Bernoulli Distribution

Bernoulli Principle fluid

Bernoulli Distribution

Bernoulli Distribution. In our everyday life, Bernoulli's principle can be used to explain the work of carburetors, venturi meters, Pitot tubes, perfume sprayer bottles and insecticide spray guns. The Bernoulli's principle is also used to calculate the leakage of the water tank.

The Application of Bernoulli's Principle

a.  Flow from a Tank Hole

A drum that is filled with water to a height h1. There is a hole in the tank which is located at a height h2. measured from the bottom of the tank. What is the speed of water flowing from that hole? The speed of the liquid in the tank (v₁) is very small compared to the speed of the fluid through the hole (v₂). Thus we can assume that v₁ = 0. Both the water surface and the hole are affected by the atmospheric pressure, therefore P1 = P2 = P0. Based on equation

P₁ + ½ pv₁² + ρgh₁ = P₂ + ½ ρv₂² ρgh₂ 

we have:

 ρgh₁ = ½ ρv₂² - ρgh₂

Example
A cylindrical container having a large cross-sectional area is filled with water to a height of 170 cm. There is a narrow hole at a height of 125 cm from the bottom of the container. If g = 10m/s². calculate (a) the escape speed of water through the hole, and (b) the distance from the container where the water stream strikes the floor.

b. The Venturi meter

A venturi pipe or venturi meter can be used to measure the flow speed of a fluid in a pipe. There are two types of venturi meter; venturi meter without manometer and venturi meter with the manometer.

1) Venturimeter without Manometer

There is no difference in height in the flow of fluid which is going be measured so the Bernoulli's equation, in this case, will be in the form of:

P₁ + ½ pv₁² = P₂ + ½ ρv₂²     or

P₁ - P₂ = ½ ρ (v₂² - pv₁² )   

c. Perfume Sprayer

When we press the sprayer button of a perfume bottle, a rubber ball is compressed and the air is forced to come out through a narrow hole above a cylinder tube which has enough length to reach the perfume. The air which is moving faster causes the air pressure at the upper part of the tube to decrease, as a result the atmospheric pressure forces the liquid to rise to the upper part of the tube. Finally, a fast moving air spray blows the perfume liquid, and the perfume is flowed out as a fine mist.

d.  Insecticide Sprayer

Basically, the principle of an insecticide sprayer is the same with a perfume sprayer works by pressing the sprayer button, meanwhile, an insecticide sprayer works by pushing the pump-rod.

e.  The airplane wing

The wing is shaped is such a way that air flows more rapidly over the top of the wing than along the bottom. As a result, the pressure on top of the wing is smaller that the pressure on the bottom, therefore the net upward force (lift) is generated.

Exercise 6.2

1. A fluid through a pipe having a cross-sectional area of 3 cm², with a volume flow rate of 15 cm³/s. Calculate: (a) the amount of fluid flowing in one hour, and  (b) the flow speed of the fluid.
2. A tube having different cross-sectional areas is placed in a horizontal position. The radius of each portion 2 cm and 1 cm. If the flow speed through the wider portion is 20 m/s. find out the flow speed at the narrow portion.
3. The volume flow rate of water flowing through a hole is 100 cm³/s. The hole is 10 m deep from the water surface in a large container. If there is an additional pressure of 3 x 10⁴ N/m² on the surface, calculate: (a) the flow's speed of water through the hole, and (b) the volume flow rate of water through the hole.
4. When the air flows through a Pitot tube, the difference in height between mercury columns in manometer is 2 cm. Determine the flow speed of the air.  and ρ_air = 1,29 kg/m³ and ρ_mercury = 13,6 kg/m³)
5. An airplane with a mass of 500 kg has a total area of wings A = 50 m². The air flows through the upper part of the wing a speed of 40 m/s ( ρ = 1.29 kg/m³). Find out the flow speed of air below the wing. ( g = 10m/s²)

