torque and equilibrium physics 12. jokes of the day: joke from sheldon… m/watch?v=rqvt5cjw8 hy ...
TRANSCRIPT
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.
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 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 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?
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