Equilibrium and Torque

Equilibrium and Torque

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°


∑Fx = 0 and ∑Fy = 0

mg = 20 N

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


∑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


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


∑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


∑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


∑Fx = 0 and ∑Fy = 0

30°

20 N

T1 = 40 N

T2 = 35 N

∑Fy = 0T1y - mg = 0T1y = mg = 20 N

T1y/T1 = sin 30T1 = Ty/sin 30 = 40 N

∑Fx = 0T2 - T1x = 0T2 = T1x= T1cos30T2 = (40 N)cos30

T2 = 35 N


∑Fx = 0 and ∑Fy = 0

Note: The x-components cancelThe y-components cancel

30°

20 N

T1 = 40 N

T2 = 35 N

30°60°


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?

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


a v

t0

t

0

What is Torque?

Torque is like “twisting force”

The more torque you apply to a wheel the more quickly its rate of spin changes

Math Review:

1. Definition of angle in “radians”

2. One revolution = 360° = 2π radiansex: π radians = 180°ex: π/2 radians = 90°

s /r

arc lengthradius

s

r

Linear vs. Rotational Motion

• Linear Definitions • Rotational Definitions

x

v x

t

a v

t

in radians

t

in radians/sec or rev/min

t

in radians/sec/sec

x

x i

x f

i

f

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

v x

t

400m

20s20m /s

t

=400 radians

20s20 rad/sec

10 rev/sec

600 rev/min

x

x i

x f

Linear vs. Rotational

Net Force ==> linear accelerationThe linear velocity changes

a Fnet

Net Torque ==> angular accelerationThe angular velocity changes(the rate of spin changes)

net

Torque


The more torque you apply to a wheel, the more quickly its rate

of spin changes

Torque = Frsinø


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!


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 force3. The angle of the force

rø

F

F// to r

F to r

ø



What affects the torque?1. The distance from the axis rotation “r” that the force is applied2. The component of force perpendicular to the r-vector

rø

F

F to r

ø


Torque = (the component of force perpendicular to r)(r)

rø

F

F F sinø

ø

(F)(r)

(F sinø)(r)

Frsinø

Torque = Frsinø

Torque (F)(r)



rø

F

F F sinø

ø

(F)(r)

(F sinø)(r)

Frsinø

(F)(r)

Frsinø

rø

F

F F sinø

ø

F

rø

Since and ø are supplementary angles

(ie : + ø = 180 )sin = sinø

Cross “r” with “F” and choose any angleto plug into the equation for torque

Torque = (Fsinø)(r)

Two different ways of looking at torque

rø

F

F F sinø

ø

Torque = (F)(rsinø)

r ø

F

rø

r

F F sinø

r

F

Torque (F)(r)

Torque (F)(r)


Torque = (F)(rsinø)

r ø

F

rø

r

F

r is called the "moment arm" or "moment"

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


a v

t0

t

0

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

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 ??


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

20 N ??


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)


20 N 20 N


20 N ??



∑T’s = 0


F2 = 40 NF2 =??

F1 = 20 N

r1 = 4 m r2 = 2 m

(force at the fulcrum is not shown)


20 N 40 N


20 N ??



F2 =??

F1 = 20 N

r1 = 3 m r2 = 2 m∑T’s = 0


F2 = 30 N



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


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

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

TTy = mg = 200N

mg = 200N

Ty/T = sin30T = Ty/sin30 = 400N

We don’t need to use torque if the rod is “massless”!

(force at the pivot point is not shown)

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


Sign on a wall #2A 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 = 0TFT = Tcm + Tmg

FT(2)sin30 =100(1)sin90 + (200)(2)sin90FT = 500 N


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.

Diving boardA 4 meter long diving board with a mass of 40 kg.a. Determine the downward force of the bolt.

(Balance Torques)

bolt

Rcm = 1R1 = 1

Fcm = 400 NFbolt = 400 N

∑T = 0


Diving boardA 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 NFbolt = 400 N

F = 800 N ∑F = 0

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

Push-ups #1A 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 NFfeet = 333 N

A 100 kg man does push-ups as shown

Find the force on his hands and his feet

∑T = 0TH = Tcm

FH(1.5)sin60 =1000(1)sin60

FH = 667 N

∑F = 0Ffeet + Fhands = mg = 1,000 NFfeet = 1,000 N - Fhands = 1000 N - 667 N

FFeet = 333 N

1 m

0.5 m

30°

Fhands

Ffeet

CM

mg = 1000 N


Find the force acting on his hands

1 m

0.5 m

30°

Fhands

CM90°


∑T = 0TH = Tcm

FH(1.5)sin90 =1000(1)sin60

FH = 577 N

1 m

0.5 m

30°

Fhands

CM90°

mg = 1000 N

Force on hands:

(force at the feet is not shown)

Sign on a wall, againA 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!

Find the force exerted by the wall on the rod

FWx = FTx = 500N(cos30) FWx= 433N

FWy + FTy = Fcm +mgFWy = Fcm + mg - FTy FWy= 300N - 250FWy= 50N

(forces and angles NOT drawn to scale!)

Eat at Joe’s

30° 30°

FT = 500N

mg = 200NFcm = 100N

FW

FWy= 50N

FWx= 433N

FW= 436N

FW= 436N

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Equilibrium and Torque