Multiple Choice

1)  A pipe of a barometer is replaced with another pipe having twice the cross-sectional area the previous pipe. If the atmospheric pressure is 1 atm, the height of the mercury column is.....

a.  19 cm

b.  38 cm

c.  76 cm

d.  114 cm

e.  152 cm

2)  The water level at a dam is 5 m. If the lenght of the dam wall is 40 m, the force required by the wall to hold the water is ..........

a.  1.00 x 10⁴ N

b.  2.50 x 10⁵ N

c.  5.00 x 10⁶ N

d.  2.50 x 10⁷ N

e.  3.00 x 10⁸ N

3)  The pressure of a liquid depends on ......

a.  The density of the object

b.  The density and volume of the object

c.  The density and depth of the liquid

d.  The volume and depth of the liquid

e.  The density, volume and depth of the liquid

4)  The pressure applied to an enclosed fluid is transmitted equally to every part of the fluid. That statement is....

a.  The principal law of hydrostatic

b.  Archimedes' principle

c.  Pascal's law

d.  Boyle's law

e.  Hooke's law

5)  The volume flow rate is defined as the amount of fluid which flows through a section of a tube in every unit if ......

a.  volume

b.  area

c.  length

d.  time

e.  mass

6)  Water flows through a pipe with a cross-sectional area of 10 m². The time required of fully fill a container with a volume of 1 m3 is five minutes. The flow speed of the water is ..... m/s

a.  0.1

b.  0.33

c.  1/300

d.  3.33

e.  6.66

7)  Water flows into a container with a constant volume flow rate of 0.5 liters/s. If te volume of the container is 1 m³, it will be full in.....

a.  33.3

b.  3.33

c.  333

d.  0.33

e.  0.033

8)  The following devices work based on Bernoulli's principle, except ......

a.  airplane wing design

b.  hydraulic jack

c.  perfume sprayer

d.  venturi tube

e.  pitot tube

9)  A container is 20 m high. Two  holes are made in the container which is located 2 m from the surface and 2 m from the bottom of the container. The ratio between the distances where the water stream from hole 1 and hole 2 strike the floor is ......

a.   1 : 1

b.   2 : 1

c.   1 : 2

d.   3 : 2

e.   2 : 3

10)  There is a hole in a container. The hole is h₁ deep from the surface and h₂ high from the bottom of the container. The water strikes the floor at a maximum horizontal distance if the ratio between h₁ : h₂ is ......

a.   1 : 2

b. √2 : 1

c. 2 : 1

d. 1 : √2

e. 1 : 1 

Activity

Observing Capillarity and Mass Density Of Fluid

Tools :

kerosene lamp, kerosene, water, lighter

Procedure:

1. Prepare a kerosene lamp
2. Fill the lamp with kerosene. Make sure that the wick is not immersed.
3. Add some water in a way that the wick is then immersed
4. Light it on.

Question 

1. Can you turn on the lamp? Can water be used as fuel for the lamp?
2. Can the lamp be turned on if the wick is not wet yet by the kerosene?
3. Explain your finding.

You Need To know

We are familiar with egg; we may even consume it every day. Do you know how to differentiate between a good and fresh egg and the rotten one?

It is very easy and simple. We can do it by putting the eggs in a basin of water. A good egg will sink to the bottom of the basin because it has a fresh yolk and album en. The density of a fresh egg is higher than the density of water (ρ_aegg = ρ_mwater). Meanwhile, the yolk and the albumen of a rotten egg are dehydrated, thus its density becomes smaller than the density of water (ρ_egg < ρ_water). Therefore, if we put a rotten egg into the water, the egg will float.

Case Study

Mini Submarines (Mini-Sub)

A mini submarine is a specially designed submarine having different size and shapes. Inside its hull, there is a round-shaped cabin made of very thick metals. A crew man can look outside through a window. It is also equipped with very strong lamps, the sunlight could not reach the bottom of the sea. A mini-submarine is very useful for repairing the offshore oil drilling instruments, investigating the wrecked ships, and finding some valuable minerals that might exist in the seabed. A mini-sub is equipped with an arm which can be moved and controlled from the hull. The arm is able to hold or take some objects from the sea floor.

