Home PHYSICS TOPIC 2: FRICTION | PHYSICS FORM 3

TOPIC 2: FRICTION | PHYSICS FORM 3

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TOPIC 2: FRICTION | PHYSICS FORM 3 APPLICATION OF VECTORS PHYSICS FORM THREE NOTES

FRICTION​​

​​ Friction Is the force which opposes the relative motion or tendency of  ​​ ​​ ​​ ​​​​ motion between two surfaces in contact​​ 

OR  ​​ ​​ ​​ ​​​​ 

Is the force that opposes motion between any surfaces that are in contact ​​ ​​ Friction force which occurs in fluids is known as​​ viscosity​​ ​​ 

​​ How friction happens​​ ​​ 

Friction is caused by molecular adhesion, surface roughness and deformations  ​​ ​​ ​​ ​​ ​​ ​​​​ (Adhesive bond, mechanical bond and deformation)​​ 

​​ Advantage of friction​​ ​​ 

It aids in walking and movement ​​

Helps moving body to stop by applying brakes ​​

Used to wear unneeded layers of some materials ​​

Causes lighting in match stick ​​

Supports life on the earth by preventing burning asteroids​​

Causes nail to stick on the wood ​​

Enables bottle stopper to stick on the bottle neck ​​

​​Disadvantage of Friction​​ ​​ 

It causes wear and tear ​​

It produces heat in various machine parts causing efficiency to decrease  ​​​​

Due to friction noise is produced in machine​​

Causes loss of energy in form of heat and sound  ​​​​

Causes motion of a body to slow down  ​​​​

Heat produced by friction can cause appliance to burn ​​

It causes wounding, when skin wearing ​​

​​Methods of increasing Friction ​​ 

Increasing the normal force by increasing the weight of the body ​​

Increasing the roughness of the surface ​​

Use materials of high coefficient of friction. Example, rubber band ​​

Scrubbing equipment is made rough to increase friction e.g. Steel wire for scrubbing “surface”​​

​​Methods of reducing Friction ​​ 

Place roller between the two rough surfaces ​​

Use ball bearings ​​

Use of lubricants e.g. oil, water ​​

Speedy material​​ (eg Teflon)​​ which have low coefficient of friction and thus slide easily. ​​ 

Make surface soft or Polishing​​

Using streamlined bodies eg airplane, boat etc​​

Normal Force and Limiting Friction ​​ 

​​ Consider the fig. below​​ 

​​Normal​​ force, ​​ is the force which is equal and opposite to the weight of the body. ​​ 

The Normal force is always ​​ perpendicular to the surface on which it rests ​​

Limiting​​ friction:  ​​ ​​​​ Is the maximum possible value of static friction ​​ NB:​​ ​​ 

From the fig above.​​ R = mg ​​​​ 

When the body is at rest​​ FF​​ = FA ​​​​ 

When the body is at motion​​ FF​​ ≠ FA​​ so we have to find net force, F ​​ 

When the body starts to move static friction force is equal to limiting friction then the minimum force applied tends to start the motion ​​

When the body starts to move kinetic friction force is not equal to the minimum force applied, then the body tends to start the motion ​​

If limiting friction is less than the force applied, the body will move ​​

If limiting friction is greater than the force applied, then the body cannot move ​​

​​ Laws of friction forces​​ ​​ 

Frictional force is directly proportional to the normal force between the two surfaces  ​​ ​​ ​​ ​​ ​​​​ in contact. ​​ (Fr α R)​​ ​​ 

Friction depends on the nature (roughness) of surfaces in contacts. ​​

Friction does not depend on the surface areas in contact. ​​

Frictional force is independent of the speed once an object has been set in motion ​​

 ​​​​Types of Friction​​ ​​ 
o​​ Static friction  ​​​​ 

o​​ Dynamic friction ​​ 

Static Friction​​ ​​ 

Is the friction which occurs when the two objects are not moving relative to each other​​

This force causes some bodies to be stationary .Example, A book can be kept on top of desk without dropping down  ​​​​

