Force Definition And Its Calculations

During our last class session in physics, we discussed about types of frictional forces and proposed to what are forces in next class, well this is the proposed what are forces class session.

what are forces

Previous Topic : Viscosity And Surface Tension

Forces

What are forces or What is a force? Force is an act which changes the state of rest of a body or its motion in a straight line.
Simply, force can make a body to change its position, by making it to move or it can bring a body in motion to rest, force can also mean a push or a pull. This implies that, there is a relationship between force and motion,
A force is a vector quantity expressed in Newton which is represented by N, a vector quantity is a quantity that has both magnitude and direction.top↑

Newton of a force is defined as what is required to give a unit kilogram mass, and an acceleration of one metre per second square, which has a relation of;
F = ma
F = force (N)
m = mass (kg)
a = acceleration (m/s²)

Types of force

a] Contact force

b] Force field

c] Nuclear force.

Contact Force

a] Contact force : Contact force is a type of force which occurs when two surfaces are in contact.

E.g Frictional force and reactional force.

Frictional Force

1] Frictional force : Frictional force is the force that opposes relative motion, when two bodies are in contact, that is, it is a force that opposes a surface to move freely without slipping.top↑

horizontal force

In the above image, a block is placed on an horizontal plane, a force F is applied to the block, the force F is supposed to be able to move the block W but there is a force F° holding the block  causing the force F to not be sufficient to move the block, that force F° is called Frictional force. Also, in an iinclined plane;top↑

inclined Force

Frictional Force

Types Of Frictional Force

a] Solid frictional force.
b] Liquid frictional force.

Solid Frictional Force

a] Solid friction: Solid friction is a frictional force that occurs between the surface of two solid in contact.

Types Of Solid Friction

a] Static or limit friction
b] Dynamic or kinetic friction.

Static Or Limit Friction

a] Static friction: Static friction or limit friction is the frictional force that occurs when a body is at rest.

E.g
1) Frictional force acting on a stationary vehicle.

2) Frictional force acting on a person standing, etc.

Dynamic Friction

b] Dynamic friction;

Dynamic friction or kinetic friction is the frictional force that occurs when a body is in motion.top↑
E.g
1) A moving vehicle

2) when a person is moving, Generally when a body is in movement.

Advantages of friction

1] It enables a person to walk freely without slipping, because it offers a firm grip between the sole of the shoes and the ground,

2] it enables rolling of a car tyre,

3] it aid rolling of a fan belt of a machine,

4] it enables a nail to hold a wood onto a wall,

5] it is also use as domestic use, in some countries where they use hand to grind, where they use a flat shaped rock and a small size rock to grind, e.g pepper and tobacco, etc.top↑

Disadvantages of fiction

1] it causes tear and wear, it is the friction that causes the sole of a shoe to tear.

2] it reduces the efficiency of a machine, that is why, the efficiency of a machine is not up to 1, Efficiency < 1 But when there is frictionless, that is, when there is no friction, Efficiency = 1

Reduction of friction

Friction can be reduced by;
1] by lubricating,

2] applications of rollers and gliders,

3] by streamline shape, etc.top↑

Law Of Frictional Force

1] Frictional force opposes the relatively motion.

2] Frictional force is directly proportional to the normal reaction,
F = µR
F = Frictional force = mg
µ = constant = coefficient of friction
R = normal reaction = mg

3] Frictional force is dependant on the nature of the surface of the bodies in contact

4] Frictional force is independant on the area of the surface of the bodies in contact

5] Frictional force is independant on the relative motion (velocity) of the body in contact.

Formulae of friction

inclined Force

Note : weigth acts downward top↑

force.jpg

coefficient of friction expression

In the above triangle, looking for θ,
Using SOH CAH TOA

Sin θ = Opp
             Hyp
 

Sin θ =
             W

Making R the subject of the formula,
F° = WSin θ

Back to the triangle ;
Making use of CAH to find θ

Cos θ = R
              W

Making R the subject of the formula;
R = WCos θ

Remember in the 2nd of law of friction,
F° = µR

µ =
      R
If F° = WSin θ and R = WCos θ,
then;

µ = WSin θ
       WCos θ
In mathematics under trigonometry;

Tan θ =  Sin θ
               Cos θ
Hence;
µ = Tan θ top↑

2nd formula;

Frictional force = Applied force – net reaction

F° = Frictional force
F = Applied force
Net reaction = mgSin θ
Hence;

F° = F – mgSin θ
F° + mgSin θ = F
No angle, F° + mg = F

3rd formula

When tension is apply

Two forces

M2g – T = M2a …. … (eqn 1)
T – M1g = M2a ….. .. (eqn 2)

