ap equilibrium torque

8/22/2019 AP Equilibrium Torque

1/61

Equilibrium and Torque


2/61

EquilibriumAn object is in Equilibrium when:

1. There is no net force acting on the object

2. There is no net Torque (well get to this later)

In other words, the object is NOT experiencing

linearacceleration or rotational acceleration.

a v

t 0

t 0 Well get to this later


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Static Equilibrium

An object is in Static Equilibrium when it is

NOT MOVING.

v x

t= 0

a v

t 0


4/61

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


5/61

EquilibriumLets focus on condition 1: net force = 0

The x components of force cancel

The y components of force cancel

F 0

Fx 0

Fy 0


6/61

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

3030 30


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Fx = 0 and Fy = 0

mg = 20 N

N = 20 NFy = 0

N - mg = 0N = mg = 20 N


8/61


Fx = 0 and Fy = 0

Fy = 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


9/61


60

mg =20 N

f = 17.4 N N = 10 N

Fx - f = 0

f = Fx = mgsin 60

f =17.4 N

N = mgcos 60

N = 10 N


10/61


Fx = 0 and Fy = 0

3030

mg = 20 N

T1 = 20 N T2 = 20 N

Fy = 0

T1y + T2y - mg = 02Ty = mg = 20 N

Ty = mg/2 = 10 N

T2 = 20 N

T2x

T2y = 10 N

30

Ty/T = sin 30

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

T = 20 N

Fx = 0

T2x - T1x = 0

T1x = T2x

Equal angles ==> T1 = T2

Note: unequal angles ==> T1 T2


11/61


Fx = 0 and Fy = 0

Note:

The y-components cancel, so

T1y and T2y both equal 10 N

3030

mg = 20 N

T1 = 20 N T2 = 20 N


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Fx = 0 and Fy = 0

30

20 N

T1 = 40 N

T2 = 35 N

Fy = 0

T1y - mg = 0T1y = mg = 20 N

T1y/T1 = sin 30

T1 = Ty/sin 30 = 40 N

Fx = 0

T2- T1x = 0T2= T1x= T1cos30

T2 = (40 N)cos30

T2 = 35 N


13/61


Fx = 0 and Fy = 0

Note:

The x-components cancel

The y-components cancel

30

20 N

T1 = 40 N

T2 = 35 N


14/61

3060


A Harder Problem!

a. Which string has the greater tension?

b. What is the tension in each string?


15/61

a. Which string has the greater tension?

Fx = 0 so T1x = T2x

3060

T1 T2

T1 must be greater in order to have the same x-component as T2.


16/61

Fy = 0

T1y + T2y - mg = 0

T1sin60 + T2sin30 - mg = 0

T1sin60 + T2sin30 = 20 N

Solve

simultaneous

equations!

Fx = 0

T2x-T1x = 0

T1x = T2xT1cos60 = T2cos30

3060

T1 T2

Note: unequal angles ==> T1 T2

What is the tension in each string?


17/61



2. There is no net Torque


linear accelerationorrotational acceleration.

a v

t 0

t 0


18/61

What is Torque?

Torque is like twisting force

The more torque you apply to a wheel the

more quickly its rate of spin changes


19/61

Math Review:

1. Definition of angle in radians

2. One revolution = 360 = 2 radians

ex: radians = 180ex: /2 radians = 90

s /r

arc length

radius

s

r


20/61

Linear vs. RotationalMotion

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

xi xf

i

f


21/61

Linear vs. RotationalVelocity

A car drives 400 m in20 seconds:

a. Find the avg linear velocity

A wheel spins thru an angle of400 radians in 20 seconds:

a. Find the avg angular velocity

v x

t

400m

20s 20m /s

t=

400 radians

20s

20 rad/sec

10 rev/sec

600 rev/min

x

xi xf


22/61

Linear vs. Rotational

Net Force ==> linear acceleration

The linear velocity changes

aFnet

Net Torque ==> angular acceleration

The angular velocity changes

(the rate of spin changes)

net


23/61

Torque


The more torque you apply to a

wheel, the more quickly its rateof spin changes

Torque = Frsin


24/61


Imagine a bicycle wheel that can only spin about its axle.

If the force is the same in each case, which case producesa more effective twisting force?

This one!


25/61



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

F

F// to r

F to r


26/61



What affects the torque?