​​​From:​​  ​​ ​​ ​​ ​​​​ 

​​ Coefficient of static friction​​ is the ratio of limiting friction force to the normal reaction​​ ​​ ​​Dynamic (kinetic) Friction ​​ ​​ 

​​ Is the friction that occurs when objects are moving relative to each other​​ and rub against each other​​ 

From:​​  ​​ 

​​ ​​​​ Coefficient of kinetic friction is the ratio of kinetic friction force to the normal reaction. against each other​​ 

Whereby:​​ ​​ 

Fr = kinetic/static friction force ​​

R = normal reaction ​​

μ = coefficient of static/kinetic friction force​​

NB:​​ ​​ When a body rolls on the surface of another ,the form of kinetic friction that exists between the surfaces is called​​ ‘ROLLING FRICTION’’​​ 

Sliding Friction​​ is the kind of kinetic friction that ​​ is caused by two bodies rubbing or sliding against each other ​​​​ 

It is easy to roll a body than to slide it on the ground ,This is because Rolling friction is always less than Sliding friction​​ 

The coefficient of kinetic friction is always less than the coefficient of static friction ​​ 

​​ Example:​​ 

1. A block of mass 500g is pulled along a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.8. What is the friction force acting on the block as it slides?  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​

Soln:​​ 

Given: m = 500 g = 0.5 kg,​​ 𝝁𝒌​​ = 𝟎.𝟖, 𝑭 =?​​ 

From:​​ 

Individual Task​​ – 2:1 ​​ ​​​​ 

A block of mass 270kg is pulled along a horizontal surface. If the coefficient of kinetic friction between the block and the surface is 0.4. What is the friction force acting on the block as it slides?  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ (ANS:  ​​​​ Fr = 1, 080N)​​ ​​ 

A box of mass 2kg rest on a horizontal surface, a force of 4.4 N is required to just start the box moving. What is the coefficient of static friction between the block and the surface?  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ (ANS: ​​ μ = 0.22) ​​ 

An alluminium block of mass 2.1kg rests on a steel platform. A horizontal force of 15N is applied to the block ​​

Given that coefficient of limiting friction 0.6, will the block ​​ move? ​​

If will move, what will be its acceleration. Given that coefficient of kinetic  ​​ ​​ ​​​​   ​​​​   ​​ ​​ ​​​​ friction is 0.47 (ANS: (a) Since:​​ mg > Fr, hence the car will move ​​ (b) a=2.44 m/s2) ​​ 

A brick starts sliding with 6m/s across a concrete horizontal surface floor and the coefficient of friction between the two surfaces is 0.4. How far will it travel before coming to rest?  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ (ANS:  ​​ ​​ ​​​​ S = 4.5 m)​​ 

Find the static friction between a block of wood of mass 10kg placed on a table. A minimum force of 50N is required to make the block just move on the top.  ​​​​ (ANS: μ = 0.5) ​​ ​​​​ ​​ 

Friction force at Inclined Plane​​ ​​ 

Consider the diagram below, a body of mss,​​ m​​ sliding down the inclined plane at​​ 𝜽​​ 

​​

​​ When the object begins to slide (from rest) there will be kinetic friction between the object and the incline .Thus the net force will be: ​​ ​​ ​​​​ 𝑭𝒏𝒆𝒕​​ = 𝑭𝑨​​ − 𝑭𝑹​​ 

Whereby:​​  ​​ ​​​​ 𝑭𝑨​​ = 𝒎𝒈𝒔𝒊𝒏𝜽 ​​ ​​​​ and ​​ : ​​ ​​ ​​​​ 𝑭𝑹​​ = 𝑭𝒌​​ = 𝝁𝒌𝑹 = 𝝁𝒌𝒎𝒈𝒄𝒐𝒔𝜽 ​​​​ ​​ 

𝑭𝒏𝒆𝒕​​ = 𝒎𝒈𝒔𝒊𝒏𝜽 −​​ 𝝁𝒌𝒎𝒈𝒄𝒐𝒔𝜽 = 𝒎𝒈(𝒔𝒊𝒏𝜽 − 𝝁𝒌𝒄𝒐𝒔𝜽) ​​​​ ​​ 

​​Thus acceleration will be given as: From  ​​ ​​​​ 𝑭𝒏𝒆𝒕​​ = 𝒎𝒂 = 𝒎𝒈(𝒔𝒊𝒏𝜽 − 𝝁𝒌𝒄𝒐𝒔𝜽) ​​​​  ​​ ​​ ​​​​ 

