Friction and Equilibrium in Physics
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KUS objectives
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BAT Understand the directly proportional relationship between
Friction force and Reaction force
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BAT Solve equilibrium problems using coefficient of friction
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Weight
Tension
Reaction (or Normal contact force)
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What is Friction?
‘The resistance that one surface or object encounters when moving over
another’.
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Limiting Equilibrium:
This is when the magnitude of the frictional force is just
sufficient to prevent relative motion
Remember:
An object moving at a
constant velocity
will still be in equilibrium
An
accelerating
object is subject to F=ma
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
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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?
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 25
0
a) What is the Reaction Force? b) What is the Friction force?
c) Find the coefficient of friction
if friction is at its maximum
Measuring the coefficient of Friction
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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
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Push
28
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Reaction
Friction
Weight
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ii) Friction =
R
= 0.3 x 25.96 = 7.79
N
iii)
Push =friction + component of weight downslope
Push = 7.79 + 29.4 sin 28 = 21.59
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Friction
28
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Reaction
Weight
Friction will be up the slope
A Block with mass 8 kg is at the point of slipping down a rough slope. The
slope is at 28
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to the horizontal. Find the value of the coefficient of friction
of the slope
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ii) Equilibrium
Friction
= 78.4 sin 28 = 36.8
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Push
30
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Reaction
Friction
Weight
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
30
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to the horizontal. Find the value of the Push force
Fr
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Reaction
30
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30
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Weight = 3 x 9.8=29.4
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Friction
30
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Reaction
Weight
Push
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 30
0
to the horizontal. Find the value of the Push, Reaction and
friction forces
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R =
10cos30 + Psin30 (1)
F
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+ Pcos30 = 10sin30 (2)
F
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=
R =
0.3 R
(3)
Solve simultaneously
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self-assess
One thing learned is –
One thing to improve is –
•
KUS objectives
•
BAT Understand the directly proportional relationship between
Friction force and Reaction force
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BAT Solve equilibrium problems with the coefficient of friction
END
Concepts of friction and equilibrium in physics, including definitions of forces like weight, tension, and reaction force. Learn about the relationship between friction force and reaction force, solving equilibrium problems, and the significance of limiting equilibrium. Find practical examples and calculations to understand the direct relationship between friction and reaction forces, as well as measuring the coefficient of friction on different surfaces."
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Presentation Transcript
Friction Equilibrium
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? ?) ? = 50? 150sin25 = 426.6 ? 150 cos 25 150 sin 25 150 R Fr 250 ?) ?? = 150cos25 ?? = 135.9 ? W = 50 9.8 = 490 ?? = ? =?? ?=135.9 426.6= 0.319
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 if friction is at its maximum ?) ? = 50? 150sin25 = 426.6 ? 150 cos 25 150 sin 25 150 R ?) ?? = 150cos25 = 135.9 ? Fr 250 ?) ?? = ? =Fr R=135.9 W = 50 9.8 = 490 426.6= 0.319
Measuring the coefficient of Friction Aim: To find values for the coefficient of friction of different surfaces Board calculator Equipment: 1 kg weight Carpet, smooth board, sandpaper, Big protractor 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
WB3. 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 280 Weight
WB3 Object on a slope about to go up the slope solution i) R = 29.4cos 28 = 25.96 ii) Friction = R = 0.3 x 25.96 = 7.79 N 280 29.4 sin 28 29.4 cos 28 280 iii) Push =friction + component of weight downslope Push = 7.79 + 29.4 sin 28 = 21.59 N
WB4 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 280 Weight
WB4 Object on a slope about to go down the slope solution i) R = 78.4cos 28 = 69.2 ii) Equilibrium Friction= 78.4 sin 28 = 36.8 N 78.4 sin 28 78.4 cos 28 280 280 iii) Friction = R =36.8 69.2= 0.532 Weight = 8 x 9.8=78.4
WB5. 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 Push 300 Weight
WB5 (cont). Simultaneous equations question solution part 1 P cos 30 P sin 30 300 300 300 Weight = 3 x 9.8=29.4 29.4 sin 30 29.4 cos 30
WB5 (cont). Simultaneous equations question solution part 2 P cos 30 P sin 30 R = 29.4cos30 + Psin30 (2) Friction = R = 1 3R (3) 29.4 sin 30 29.4 cos 30 29.4sin30 + Fr Pcos30 = 0 (1) 300 So 29.4sin30 +1 3R Pcos30 = 0 (1) solve simultaneously gives So 29.4sin30 +1 329.4cos30 + Psin30 Pcos30 = 0 P = 33.2 N (and R = 42.06 N)
WB6. 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 Push 300 Weight
WB6. rough slope in equilibrium, object is about to go down the slope solution P cos 30 P sin 30 300 300 300 10 sin 30 10 cos 30
WB6. rough slope in equilibrium, object is about to go down the slope solution R = 10cos30 + Psin30 (1) P cos 30 P sin 30 300 Fr + Pcos30 = 10sin30 (2) 300 Fr = R = 0.3 R (3) 300 Solve simultaneously 10 sin 30 10 cos 30 0.3 10cos30 + Psin30+ Pcos30 = 10sin30 (2) 0.3(8.66 + ?) + 0.866 ? = 5 ? = 2.36 ? = 9.84 ?? = 2.95
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
