1. Torque and Equilibrium
Physics 12

2. Jokes of the day: Joke from Sheldon…

3. 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 !!!! FN Ff Fa Fg

4. 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 FN Fa Ff Fg

5. 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
θ is the angle between the position/lever and force vectors.
Units = N • m

6. Common Example:
Which place to you have to apply more force at A or B?

7. 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

8. Torque m1 m1 m1 m2 m2 m2 r1 r1 r2 r2 r2
Fg1
Fg1
Fg2
Fg2

9. Torque
Torque is positive, if alone it would turn the object counter clockwise m1 m2 r1 r2
Fg1
Fg2
At equilibrium, the net torque would be equal to zero
This will result in positive torque
This will result in negative torque

10. 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

11. Torque Example m2 m1 r1 r2
Fg1
Fg2

12. 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?

13. Step 1 – Sum of the Forces
F2=?
F1=60.N
Fg=160N

14. Step 2 - Torque
F2=100N
F1=60.N
r2=?
r1=1.50m
Fg=160N

15. 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?

16. FBD: Ladder Problem
FNw
Fgp
FNg
Fgl Ff

17. Ladder Problem

18. Ladder Problem

19. Try it :
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