Therefore,​​ the acceleration is given by ,​​ 

𝒂 = 𝒈(𝒔𝒊𝒏𝜽 − 𝝁𝒌𝒄𝒐𝒔𝜽),​​ For downward motion​​ 

​​ 𝒂 = 𝒈(𝒔𝒊𝒏𝜽 + 𝝁𝒌𝒄𝒐𝒔𝜽),For upward motion (motion is against the gravity)​​ 

When Fr​​ = 0 (when the incline is frictionless) ​​ ,​​ 𝒕𝒉𝒆𝒏,  ​​​​ 𝒂 = 𝒈𝒔𝒊𝒏𝜽 ​​​​ 

At constant speed (at rest), a = 0 m/s2,  ​​​​ 𝑭 = 𝒎𝒈𝒔𝒊𝒏𝜽 𝑎𝑛𝑑 𝑹 = 𝒎𝒈𝒄𝒐𝒔𝜽 ​​​​ 

Example:​​ 

1. A block of wood of 4kg just slides without acceleration down an inclined plane of 400​​ to the horizontal. What is the coefficient of dynamic friction?​​ 

Soln:​​ 

Given: m = 4 kg,​​ ​​ = 400​​ 

Consider the fig below​​

​​ From:​​ 𝑨𝒕 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒔𝒑𝒆𝒆𝒅 (𝒂 = 𝟎), ​​​​ ​​ 𝝁 = 𝒕𝒂𝒏𝜽​​ 

Consider the fig below​​ 

​​ From:​​ 𝑨𝒕 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒔𝒑𝒆𝒆𝒅 (𝒂 = 𝟎), ​​​​ ​​ 𝝁 = 𝒕𝒂𝒏𝜽​​

​​ 𝝁 = 𝒕𝒂𝒏𝜽 = 𝒕𝒂𝒏𝟒𝟎 = 𝟎. 𝟖𝟒​​ 

​​ Individual task – 2:2​​ 

A mass is placed on an inclined plane such that it can move at constant speed, when slightly tapped. If the angle of the plane makes with the horizontal plane is​​

300. Find the coefficient of kinetic friction. ​​ (ANS: ​​ μ = 0.56) ​​ ​​ 

A mass of 5 kg is placed on a plane inclined at an angle of 300​​ to the horizontal. What is the accelerating force required to pull the mass up the plane if the coefficient of friction is 0.5?​​ (ANS: FA​​ = 46.65N)​​ 

A block of wood of mass 5kg is placed on a rough plane inclined at 600. Calculate its acceleration down the plane if coefficient of friction between the block and the plane is 0.32  ​​​​ (ANS:  ​​​​ a = 7.1 ms-2)  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 

Class Activity–2 ​​ 

Calculate the coefficient of kinetic friction between the surface of a table and a block of wood when 5 kg block of wood is moving on the table and experiencing a frictional of 5 N. (ANS:​​ 𝝁 = 𝟎.𝟏)​​ 

A box weighing 2 kg is at rest on a wooden floor. The coefficient of static friction is 0.6 and the coefficient of kinetic friction is 0.35.​​

  • What minimum force is required to start the box sliding?​​
  • What minimum force is required to keep it sliding at a constant velocity?​​

A 12 kg box is being pulled across a level floor by a force of 60 N. If the acceleration ​​ of the box is ​​ 2 ms-2​​ ,What is the force of friction between the box ​​ and the floor ​​ 

A 0.5 kg object is given an initial velocity of 3 m/s after which it slides a distance of 8 m across a level floor. What is the coefficient of kinetic friction between the object and the floor?​​

The coefficient of kinetic friction between a block of wood and a wooden inclined plane at an angle of 400​​ is 0.126. If the friction acting on the sliding prism is 42 N, calculate the mass of the prism.(ANS: mass = 43.4 kg)​​ 

