Multiple choice
Have a go at the two multiple choice questions at the top
FrictionAims:
•To know when friction acts.•To know how to calculate friction in different situations.
•To consider friction problems on slopes.
When does friction act?
• On any object that is moving on a rough surface.
• On any object you are trying to move on a rough surface.
• No friction acts on a smooth surface.
What affects the size of friction?
• What the two materials in contact are made from.
• The reaction force between the object and the surface.
• The size of the force you are applying to the object to make/try to make it move.
Static Friction• We have already said friction acts when you
try to move an object.• This can take any value up to a certain point to
try and keep the object in equilibrium.
• When an object is on the point of moving it is said to be in limiting equilibrium.
f Rf = Friction; R - normal Reaction; is the coefficient of friction
(material dependant)
F = R
Values for
>00
No friction Very slippery
Rough
ExamplesSteel = 0.75 Teflon = 0.04Wood = 0.6
Example 1
(i) Draw a force diagram.
(ii) State the set of values for P when the block is stationary.
(iii) What is the value of P when the block is in limiting equilibrium?
(iv) State the set of values for P when the object is in motion.
Example 1
(i) Draw a force diagram.
(ii) State the set of values for P when the block is stationary.
(iii) What is the value of P when the block is in limiting equilibrium?
(iv) State the set of values for P when the object is in motion.
Friction Force P
Weight
Reaction
Example 1
(i) Draw a force diagram.
(ii) State the set of values for P when the block is stationary.
(iii) What is the value of P when the block is in limiting equilibrium?
(iv) State the set of values for P when the object is in motion.
RF
4.29
8.956.0
4.29P4.29P
4.29P
Example 1
(i) Draw a force diagram.
(ii) State the set of values for P when the block is stationary.
(iii) What is the value of P when the block is in limiting equilibrium?
(iv) State the set of values for P when the object is in motion.
RF
4.29
8.956.0
4.29P
Example 2
gRF 1071
14F 12.1230cos14
No, requires a force above 14N to overcome friction
Example 3
Example 3
R45cos49
8.910
45cos493535.0
Kinetic Friction
• When an object is moving then friction also acts.
• Friction is constant and has the maximum possible value when in motion.
F = R
Example 4A particle on a rough surface weighing 5kg is being pulled by force of 80N at 20o the horizontal.The coefficient of friction is 0.3 between the particle and the surface.(i) What is the normal reaction force?(ii) How big is the force of friction?(iii) At what rate does the particle accelerate?
Example 4A particle on a rough surface weighing 5kg is being pulled by force of 80N at 20o the horizontal.The coefficient of friction is 0.3 between the particle and the surface.(i) What is the normal reaction force?(ii) How big is the force of friction?(iii) At what rate does the particle accelerate?
NmgR 497.14493.0 RF N5.607.1420cos80
ma5.60 1.12a
Friction and Slopes
• When an object is placed on a slope then friction will act. (unless the slope is smooth)
• Usually friction will act up the slope as the object will be trying to slide down.
• When will this not be the case?
• When the object is moving up the slope.
Example 5A 5kg sledge is sliding down a 600 slope, the coefficient of friction between the sledge and the slope is 0.3.(i) Find the normal reaction force.(ii) Find the size of the frictional force.(iii) Find the acceleration of the sledge.
Example 5A 5kg sledge is sliding down a 600 slope, the coefficient of friction between the sledge and the slope is 0.3.(i) Find the normal reaction force.(ii) Find the size of the frictional force.(iii) Find the acceleration of the sledge.
5.2460cos mgR35.7 RF
1.3535.760sin mg
ma1.35 7a
Example 6An object of mass 2.8kg is pulled up a rough slope inclined at 40o by a light string parallel to the slope. The coefficient of friction between the object and slope is 0.3.Find : the normal reaction and the tension in the string if … (a) it moves up the slope with constant speed(b) it accelerates at 1.5 ms-1
2140cos mgR3.6 RF 3.6T
2.48.25.1 ma
3.62.4 T 5.10T
Questions
• Draw a sketch for each question.
• You should find your answer on the bottom of the sheet.
• Work carefully.
PlenaryA force of magnitude 20N is acting downwards at 25˚ to the horizontal on a block of mass 4kg, which is at rest in limiting equilibrium on a horizontal surface. Calculate the coefficient of friction between the block and the surface. The direction of the force of magnitude 20N is now reversed. Calculate the acceleration with which the block starts to move.
Independent Study
Exercise F p95 (solutions p155)