tcm = 0
T2 = T1 + Tcm
F2r2sinø2 = F1r1sinø1 + FcmRcmsinøcm
(F2)(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90)
F2 = 160 N
(force at the fulcrum is not shown)

43. Other problems:
Sign on a wall#1 (massless rod)
Sign on a wall#2 (rod with mass)
Diving board (find ALL forces on the board)
Push ups (find force on hands and feet)
Sign on a wall, again

44. Sign on a wall #1
Eat at Joe’s
30°
A 20 kg sign hangs from a 2 meter long massless rod supported by a cable at an angle of 30° as shown. Determine the tension in the cable.
30°
Pivot point
T
Ty = mg = 200N
mg = 200N
Ty/T = sin30
T = Ty/sin30 = 400N
We don’t need to use torque if the rod is “massless”!
(force at the pivot point is not shown)

45. Sign on a wall #2
Eat at Joe’s
30°
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kg and is supported by a cable at an
angle of 30° as shown. Determine the tension in the cable “FT”
30°
Pivot point
FT
mg = 200N
Fcm = 100N
(force at the pivot point is not shown)

46. Sign on a wall #2
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kg and is supported by a cable at an
angle of 30° as shown. Determine the tension in the cable.
30°
Pivot point
FT
mg = 200N
Fcm = 100N
ΣT = 0
TFT = Tcm + Tmg
FT(2)sin30 =100(1)sin90 + (200)(2)sin90
FT = 500 N
(force at the pivot point is not shown)

47. Diving board
bolt
A 4 meter long diving board with a mass of 40 kg.
a. Determine the downward force of the bolt. b. Determine the upward force applied by the fulcrum.

48. Diving board
A 4 meter long diving board with a mass of 40 kg.
a.Determine the downward force of the bolt. (Balance Torques)
bolt
Rcm = 1
R1 = 1
Fcm = 400 N
Fbolt = 400 N
ΣT = 0
(force at the fulcrum is not shown)

49. Diving board
A 4 meter long diving board with a mass of 40 kg.
a.Determine the downward force of the bolt. (Balance Torques)
b. Determine the upward force applied by the fulcrum. (Balance Forces)
bolt
Fcm = 400 N
Fbolt = 400 N
F = 800 N
ΣF = 0 upward force is
equal to
downward force

50. Remember:
An object is in “Equilibrium” when:
a.There is no net Torque
b. There is no net force acting on the object
 F0  0

51. Push-ups #1
A 100 kg man does push-ups as shown
Find the force on his hands and his feet
1 m
0.5 m
30°
Fhands
Ffeet
CM
Answer:
Fhands = 667 N
Ffeet = 333 N

52. A 100 kg man does push-ups as shown
Find the force on his hands and his feet
ΣT = 0
TH = Tcm
FH(1.5)sin60 =1000(1)sin60
FH = 667 N
ΣF = 0
Ffeet + Fhands = mg = 1,000 N
Ffeet = 1,000 N - Fhands = 1000 N - 667 N
FFeet = 333 N
1 m
0.5 m
30°
Fhands
Ffeet
CM
mg = 1000 N

53. Push-ups #2
A 100 kg man does push-ups as shown
Find the force acting on his hands
1 m
0.5 m
30°
Fhands
CM
90°

54. Push-ups #2
A 100 kg man does push-ups as shown
ΣT = 0
TH = Tcm
FH(1.5)sin90 =1000(1)sin60
FH = 577 N
1 m
0.5 m
30°
Fhands
CM
90°
mg = 1000 N
Force on hands:
(force at the feet is not shown)

55. Sign on a wall, again
A 20 kg sign hangs from a 2 meter long rod that
has a mass of 10 kg and is supported by a cable at an
angle of 30° as shown
Find the force exerted by the wall on the rod
Eat at Joe’s
30°
30°
FT = 500N
mg = 200N
Fcm = 100N
FW = ?
(forces and angles NOT drawn to scale!

56. Find the force exerted by the wall on the rod
FWx = FTx = 500N(cos30)
FWx= 433N
FWy + FTy = Fcm +mg
FWy = Fcm + mg - FTy
FWy= 300N - 250
FWy= 50N
(forces and angles NOT drawn to scale!)
Eat at Joe’s
30°
30°
FT = 500N
mg = 200N
Fcm = 100N
FW
FWy= 50N
FWx= 433N
FW= 436N
FW= 436N