Friction Equilibrium KUS objectives BAT Understand the directly
Friction: Equilibrium • • • KUS objectives BAT Understand the directly proportional relationship between Friction force and Reaction force BAT Solve equilibrium problems using coefficient of friction Starter: Write a definition for each of these forces Weight Tension Reaction (or Normal contact force)
What is Friction? ‘The resistance that one surface or object encounters when moving over another’. Friction is best given by Fr or Frmax Do not confuse Fr with F for Resultant Force
Limiting Equilibrium: This is when the magnitude of the frictional force is just sufficient to prevent relative motion Friction is a Force on objects caused by a touching surfaces resistance to a direction. ‘Rougher’ surfaces cause greater friction The coefficient of friction is a measure of this ‘roughness’ of different surfaces. There is a direct relationship between friction and the reaction force when an object is touching a surface when a particle is in equilibrium and static when a particle is moving Remember: An object moving at a constant velocity will still be in equilibrium An accelerating object is subject to F=ma
WB 1 max friction: Finding coefficient friction A Trolley of mass 50 kg is being dragged along the ground at a constant velocity by a force of 150 N at an angle of elevation of 25. Find the coefficient of friction between the ground and the trolley a) What is the normal Reaction force? b) What is the Friction force? R Fr 150 250 W = 50 9. 8 = 490
WB 2 max friction: Finding the coefficient of friction A Trolley of mass 50 kg is being dragged along the ground at a constant velocity by a force of 150 N at an angle of elevation of 250 a) What is the Reaction Force? b) What is the Friction force? c) Find the coefficient of friction is at its maximum R Fr 150 250 W = 50 9. 8 = 490
Measuring the coefficient of Friction Aim: To find values for the coefficient of friction of different surfaces Equipment: Board calculator 1 kg weight Carpet, smooth board, sandpaper, OUTLINE METHOD: Place the 1 kg weight on the board Raise the board to the point where the weight is about to move (point of slipping) Measure the angle of the slope using the Protractor Repeat with different surfaces Find the coefficient of friction for each surface Comment on the validity of your results Big protractor
WB 3. Object on a slope about to go up the slope A Block with mass 3 kg is pushed by a force parallel to the direction of the slope. The block is at a tipping point where it is about to go up the slope. The slope surface has coefficient of friction of 0. 3 and is at 280 to the horizontal. Find the values of friction and the Push force n o i t ac Re h s u P j n o i ict Fr 280 n o i t a t o n , j i Weight i
WB 3 Object on a slope about to go up the slope solution h n o i t c a e R on i t c Fri 280 Pus ii) Friction = R = 0. 3 x 25. 96 = 7. 79 N 280 iii) Push =friction + component of weight downslope Push = 7. 79 + 29. 4 sin 28 = 21. 59 N
WB 4 Object on a slope about to go down the slope A Block with mass 8 kg is at the point of slipping down a rough slope. The slope is at 280 to the horizontal. Find the value of the coefficient of friction of the slope Friction will be up the slope n o i t c a e R n o i t ric F 280 Weight
WB 4 Object on a slope about to go down the slope solution n o i t Fric on ti c a e R 280 ii) Equilibrium Friction= 78. 4 sin 28 = 36. 8 N 280 Weight = 8 x 9. 8=78. 4
WB 5. Simultaneous equations question A Block with mass 3 kg is held at rest on a rough slope by a horizontal force (Push). The slope has coefficient of friction of 1/3 and the slope is at 300 to the horizontal. Find the value of the Push force n tio c a e R Push n o i ict Fr 300 Weight
WB 5 (cont). Simultaneous equations question solution part 1 n o i t c a e R 300 Fr 300 j 300 Weight = 3 x 9. 8=29. 4 o n j i, n o i t ta i
WB 5 (cont). Simultaneous equations question solution part 2 n R = 29. 4 cos 30 + Psin 30 (2) io t c a Re Fr 29. 4 sin 30 + Fr – Pcos 30 = 0 (1) 300 solve simultaneously gives P = 33. 2 N (and R = 42. 06 N)
WB 6. horizontal push, object is about to go down the slope A Block with Weight 10 Newton’s is held at rest on a rough slope by a horizontal force (Push). The slope has coefficient of friction of 0. 3 and the slope is at 300 to the horizontal. Find the value of the Push, Reaction and friction forces n o i t c a e R n o i ict Fr j n o i t a Push t o n , j i 300 Weight i
WB 6. rough slope in equilibrium, object is about to go down the slope solution Fr R 300 300
WB 6. rough slope in equilibrium, object is about to go down the slope solution Fr R 300 Fr + Pcos 30 = 10 sin 30 (2) Fr = R = 0. 3 R 300 R = 10 cos 30 + Psin 30 (1) Solve simultaneously (3)
• • • KUS objectives BAT Understand the directly proportional relationship between Friction force and Reaction force BAT Solve equilibrium problems with the coefficient of friction self-assess One thing learned is – One thing to improve is –
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