Torque and Equilibrium

Physics 12

Jokes of the day: Joke from Sheldon…

http://www.youtube.com/watch?v=RQVt5cjw8HY

Torque In the past we have looked at

forces that have acted through the centre of mass of an object

These leads to forces that will act against each other but not result in a net torque on an object

This leads to an object that is either at equilibrium or accelerating but everything is linear

Fnet =0 !!!!

Ff Fa

FN

Fg

Torque We will now decouple forces so that forces

may act at a distance from the centre of mass

This can result in a net torque which will result in the object twisting due to the applied forces

You will need to pick a convenient pivot point

Ff

Fa

FN

Fg

Torque Torque is calculated as

the cross product of the applied force and the distance away from a pivot point that the force is applied

If a system is at equilibrium, the net torque and net force must be equal to zero

sinrF

Fr

Units = N • m

θ is the angle between the position/lever and force vectors.

Common Example: Which place to you have to apply more

force at A or B?

Torque and direction: As torque is a vector, it

must have direction When a cross product is

used, the direction must be perpendicular to both vectors used in the product

Torque

m1 m2

Fg1 Fg2

r1 r2m1 m2m1 m2

Fg1 Fg2

r2r1 r2

Torque Torque is positive, if alone it would turn the object counter

clockwise

m1 m2

Fg1 Fg2

r1 r2

This will result in positive torque

This will result in negative torque

At equilibrium, the net torque would be equal to zero

Example #1: Two masses are placed on a seesaw of length

2.0m with the fulcrum in the middle. The first mass is 10.kg and the second mass is 15kg. If the first mass is placed so that its mass acts through the end of the seesaw, where must the second mass be placed?

fulcrum = the point on which an

object balances or turns

Torque Example

m2

Fg1

Fg2

r1 r2

Nm

Nmnet

net

98

980

0

2

2

21

mr

smkgrNm

Fr

Fr

o

g

g

67.

90sin)81.9)(.15)((98

sin

2

22

222

222

Nm

smkgm

Fr

Fr

o

g

g

98

90sin)81.9)(.10)(0.1(

sin

1

21

111

111

m1

Example# 2: Two people are carrying a uniform wooden board

that is 3.00 m long and weighs 160 N. if one person applies an upward force of 60. N at one end of the board, where must the other person be standing in order for the board to be in static equilibrium and with what force must they apply?

Step 1 – Sum of the Forces

Fg=160N

F1=60.N

F2=?

NF

NFN

FFF

F

g

net

22

2

21

100.1

0)160(.60

0

0

Step 2 - Torque

Fg=160N

F1=60.N

F2=100N

r2=? r1=1.50m

mr

NmNr

NrNm

FrFr

net

90.

.90)100(

0)90sin()100()90sin().60(50.1

0sinsin

0

0

2

2

2

2211

21

Example #3: Ladder Question A painter (mass = 75kg) stands 2.0m up a

3.0m ladder with a mass of 15kg. The ladder makes an angle of 60.° with the ground and there is no friction between the ladder and the wall. What force does the wall apply to the ladder? What is the minimum coefficient of friction that

must exist between the ladder and the ground?

FBD: Ladder Problem

Fgl

Fgp

FNw

FNg

Ff

Ladder Problem

NF

NmmF

mFNmNm

FrFrFr

NF

NNF

FFFF

F

Nw

Nw

Nw

Nwwgppgllnet

wplnet

net

N

N

gpglNnety

net

328

85060sin0.3

60sin0.330sin)740(0.230sin)150(5.10

sinsinsin

0

880

7401500

0

Ladder Problem

37.

3278800

327

880

NN

FFF

NF

NF

Nwfnetx

Nw

N

Try it : Page 495

29-30

Page 501 31, 32, 34