Discussion Material

A driver cannot work a very deep place below the surface without using additional equipment to protect his body. Why is that so? Is it related to the hydrostatic pressure?

Work Physics Definition

work and energy Work Physics Definition

Work Physics Definition

Work Physics Definition. Pay attention to the arrow that goes off from the archer's bow. The event involves several concepts of physics. Force (in the pulling) is done by the hand of the athlete on the bow in such a way that the arrow could go off from the archer's bow.

When a force interacts with a matter (body) so that moves, then the force said does work on the body. What does it mean by work? Does the meaning of work equal to the one used in daily lives?

A. The Meaning Of Work

In daily lives, the term "work" is used for all efforts or activities done to reach certain objectives. For example, to gain knowledge, someone does the study activity. Then, what is the meaning of work in physics?

In Physics, work always involves force and displacement. Work exists if the force done to the body results from displacement. Although a huge force is applied to a body, but if it does not move, then there is no work done.

Work done by the constant force F is the dot product between force and displacement of s, which is mathematically written as

W = F.s

In SI, the unit of work is joule, abbreviated by J. If the unit of force is Newton and the unit of displacement is a meter, the one joule equals to one newton-meter (Nm).
1 Joule = 1 newton meter = 10⁷ erg
Base on the relationship, we can conclude that:

The one Joule work is work done by one newton of force to displace a body one matter in the direction of the force. 

1. Work of Constant Force 

The work done by force F in figure 3.4 is the product of the component of force in the direction of the displacement. Mathematically it can be written as follows.

W = (F cosӨ) (s),

Where Ө is the angle between the force and the displacement.

The value of cos Ө can be positive, negative, or zero; thus W can be positive, negative, or zero as well. Work is positive if the force that caused the displacement has the same direction with the displacement. If the force is perpendicular to the displacement, the work is zero.

2. Work of several Forces

Work is a scalar quantity. If several forces do work on a body each with the amount of W₁, W₂, W₃,…., Wn then the total work done by the forces equals to the sum of all works done by each force; that is:

W₁+ W₂ + W₃ +….+ W_n  ....... [3.3]

In a system shown in Figure 3.5, the working forces are F and, thus the total work is

W = (F - fk).s  ....... [3.4]

Figure 3.7 shows the graph of constants force F, which works on a body so that it moves/displaces with the distance of s. The works done by the force equals to the area of Fs under the graph.

Example 3.1

1.  A body with the mass of 2 kg, is placed on a slippery level floor. The force done to the body is 40 N, which direction forms an angle of 60^o toward the horizontal line. If the displacement is 5 m, then determine the work done by the force.
Answer:
W = (F cosӨ) (s)
    = (40 N) (cos 60^o ) ( 5 m)
    = 100 joules
Thus, the work done by the force to the body is 100 joules.

Exercise 

1. A Horizontal force of 100 N is used to push a box of 20 kg on the surface of a rough floor. The box is moving with constant velocity as far as 10 m. What is the work done by the force?
2. A body of 4 kg mass is placed on a smooth level plane. If a pulling force of 60 Newton, which makes an angle of 60 toward horizontal, work on the body for 2 seconds; then determine the work done by the force.

B. Energy

Energy is the ability to do work. We will learn two kinds of energy, potential, and kinetic energies. The potential energy we will learn first is the gravitational potential energy.