Calculate the friction force acting on a carton box of mass 9 kg which is moving over a surface .The coefficient of kinetic friction between the two surfaces is given as 0.45. (ANS: FR​​ = 40.5 N)​​ 

The coefficient of friction between a particle of mass 8 kg, and a rough horizontal plane is 0.4. Given that a horizontal force of 29 N acts on the particle as shown in the figure below. Would it start to move​​ (ANS: Since FA(29 N) < FR(32N), No motion)​​ 

​​A wooden block of mass 8 kg is resting on a wooden table. If the coefficient of static friction between the pair is 1.3 , Calculate the minimum horizontal force required to just slide the ​​ box .Given that g = 10 N/ kg​​ 

A 3 tones lorry is resting on a tarmac road .The lorry requires a minimum force of 12000 N in order for it to just move .Determine the coefficient of static friction between the lorry’s tyres and the road​​

A crate of soda with mass 40 kg will just begin to slide with constant speed down a rough ramp (slope) at 300​​ to the horizontal .What is the coefficient of static friction. (ANS:​​ 𝝁 = 𝟎.𝟓𝟕𝟕𝟒)​​ 

A boy applies a horizontal force of 12 N on a metal solid block of mass 3.4 kg resting on a concrete floor .Given that the coefficient of static friction between the pair is 0.56 and ​​ g = 10 N/ kg​​ 

A box of mass 5 kg is at rest on a wooden floor. If the coefficient of static friction between the box and the floor is 0.6, what minimum external force is required to set the box sliding?​​

Define the following (a)  ​​​​ Rolling friction  (b) Sliding Friction​​ 

A 0.5 kg object is given an initial velocity of 3 m/s after which it slides a distance of 8 m across a level floor, What is the coefficient of kinetic friction between the object and the floor?​​

A box weighing ​​ 2 kg is at rest on a wooden floor .The coefficient of static friction is 0.6 ​​ and the coefficient of kinetic friction is 0.35​​

  • What minimum force is required to start the box sliding ​​
  • What minimum force is required to keep it sliding at a constant velocity?​​

In a car, The brakes stop the tyres while friction between the tyres and the road surface stops the car .On a wet road the coefficient of kinetic friction between the road surface and the tyre is 0.1 .Two cars, A and B, are travelling at a speed of​​

15 m/s and 30 m/s ,respectively .Brakes are suddenly applied on each of the cars .How far will each of the cars travel before coming to rest?​​

A rectangular box of mass 10 kg rests on an incline with a coefficient of static friction of 0.55 ​​ and coefficient of kinetic friction of 0.25 (a) At what angle will the box begin to slide?​​ 

(b) If the incline is kept at that angle after the box begins to slide, what will be the box‘s acceleration?​​ 

The coefficient of kinetic friction between the tyres of a car and the road is 0.7. The car brakes are applied and it travels a distance of 120 m before stopping .What was the car’s velocity just before the brakes were applied?​​

A box of mass 5 kg is at rest o a wooden floor. The coefficient of static friction is 0.42 and the coefficient of dynamic friction is 0.15. Find its acceleration if a force of :  ​​ ​​ ​​ ​​​​ (a) 15 N is applied to the box  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ (b) 25 N is applied to the box ​​ 

A 42 kg refrigerator is sitting on the back of a stationary pick – up .The coefficient of static friction between the refrigerator and the pick – up bed is 0.44 .At what rate can the pick – up accelerate without the refrigerator sliding off the back?​​

A 6 kg mass is resting on a horizontal surface .It is determined that a force of 20 N will start the object sliding and keep it sliding with an acceleration ​​ of ​​ 0.83 m/s2​​ .What are the coefficients of static and kinetic friction between the mass and the surface ?​​ 

What is the normal normal reaction of the body of mass 10kg placed on an inclined plane of angle 30˚c?​​

A concrete block of mass 10kg rests on a table. It is found that when a horizontal force of 4kg weight pulls the mass, it is just begins to slide on the table. Find the coefficient of static friction​​ 