1. The distance from the axis rotation rthat the force is applied

2. The component of force perpendicular to the r-vector

r

F

F to r


27/61


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

r

F

F Fsin

(F

)(r)

(Fsin)(r)

Frsin

Torque = Frsin

Torque (F)(r)


28/61



r

F

F Fsin

(F

)(r)

(Fsin)(r)

Frsin


29/61

(F )(r) Frsin

r

F

F Fsin

F

r

Since and are supplementary angle

(ie : + = 180)sin = sin

Cross r with F and choose any angle

to plug into the equation for torque


30/61

Torque = (Fsin)(r)

Two different ways of looking at torque

r

F

F Fsin

Torque = (F)(rsin)

r

F

r

r

F Fsin

r

F

Torque (F)(r)

Torque (F)(r)


31/61


Torque = (F)(rsin)

r

F

r

r

F

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


32/61



2. There is no net Torque


linear accelerationorrotational acceleration.

a vt

0

t 0


33/61

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


34/61

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 ??


35/61


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

20 N ??


36/61


F1 = 20 N

r1 = 4 m r2 = 4 m

F2 =??

Ts = 0

T2 - T1 = 0

T2 = T1F2r2sin2 = F1r1sin1

(F2)(4)(sin90) = (20)(4)(sin90)

F2= 20 N same as F1

Upward force

from the fulcrum

produces no torque

(since r = 0)


37/61


20 N 20 N


38/61


20 N ??



39/61


Ts = 0

T2 - T1 = 0T2 = T1

F2r2sin2 = F1r1sin1

(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)


40/61


20 N 40 N


41/61


20 N ??



42/61


F2 =??

F1 = 20 N

r1 = 3 m r2 = 2 m

Ts = 0

T2 - T1 = 0

T2 = T1F2r2sin2 = F1r1sin1

(F2)(2)(sin90) = (20)(3)(sin90)

F2 = 30 N



43/61


20 N 30 N


44/61

In this special case where

- the pivot point is in the middle of the lever,

- and 1 = 2

F1R1sin1 = F2R2sin2

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)


45/61

More interesting problems

(the pivot is not at the center of mass)

Masses are attached to an 8 meter long leverat rest.

The lever has a mass of 10 kg.

Determine the unknown weight below.

20 N ??

CM


46/61

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




47/61



Determine the unknown weight below.

F1 = 20 N

CM

R1 = 6 m R2 = 2 m

F2 = ??

Rcm = 2 m

Fcm = 100 N

Ts = 0

T2 - T1 - Tcm = 0T2 = T1 + Tcm

F2r2sin2 = F1r1sin1 + FcmRcmsincm(F2)(2)(sin90)=(20)(6)(sin90)+(100)(2)(sin90)

F2 = 160 N



48/61

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


49/61

Sign on a wall #1

Eat at Joes

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 dont need to use torque

if the rod is massless!

(force at the pivot point is not shown)


50/61

Sign on a wall #2

Eat at Joes

30

A 20 kg sign hangs from a 2 meter long rod that

has a mass of 10 kgand is supported by a cable at an

angle of 30 as shown. Determine the tensionin the cable FT

30

Pivot point

FT

mg = 200N

Fcm= 100N



51/61

Sign on a wall #2



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



52/61

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.


53/61

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 = 1R1 = 1

Fcm = 400 NFbolt = 400 N

T = 0



54/61

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


55/61

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


56/61

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

A 100 kg man does push ups as shown


57/61


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 - 667 N

FFeet = 333 N

1 m

0.5 m

30

Fhands

Ffeet

CM

mg = 1000 N


58/61

Push-ups #2


Find the force acting on his hands

1 m

0.5 m

30

Fhands

CM90


59/61

Push-ups #2A 100 kg man does push-ups as shown

T = 0

TH = TcmFH(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)


60/61

Sign on a wall, again



angle of 30 as shown

Find the force exerted by the wall on the rod

Eat at Joes

30 30

FT = 500N

mg = 200N

Fcm = 100N

FW = ?

(forces and angles NOT drawn to scale!


61/61

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

FW 50N (forces and angles NOT drawn to scale!)

Eat at Joes

30 30

FT = 500N

mg = 200NFcm = 100N

FW

FWy= 50NFWx= 433N

FW= 436N

FW= 436N

ap equilibrium torque

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