PHYSICS S.S ONE (FRICTIONAL FORCES)

FIRST TERM: WEEK 4
TOPIC: FRICTIONAL FORCE

Friction  is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose the motion of one over the other.

Laws of solid frictions

1.       Friction opposes the relative motion between two surfaces in contact. It acts in a direction opposite to that of the motion.

2.       The force of friction increases to the same extent as the force which tends to start the motion.

3.       Frictional force depends on the nature of the two surfaces in contact.

4.       It is independent of the area of the surface in contact.

5.       It varies directly with the normal force pressing the surface together i.e it is proportional to the normal reaction R.

STATIC AND DYNAMIC FRICTION

https://www.blogger.com/img/img-grey-rectangle.png

Fs- frictional force;  N- normal reaction; w- weight (mg) and F- applied force.

From the above, frictional force is directly proportional to the normal reaction. i.e fs ; fs = u N. where u is the coefficient of static friction.

Also at equilibrium N = w = mg

COEFFICIENT OF STATIC FRICTION (u) IN AN INCLINED PLANE.

                U = tan θ =

                                Θ = angle inclined to the horizontal.

 

 

In diagram a downward motion

W > T

So  w = T + something

      W = T + ma

Also upward motion write fs

T > fs

T = fs + something

T = fs + ma

T = uN + ma

EXAMPLES

1. A car weighs 200N rest on rough surface, when the car is about to move with a force 40N from the engine. Calculate the coefficient of friction.

Solution

W= 200N; fs= 40N; u = ?

U =   =  = 0.2

2. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?
Solution
First let us find the forces acting on the block
(i) the weight mg acting vertically downwards;  (ii) the normal force N of the plane on the block
(iii) the static frictional force fs opposing the impending motion.
In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have
mgsinθ=fs -(1);                                 mgcosθ=N -(2)
As θ increases, the self-adjusting frictional force fs increases until at θ=θmax, fs achieves its maximum value, fsmax=μsN .
Now dividing equation (1) by (2)
tanθmax=     or       θmax=tan−1μs
When θ becomes just a little more than ?max , there is a small net force on the block and it begins to slide. Note that θmax depends only on μs and is independent of the mass of the block.
For θmax=150,
μs=tan15
=0.27
3. What is the acceleration of the block and trolley system shown in below figure if the coefficient of kinetic  friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g =10 m/s-1). Neglect the mass of the string.
Solution
in a diagram a downward motion

W > T

    W = T + ma

   3 X 10  = T  + 3a   ----- (1)

Also upward motion write fs

T > fs

T = fs + ma

T = uN + ma

T = (0.04X20X10) + 20a

T= 8 + 20a    --------(2)
equate (1) and (2)

30 = 8 +20a +3a

30-8 =23a

a = 22/23 = 0.96ms-2.

T= 8 + (20X0.96);   T = 8+19.1; T= 27.1N

ADVANTAGES OF FRICTION

1.       Friction allows us to work or stop walking after getting started.

2.       It enables the automobile tyres to make a firm grip with the roadway.

3.       Used in fan belts used over wheels or pulley in machinery.

4.       It allows the brake to stop the car.

5.       Used in grindstone to sharpen knives and chisels.

6.       helps in Lighting a match stick etc

DISADVANTAGES OF FRICTION

1.       It brings about wear and tear on the moving part of machinery.

2.       Friction causes loss of energy in machinery, therefore reduce their efficiency.

3.       It causes heating of engines.

4.       Produce unwanted sound

METHODS REDUCING FRICTION

1.       The use of lubricants like grease, oil, air and graphite.

2.       Use of ball or roller bearing

3.       The streamlining of body shapes of moving objects.

PRESENTATION

Step I: The teacher introduces the topic to the students

Step II: The teacher states the law of solid friction

Step III: The teacher leads the students to solve problems on solid friction.

Step IV: The teacher states the advantages and disadvantages of solid friction

Step V: The teacher allows the students to ask questions.

EVALUATION

The teacher evaluates the lessons by asking the following questions:

1.       Define friction.

2.       State the law of solid friction.

3.       State some advantages and disadvantages of friction.

4.      Mention ways of reducing friction

  ASSIGNMENT

A car weigh 300N rest on a smooth surface, when the car is about to work a force 25N from the engine. Calculate the coefficient of the friction.

 

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