1. Equilibrium and Torque

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

3. An object is in “Static Equilibrium” when it is
NOT MOVING.

4. Dynamic Equilibrium An object is in “Dynamic Equilibrium” when it is
MOVING with constant linear velocity
and/or rotating with constant angular velocity.

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

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°

7. Condition 1: No net Force
∑Fx = 0 and ∑Fy = 0
mg = 20 N
N = 20 N
∑Fy = 0
N - mg = 0
N = mg = 20 N

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

9. Condition 1: No net Force
N = mgcos 60
N = 10 N
60°
mg =20 N
f = 17.4 N
N = 10 N
Fx - f = 0
f = Fx = mgsin 60
f = 17.4 N

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

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

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

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

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

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

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

17. Equilibrium An object is in “Equilibrium” when:
There is no net force acting on the object
There is no net Torque
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.

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

19. Math Review:
Definition of angle in “radians”
One revolution = 360° = 2π radians
ex: π radians = 180°
ex: π/2 radians = 90° s r

20. Linear vs. Rotational Motion
Linear Definitions
Rotational Definitions

21. Linear vs. Rotational Velocity
A car drives 400 m in
20 seconds:
a. Find the avg linear velocity
A wheel spins thru an angle of 400π radians in 20 seconds:
a. Find the avg angular velocity

22. Linear vs. Rotational Net Force ==> linear acceleration
The linear velocity changes
Net Torque ==> angular acceleration
The angular velocity changes
(the rate of spin changes)

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

24. 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!

25. Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle.
What affects the torque?
The place where the force is applied: the distance “r”
The strength of the force
The angle of the force r ø F

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

27. Torque = Frsinø
Imagine a bicycle wheel that can only spin about its axle.
Torque = (the component of force perpendicular to r)(r) r ø F

28. Torque is like “twisting force”
Imagine a bicycle wheel that can only spin about its axle. r ø F

29. Cross “r” with “F” and choose any angle
to plug into the equation for torque F r ø r ø F

30. Two different ways of looking at torque
Torque = (Fsinø)(r)
Torque = (F)(rsinø) r F r ø F r ø F

31. Imagine a bicycle wheel that can only spin about its axle.
Torque = (F)(rsinø) r ø F

32. Equilibrium An object is in “Equilibrium” when:
There is no net force acting on the object
There is no net Torque
In other words, the object is NOT experiencing
linear acceleration or rotational acceleration.

33. 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

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

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

36. Condition 2: No net Torque
Upward force from the fulcrum produces no torque (since r = 0)
∑T’s = 0
T2 - T1 = 0
T2 = T1
F2r2sinø2 = F1r1sinø1
(F2)(4)(sin90) = (20)(4)(sin90)
F2 = 20 N … same as F1
r1 = 4 m
r2 = 4 m
F1 = 20 N
F2 =??

37. Condition 2: No net Torque

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

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

40. Condition 2: No net Torque

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

42. Condition 2: No net Torque
F2 =??
F1 = 20 N
r1 = 3 m
r2 = 2 m
∑T’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)

43. Condition 2: No net Torque

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

45. 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

46. More interesting problems (the pivot is not at the center of mass)
Trick: gravity applies a torque “equivalent to” (the weight of the lever)(Rcm)
Tcm =(mg)(rcm) = (100 N)(2 m) = 200 Nm
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.

47. 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
∑T’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)

48. 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

49. Sign on a wall #1 Ty/T = sin30 T = Ty/sin30 = 400N
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.
Eat at Joe’s
30°
(force at the pivot point is not shown)
30°
Pivot point T
Ty = mg = 200N
mg = 200N
We don’t need to use torque if the rod is “massless”!
Ty/T = sin30
T = Ty/sin30 = 400N

50. 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 “FT”
Eat at Joe’s
30°
30°
Pivot point FT
mg = 200N
Fcm = 100N
(force at the pivot point is not shown)

51. 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
∑T = 0
TFT = Tcm + Tmg
FT(2)sin30 =100(1)sin90 + (200)(2)sin90
FT = 500 N
(force at the pivot point is not shown)

52. Diving board 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.
bolt

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

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

55. Remember: An object is in “Equilibrium” when: There is no net Torque
b. There is no net force acting on the object

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

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

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

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

60. Sign on a wall, again Find the force exerted by the wall on the rod
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°
FT = 500N
mg = 200N
Fcm = 100N
FW = ?
(forces and angles NOT drawn to scale!

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