1. Gravitational Potential Energy

Gravitational potential energy is the energy stored in a body due to its position. Gravitational potential energy (Ep) owned by a body with mass m and situated in a height of meters above the earth's surface can be calculated by using the equation:

Ep = mgh

where :
Ep = potential energy (Joule)
m  = mass (kg)
g   = acceleration due to gravity (m/s2)
h   =  height  (m)

2.  Kinetic energy

Kinetic energy id the energy owned by moving object. Kinetic energy (Ek) of a body with m, which moves at speed of v, can be calculated by using the equation:

Ek  =  1/2 mv2

Where

Ek   = kinetic energy  (Joule)

m    = mass (kg)

v     = speed or velocity   (m/s)

Example 3.2

1.  A mango with mass 500 g is hanging 7 m high above the ground (g = 10 m/s2). Calculate the potential energy stored in the mango.

Answer :

Ep  =  m g h

      = (0,5 kg) (10 m /s2) ( 7 m)

      = 35 joules

2.  A car with mass of 2,000 kg moves at speed of 72 km/hour. Calculate the kinetic energy owned by the car.

v = 72 km/hour   = 20 m/s

Ek   = 1/2  mv2

       = (1/2)  (2,000 kg) (20 m/s2)

       =  4 x 105 joules

The kinetic energy owned by the car is 4 x 105 J

3.  Conservative Force

When we throw a ball vertically upward, the kinetic energy of the ball will decrease, but its potential energy will increase. The kinetic energy of the ball gradually changes into potential energy. When the ball reaches its highest potion, its kinetic energy equals to zero and its potential energy s maximum. On the contrary, when the ball moves downward, its kinetic energy ill increase, but its potential energy will decrease. On downward motion, the potential energy gradually changes into kinetic energy.

On the case above, the force that works on the ball is gravitational force. Gravitational force is a conservative force. If a system is subjected only to a conservative force, then the law of mechanical energy conservation will be applied to the system.

Exercise

1. A boy with the mass of 25 kg is able to climb a stair of 4 m in 15 seconds. If g = 10 m/s2, what is the power of the boy?
2. A runner engages the average power of 200 W when he covers the distance of 100 m in 80 seconds. What is the average force needed?
3. A waterfall with the height of meters flows 100,000 kg of water every second. If g = 10 m/s2, what is the force produced by the waterfall?

Multiple Choice

1. From this unit, the one that is not the unit of work energy is ............

a.  watt

b.  watt hours

c.  kilowatt hours (kWh)

d. newton meter

e. joule

2.  A body with mass of 2 kg, which moves with velocity of v, has kinetic energy of 400 J. The magnitude of velocity v is ....... m/s

a.  8                       d.  35

b.  20                     e.  37

c.  31

3. The energy kinetic possessed kinetic by a moving body is proportional to ..............

a.  the gravitational acceleration

b.  the square of velocity

c.  the square root of velocity

d.  the velocity

e.  the square root of mass

4.  A mango with the mass of 0.2 kg is falling down from its branch at the height of 5 meters. If the acceleration is 10 m/s², the kinetic energy of the mango when it hits the ground is.......

a.  16 J              d.  7 J

b.  15 J              e.  5 J

c.  10 J

5.  A bullet, which mass is 200 gram, is fired at the elevation angle of 30^o and with an initial velocity of 10 m/s. If the earth gravitation acceleration 10 m/s², the potential energy of the bullet at the highest point is....

a.  3.5 J

b.  3.0 J

c.  2.75 J

d.  2.5 J

e.  1.5 J

6. A block of 4 kg mass is initially moving with a velocity of 2 m/s. In order to stop the block at a distance of 5 meters measured from the initial position, the magnitude of friction force that should be provided is....

a.  1.2 N               d.  1.8 N

b.  1.4 N               e.  2.0 N

c.  1.6 N

7.  Below is the unit of power, except...........

a.  kWh                    

b.  watt

c.  J/s
d.  kW

e.  hp

8.  A loaded lift has a mass of 2,000 kg. The power needed to lift up the loaded lift by 50 m in 20 seconds is....

a.  1,000 kW

b.  200 kW

c.  100 kW

d.  50  kW

e.  40  kW

9.  A body with a mass of 2 kg is moving with velocity of 2 m/s. A short time later the body is moving with a velocity of 5 m/s. The total work applied to the body is....

