1. Torque and Equilibrium Physics 12

2. 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 FfFf FaFa FNFN FgFg

3. 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 FfFf FaFa FNFN FgFg Decouple — vb ( tr ) to separate (joined or coupled subsystems) thereby enabling them to exist and operate separately

4. 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  The pivot point is often the centre of mass but may be chosen to be another point based upon the question  If a system is at equilibrium, the net torque and net force must be equal to zero

5. Torque m1m1 m2m2 F g1 F g2 r1r1 r2r2 m1m1 m2m2 m1m1 m2m2 F g1 F g2 r2r2 r1r1 r2r2

6. Torque  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  The direction will be determined using the right hand thumb rule  Thumb out of the board is positive, thumb into the board is negative

7. Torque m1m1 m2m2 F g1 F g2 r1r1 r2r2 This will result in positive torque This will result in negative torque At equilibrium, the net torque would be equal to zero

8. Torque Example 1  Copy this down!!  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 15.kg. If the first mass is placed so that its mass acts through the end of the seesaw, where must the second mass be placed so that the net torque is zero?

9. Torque Example m2m2 F g1 F g2 r1r1 r2r2 m1m1

10. Torque Example 2  2 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?

11. Step 1 – Sum of the Forces F g =160N F 1 =60.N F 2 =?

12. Step 2 - Torque F g =160N F 1 =60.N F 2 =100N r 2 =? r 1 =1.50m

13. 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?

14. Ladder Problem F gl F gp FwFw F Nw FNFN FfFf

15. Ladder Problem

17. Practice Problems  Page 495 29-30  Page 501 31-34