If friction is applied
M2g – T = M2a ……… (eqn 3)
T – µM1g = M1a ……. (eqn 4)

In an inclined plane; top↑

Force Tension

M2g – T = M2a ……… (eqn 5)
T – M1gSin θ = M1a……… (eqn 6)

If friction is applied;
M2g – T = M2a ……… (eqn 7)
T – µM1gSin θ = M1a …….. (eqn 8)

Example 1

A block of mass 2kg resting on the surface of a rough horizontal plane, if the force sufficient to drag the block over the surface of the plane is 30N calculate the frictional force acting on the block,
Ii) calculate it frictional force when it is inclined at an 30°.top↑

Solution

Mass = 2kg
g = acceleration due to gravity = 10 m/s2
F = 30 N
F° = ?
F° = F – mg
mg = 2 × 10 = 20N
F° = F – mg
F° = 30 – 20
F° = 10N

ii) F° = F – mgSin θ
F° = 30 – 20Sin 30
Sin 30 = 0.5 or ½
F° = 30 – (20 × ½)
F° = 30 – 10
F° = 20

Example 2

A body of weight 20N rest on a rough horizontal plane when the body just begin to slip, the applied force is found to be 40N. Calculate the coefficient of friction between the two surfaces.top↑

Solution

Weight of the body = 20N = mg
Applied force = F = 40N
µ = ?

µ =
      R

F° = F – mg
F° = 40 – 20
F° = 20N
Hence;

µ = 20
       20
                    µ = 1
Note: coefficient of friction has no unit.

Example 3

Two force

A mass of 2 kg is connected to a second mass of 4 kg over a frictionless pully, if the acceleration due to gravity and the coefficient of friction are 9.8 m/s² and 0.5 respectively, calculate the acceleration at which the masses will accelerate.top↑

Solution

M1 = 2 kg
M2 = 4 kg
g = 9.8 m/s²
µ = 0.5
a = ?

M2g – T = M2a ……… (I)
T – µ M1g = M1a ……….. ( ii)
T is useless in this example, so make T the subject of the formula in eqn ii.
Make T the subject of the formula in eqn ii
T – µ M1g = M1a
T = M1a + µM1g
Substitute T = M1a + µM1g in eqn I
M2g – T = M2a ……. eqn I
M2g – ( M1a + µM1g) = M2a
Open the bracket
M2g – M1a – µM1g = M2a
Substitute the parameters
4 × 9.8 – 2 × a – 0.5 × 2 × 9.8 = 4 × a
39.2 – 2a – 9.8 = 4a
Collect the like terms
39.2 – 9.8 = 4a + 2a
29.4 = 6a
a = 29.4/6
a = 4.9
Ans = 4.9 m/s²

Example 4

Force T

A body of mass 10 kg on a smooth inclined plane is connected over a smooth pulley to a mass of 15 kg, calculate the acceleration of the system.top↑

Solution

M1 = 10 kg
M2 = 15 kg
g = 10 m/s²
a = ?
T = is useless ( in calculation)

No friction
M2g – T = M2a ……… (I)
T – M1gSin θ = M1a ………. (ii)

Make T the subject of the formula in equation ii
T – M1g = M1a
T = M1a + M1gSin θ
Substitute T = M1a + M1gSin θ to eqn I

M2g – T = M2a ….. eqn I
M2g – ( M1a + M1gSin 30) = M2a
Open the bracket
M2g – M1a – M1gSin 30 = M2a
Sin 30 = 0.5
Substitute the parameters
15 × 10 – 10 × a – 10 × 10 × 0.5 = 15 × a
150 – 10a – 50 = 15a
Collect the like terms
150 – 50 = 15a + 10a
100 = 25a
a = 100/25
a = 4 m/s²
Ans = 4 m/s²

top↑

Example 5

Two force

In the above diagram, the hanging mass M2 is adjusted until M1 is on the verge of sliding. Calculate the coefficient of static friction between the mass M1 and the table.

Solution

µ =
     R

F° = F – µM1g
In the question, it was stated that M1 is on the verge of sliding, that means, frictional force will not be able to act on M1 because on a verge of sliding means the object can freely fall.
Hence F° = 0
F° = F – µM1g
0 = F – µM1g
µM1g = F

In the question, it was also stated that, M2 is adjusted until m1…….., that means, M2 is the force applied to M1

F = M2g
µM1g = M2g

µ = M2g
       M1g
Hence;

µ = M2
      M1
If this article really help, then don’t forget to hit the share button and drop your comment, Thank you for always reading.top↑

An.d that is all on our today’s topic “What Are Forces” see you next class
Always feel free to ask me any question via the comment box, and I will be happy to take the session again, your comment notifications is like a bank alert to me.
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