a.  25 J                d.  22  J 

b.  24 J                e.  21  J

c.  23 J

10. The engine of an airplane is able to move the airplane with a force of 15,000 Newtons. When the airplane is moving with constant speed of 300 m/s, the power delivered by the engine is...........

a.  4,500,000 kW

b.  450,000 kW

c.  450,000 W

d.  4,500 kW

e.  4,500 W

Hydrostatic Pressure Definition

fluid

Hydrostatic Pressure Definition

A submarine can float above the water surface or dive below the water surface by emptying or filling its ballast tank. A ballast tank is cavities on a hull (larboard and starboard) of a submarine that can be filled with air or water. The submarine moves forward by using propellers and move downward by using hydroplane which looks like the tail of the airplane which pushes it downward. On the other hand, when the submarine is going to the surface, the water inside the ballast tank has to be pushed outward by the pressurized air. The working principle of a submarine agrees with Archimedes' principle. What does the Archimedes' principle state' You will study it in this article?

A. Hydrostatics

There are three states of mater, i.e.: solid, liquid, and gas. In solid, a matter retains a definite size and shape. In a liquid, a matter has a definite volume and takes the shape of its container. Liquid and gasses have an ability to flow. That is why both liquid and gasses are usually called fluid or the flowing substance. Our study or fluid will be divided into two parts, i.e. hydrostatics which discusses the static fluid or fluid at rest and hydrodynamics which discusses fluid in motion.

1. Density

Density (𝝆) of a material is defined as its mass (m) per unit volume (V). Mathematically, density is formulated as:

The SI unit of density is kg/m3; meanwhile, the cgs unit of density is g/cm³.

2. Pressure

a. Definition of Pressure

Pressure is defined as the force per unit of area.
If a force F works perpendicularly to a surface with area A, we can write the pressure P as:

The SI unit of pressure is Pascal (Pa), where 1 Pa = 1N/m2. The pressure will increase if the force applied to a given area increases or the area of a given force is reduced.

See Figure 6.3 with the same magnitude of force, the pressure given by a needle is bigger than that by a finger on a balloon. The pressure, given by a finger only causes the balloon to undergo a little deformation., whereas the same magnitude of force given by a needle is able to blast the ballon.

b. Atmospheric Pressure and Gauge Pressure

We are all used to work under the pressure called atmospheric pressure. This pressure is caused by the weight of the air above us. The value and unit of atmospheric pressure are:

Have you ever measured the air pressure in a car's tire? What do you find? Has the pressure inside the tire ever been lower than the outside air pressure? Let's examine this.

In some cases, we sometimes measure the pressure inside a certain object by calculating its difference with the outside air pressure (atmospheric pressure). Even the air pressure inside a flat tire will never be zero. A measurement of the pressure inside a flat tire will result in the value of the outside air.

To inflate a car's tire so that the pressure inside it reaches the value of 241 kPa, the pressure inside the tire must be higher than the outside air pressure in the amount of P = 241 kPa + Po = 302 kPa. The241 kPa pressure that will appear in a measurement is called gauge pressure (Pg) which is generally formulated by

Pg = P + Po

c. Pressure in a Static Fluid

We have discussed pressure as the result of the division between the working force and the surface area. We have also discussed the example of pressure comparison two equal forces acting on a different surface area (Figure 6.3). The example describes the pressure caused by mechanical force; how about the pressure caused by fluid?
The experiment result shows that fluid pushes in all directions. Thus, somebody who dives under water will feel the water pressure in whole parts of his body.
A point of fluid's particle also undergoes pressure from all directions. If the fluid is at rest, the forces acting on that point will be equal in magnitude. The direction of the compression force is always perpendicular to its contact surfaces.
Now, we area going to calculate the pressure in the bottom of a container filled with a density ρ and cross section of A (Figure 6.3). The pressure at the bottom of the container is caused by the compression force of the outside air and the weight of the fluid above it.