A block of wood rests on a sloping plank which makes an angle of 31˚ with the horizontal. If the block suddenly begins to slide down hill, what is the coefficient of static friction?​​

A box of mass 50 kg is dragged on a horizontal floor by means of a rope tied to its front. If the coefficient of kinetic friction between the floor and the box is 0.30, what is the force required to move the box at uniform speed? (ANS: F = 150 N)​​ 

A car of mass 1200 kg is brought to rest by a uniform force of 300 N, in 80 sec. What was the speed of the car? (ANS: u = 20 m/s)​​ 

A loaded trailer weighing 10kg is being towed across level ground .The coefficient of dynamic friction is 0.25. What is the frictional force of the trailer?​​ 

A block of wood just slides without acceleration down an inclined plane of 25˚ to the horizontal. What is the coefficient of dynamic friction?​​

Define the following terms  ​​ ​​ ​​ ​​​​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​(a) Limiting friction  ​​​​ (b) Normal reaction ​​ (c) Viscosity (d) Coefficient of Friction​​ 

   (a) ​​ State the laws of friction  ​​​​ (b) Explain, why Friction is friend and foe?​​

A​​ brick is sliding at 8m/s across a concrete horizontal surface floor and the coefficient of friction between the two surfaces is 0.5 How far will it travel before coming to rest?​​ ​​

Show that the acceleration of a stone sliding at a velocity ,v across a concrete horizontal surface floor is given by a =​​ 𝜇𝑔​​ where​​ 𝜇​​ is the coefficient of friction between the stone and the floor and g = acceleration due to gravity​​ 

A mass of 5 kg is placed on a plane inclined at an angle of 300​​ to the horizontal .What is the accelerating force required to pull the mass up the plane if the coefficient of friction is 0.5?​​ 

A block of metal with a mass of 20 kg requires a horizontal force of 50 N to pull it with uniform velocity along a horizontal surface. Calculate the coefficient of friction between the surface and the block. (ANS:​​ 𝝁 = 𝟎.𝟐𝟓)​​ 

A Car of weight 1000 N is moving with uniform speed .If the kinetic friction acting on the car is 500 N , calculate the coefficient of kinetic friction ​​

A wooden box of mass 30 kg rests on a rough floor. The coefficient of friction between the floor and the box is 0.6. Calculate​​

The force required to just move the box. (ANS: FF​​ = 180 N, a = 0.67 m/s2)​​ (b) If a force of 200 N is applied to the box, with what acceleration will it move?​​ 

Describe how friction is minimized by the following methods:  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 

(a)Lubrication  ​​​​(b) Use of bearings  ​​(c) Streamline flow​​ 

A boy is pulling a box of mass 10 kg. What is the normal force and the frictional force if the coefficient of static friction? ( g = 10 N/kg)​​

A 50 g mass is placed on a straight track slopping at an angle of 450​​ to the horizontal as shown from the figure below calculate ​​ 

  • Acceleration of the load as it slides down the slope​​ 
  • The distance moved from rest in 0.2 seconds​​ 

​​ A 5 kg block is resting on a horizontal surface .Given that the coefficient of static friction is 0.57 ​​ g = 10 N/kg​​ 

  • What is the frictional force required to just move the block?​​ 
  • What force must be applied to the block to keep it moving at constant velocity?​​ 
  • Determine its acceleration if a force of 35 N is applied ​​ ​​​​ 

A force of 8.0 N gives a 3.0 kg mass an acceleration of 0.6 m/s2​​ to the right ​​ (a)What is the limiting friction on the block?​​ 

(b) Determine the coefficient of static friction required to produce a net kinetic force of 6.0 N? (g =10 N/kg)​​ 

A 53.0 kg block slowed by friction has an acceleration of  ​​​​ -0.1 m/s2​​ .Determine ​​ the force of friction on the block​​ 

A 10.0 kg solid sliding along a horizontal surface is brought to rest after 30 minutes  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​​​ 

​(a) Name the force that caused it to stop  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ (b) Determine the magnitude of the force that caused it to stop  ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​ ​​(Given that:  ​​​​ 𝝁𝒌​​ = 0.45 , g = 9.8 N/kg

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