Torque 10.3 (pages 490 – 502)

Friday Fun… Unravelled What did you learn? What does it mean to balance? –For an object to balance, you must use the centre of mass/centre of gravity. –Cool balancing dible-balancing-of-objects / dible-balancing-of-objects /

Centre of Gravity Also known as centre of mass Not always the geographic centre (only if the object is of uniform density) –Use plumb lines to find – yREhttps:// yRE Can be outside the object(s) –Usually true for an open object (horseshoe) or more than 1 object (forks)

Torque Symbol: tau = A twisting or turning force that acts over a distance from the pivot Affects ROTATIONAL motion (not translational motion) PIVOT POINT: point around the rotation occurs (also called FULCRUM or AXIS of rotation)

Where have I seen torque before? Wrenches –A wrench rotates a screw. There is no translational motion but there is rotational motion. See saws –How do you have equilibrium on a see saw? Equal masses on each end If unequal masses, move the heavier child closer to the centre of the see saw

Formula Torque is the cross product of the LEVER ARM (r ┴ ) and the force (F) If the Force is on an angle, we only need one component (the perpendicular component) of the entire force (which is why we use sinθ) Units: Nm Lever Arm: perpendicular distance between the line along which the force is acting and the pivot point

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

Vector? Torque is a vector. Positive torque: if it rotates the object counter clockwise (ie, from quadrant 1 to 2, etc) Negative torque: clockwise rotation

RIGHT HAND THUMB RULE Step 1: Start at axis of rotation or pivot point or fulcrum Step 2: Point your fingers along the lever arm Step 3: Curl your fingers in the direction of the force Step 4: Your thumb points in the direction of the torque. Thumb out of the board is positive, thumb into the board is negative

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

So what about gravity??? Gravity acts on an object at its centre of mass (or centre of gravity at the surface of the Earth) To determine the centre of mass, just find the average x, y, and z “coordinate” of the object. THIS CAN BE OUTSIDE THE OBJECT(s)! We found the centre of mass in objects on Friday by trial and error.

Equilibrium For translational motion, equilibrium occurs when Fnet = 0. For rotational motion, equilibrium occurs when torque = 0. For an object to be in static equilibrium, both Fnet and net torque = 0.

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 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?

Answer Step 1: Where is the fulcrum? –At 1.0 m –Step 2: Draw a diagram labelling the forces and lever arm(s). –Step 3: tau = 0, solve for r2 –r2 = 0.67m

Torque Example m2m2 F g1 F g2 r1r1 r2r2 m1m1

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?

Step 1: Fnet = 0 Step 2: tau = 0 Step 3: solve for r2

Torque Example 2 A student exerts a force of 50.N on a lever at a distance of 0.40m from its axis of rotation. The student pulls at an angle of 60° above the lever arm. What is the torque experienced by the lever arm? Is it positive or negative? What equilibrium torque would be needed to stop the rotation of the lever arm?

Torque Example 3 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?

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

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

Remembering torque…