The top of the fluid which is in direct contact with the air will undergo a compression force of
F_bottom = F_top + mg
The bottom of the container will undergo downward force in the amount of Ftop plus the weight of the fluid, i.e. w = m.g;

P_bottom = P₀ + ρhg   ....... (6.4)

Pressure caused by a fluid at rest is called hydrostatic pressure. The hydrostatic pressure is proportional to the depth (h) and the density (P) of the fluid. Any point at the same level always has the same pressure, regardless of the shape of the container. Equation (6.4) does apply for any depths, not only the base of the container.

Example 6.1

A drum of 1 meter high is filled with kerosene ( ρ  = 0,8 g/cm³ to 3/4 of the drum's height. If g = 10m/s², find the pressure at the bottom of the drum.

Answer :
= 800 kg/m³ ; h = 0,75 m
P_bottom =  ρgh = P₀ + 800 kg/m³) (10m/s²) (0,75m) = 6,01 N/m².

3.  Pascal's Principle

As a fluid, the earth's atmosphere exerts pressure to all materials which are contact with it, including all the fluids on the earth's surface. The atmospheric pressure applied to a fluid is transmitted to every part of the fluid. This was stated by a French philosopher and scientist, Blaise Pascal (1623-1622). Pascal's principle or commonly known as Pascal's law states that:

Pressure applied to an enclosed fluid is transmitted equally to every part of the fluid.

On of the applications of Pascal's law is the hydraulic lift, as shown, If a force F₁ strikes a piston with a small area A1, the pressure of the fluid will increase the amount of P = F₁/A₁.  The increase of the pressure is then transferred to the cross-sectional area of a larges piston A₂. Since the applied pressure is the same at both pistons, it applies:

F₁/A₁ = F₂/A₂

4. Archimedes' Principle

A sand miner finds it easier to lift a bucket of sand from a river when the bucket is still on the water. However, when the bucket is taken out of the water, it will become heavier. How can this happen?

If a body is immersed in a fluid, the fluid exerts an upward force on the body. The force makes the body as though it had less weight. This fact was firstly acknowledged by Archimedes, it is therefore known as Archimedes' principle, which states that:

When a body is immersed in a fluid, the fluid exerts an upward force (buoyant force) on the body whose magnitude is equal to the weight of the fluid displaced by the body.

Archimedes' principle is mathematically formulated as:

F_A = ρ_fluid V g

where is a buoyant force, ρ_fluid is the density of the fluid, and V_t  is the volume of immersed body.

There are three possible cases of the body in a fluid, i.e floating, barely floating (completely submerged), and sinking. By using Newton's first law and Archimedes' principle, we can determine in what terms a body will be floating, barely floating, or sinking.

a)  Floating

A body is said to be floating if only some parts of the body are submerged or beneath the water surface, meanwhile the other part is above the water surface. On a floating body, the magnitude of buoyant force F_A is equal

F_A = mg
ρ_fluid g V_t = ρ_b V_b g
ρ_fluid  V_t = ρ_b V_b 

The volume of the body below the water surface V_t is always smaller than the total volume of the body V_b. Hence he density (ρ_b) of a floating body is smaller than the fluid's density ρ_b < ρ_fluid 

b) Barely Floaing (Completely Submerged)

On a barely floating body, the magnitude of buoyant force F_A is equal to the body's weight w = mg, Thus, we have:
ρ_fluid g V_t = ρ_b V_b g
However, the volume of the immersed body V_t is equal to the total volume of the body V_tb . Hence for a barely floating body, we have the relation ρ_b = ρ_fluid  . Therefore on a barely floating body, the density of the body is equal to the density of fluid.

c) Sinking

When a body is sinking, it means that the buoyant force F_A acts on the body is smaller than its weight mg. On a sinking body, the volume of the immersed body V_t is equal to the total volume of the body V_b. However, if the body stays at the bottom of a container, there will be a normal force N acting on it, so it goes:
 F_A + N = w
N = ρ_b V_b g - ρ_fluid g V_t
Since the normal force is always positive, we have ρ_b < ρ_fluid . Thus, a body sinks in a fluid when its density is greater than the density of the fluid.

