1. 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

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

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

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

5. 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°

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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°

13. 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.

14. 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

15. 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 +
Negative Torque
Positive Torque

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

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

18. 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 =??

19. Condition 2: No net Torque

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

21. 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)

22. Condition 2: No net Torque

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

24. 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)

25. Condition 2: No net Torque

26. 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)

27. 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

28. 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.

29. 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)