EquilibriumAn object is in “Equilibrium” when: 1. There is no net force acting on the object2. 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

t0

t

0 We’ll get to this later

Static Equilibrium

An object is in “Static Equilibrium” when it is

NOT MOVING.

v x

t= 0

a v

t0

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

t0

t

0

EquilibriumLet’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

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 the2 kg masses at rest below.

60°

30°30° 30°

Condition 1: No net Force

∑Fx = 0 and ∑Fy = 0

mg = 20 N

N = 20 N∑Fy = 0N - mg = 0N = mg = 20 N

Condition 1: No net Force

∑Fx = 0 and ∑Fy = 0

∑Fy = 0T1 + T2 - mg = 0

T1 = T2 = T

T + T = mg2T = 20 NT = 10 N

T1 = 10 N

20 N

T2 = 10 N

Condition 1: No net Force

60°

mg =20 N

f = 17.4 N N = 10 N

Fx - f = 0f = Fx = mgsin 60 f = 17.4 N

N = mgcos 60 N = 10 N

Condition 1: No net Force

∑Fx = 0 and ∑Fy = 0

30°30°

mg = 20 N

T1 = 20 N T2 = 20 N

∑Fy = 0T1y + T2y - mg = 02Ty = mg = 20 N

Ty = mg/2 = 10 N

T2 = 20 N

T2x

T2y = 10 N

30°

Ty/T = sin 30T = Ty/sin 30 T = (10 N)/sin30 T = 20 N

∑Fx = 0

T2x - T1x = 0T1x = T2x

Equal angles ==> T1 = T2

Note: unequal angles ==> T1 ≠ T2

Condition 1: No net Force

∑Fx = 0 and ∑Fy = 0

Note: The y-components cancel, soT1y and T2y both equal 10 N

30°30°

mg = 20 N

T1 = 20 N T2 = 20 N

Condition 1: No net Force

∑Fx = 0 and ∑Fy = 0

Note: The x-components cancelThe y-components cancel

30°

20 N

T1 = 40 N

T2 = 35 N

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?

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.

∑Fy = 0T1y + T2y - mg = 0T1sin60 + T2sin30 - mg = 0T1sin60 + T2sin30 = 20 N

Solve simultaneous equations!

∑Fx = 0T2x-T1x = 0T1x = T2x

T1cos60 = T2cos30

30°60°

T1 T2

Note: unequal angles ==> T1 ≠ T2

What is the tension in each string?

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

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 ??

Condition 2: No net Torque

Weights are attached to an 8 meter long lever at rest.Determine the unknown weight below

20 N ??

Condition 2: No net Torque

F1 = 20 N

r1 = 4 m r2 = 4 m

F2 =??

∑T’s = 0

T2 - T1 = 0T2 = T1F2r2sinø2 = F1r1sinø1 (F2)(4)(sin90) = (20)(4)(sin90)

F2 = 20 N … same as F1

Upward force from the fulcrumproduces no torque (since r = 0)

Condition 2: No net Torque

20 N 20 N

Condition 2: No net Torque

20 N ??

Weights are attached to an 8 meter long lever at rest.Determine the unknown weight below

Condition 2: No net Torque

∑T’s = 0

T2 - T1 = 0T2 = T1F2r2sinø2 = F1r1sinø1 (F2)(2)(sin90) = (20)(4)(sin90)

F2 = 40 NF2 =??

F1 = 20 N

r1 = 4 m r2 = 2 m

(force at the fulcrum is not shown)

Condition 2: No net Torque

20 N 40 N

Condition 2: No net Torque

20 N ??

Weights are attached to an 8 meter long lever at rest.Determine the unknown weight below

Condition 2: No net Torque

F2 =??

F1 = 20 N

r1 = 3 m r2 = 2 m∑T’s = 0

T2 - T1 = 0T2 = T1F2r2sinø2 = F1r1sinø1 (F2)(2)(sin90) = (20)(3)(sin90)

F2 = 30 N

(force at the fulcrum is not shown)

Condition 2: No net Torque

20 N 30 N

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)

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

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.

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 = 0T2 = T1 + TcmF2r2sinø2 = F1r1sinø1 + FcmRcmsinøcm

(F2)(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90)

F2 = 160 N

(force at the fulcrum is not shown)