Archimedes' principle is widely used in either technology or science. Some examples of devices that work based on Archimedes' principle are hydrometers, ships, dockyards, submarines, and hot air ballons.

5) Viscosity

When sliding across a rough floor, a block experiences a frictional force opposing then motion. Similarly, a fluid flowing past a stationary surface experiences a force opposing the flow. This tendency to resist flow is referred to the viscosity of a fluid. Fluid like air have low viscosity, thicker fluid like water are more viscous, and fluids like honey and molasses are characterized by their high viscosity.

A fluid's viscosity is related to the frictional force between fluid's layers when a layer slides over him other. On liquid, the viscosity is caused mainly by the cohesive force between its molecules; meanwhile, on gas, the viscosity is caused mainly by collisions between its molecules.

The ideal fluid we often use in the discussion has zero viscosity. This is because we neglect the inter-molecule interaction of the ideal fluid. The flow of ideal fluid is laminar (streamline), every point in the fluid has an equal speed.

the true fluid with the certain value of viscosity will generate a flow pattern as. Because adjacent portions of the flow pasts one another with different speed, a force mus be exerted on the fluid to maintain the flow.

The force causing a viscous fluid to flow is provided difference by the pressure, (P₁ - P₂) across a given length of tube (L). Experiments show that the required pressure difference is proportional to the length of the tube (L) and the average speed of the fluid (v), and inversely proportional to the cross-sectional area of the tube (A).

We can observe viscosity by dropping a marble into a glass of cooking oil. The marble is slowed down by the fluid due  the frictional force. The magnitude of the frictional force on the marble moving at its constant velocity v in the fluid can be calculated by using the following equation:

f = 5πηrv

where is fluid's viscosity and r is the marble's radius.
When the marble is dropped into the oil, at a certain time it will move in a constant velocity. The velocity is known as terminal velocity. There are three kinds of force acting on the marble during its movement on the cooking oil,i.e. weight (w) buoyant force (F_A), and frictional force of fluid (f).

Newton's Laws Of Motion

Newton

Newton's Laws Of Motion

If a wooden block and a block of ice stay at rest on a floor without being pushed, both of term will stay motionless. However, when they are pushed by the same magnitude of forces so that they finally move, the ice block will slide further than the wooden block. How could it be? Why don't the Why don't the wooden block and the ice block move when no push was done? Why does the ice block tend to keep moving when the push is removed?

A body moves due to the presence of force working on it. Because of the given push, the wooden block and the ice block can move before they finally stop. The relationship between the force and the motion is studied in a branch of physics called dynamics.

A) Newton's Laws of Motion

1. Newton's First law

Basically, all objects tend to keep their motion state. If they are initially at rest, they tend to stay at rest. If they are moving, they tend to keep on moving. This property is called inertia.

It takes three men to move the car from its rest condition. But, when it is already moving, it only takes one man to keep the car move, while the car move. While the car is at rest, it tends to stay at rest so that we need a great amount of force to strive against its inertia in order to make it move. On the other hand, when the car is already moving, it tends to keep its motion state. And during that time, it only takes a smaller force to keep the car move compared with the force needed to make it move from its rest condition.

Newton's First Law states that:

if the or resultant of force working on an object is zero, the object will stay at rest or keep on moving with constant velocity in a straight line.

This is also known as the law of inertia, which is mathematically written as follows
𝞢 F = 0   ...... (1)

2.  Newton's second Law

What happened if the magnitude of the resultant force working on an object is not zero? In that case, the velocity of the object will change. It will increase if the direction of the resultant force is the same as the direction of the velocity of the object. On the contrary, the velocity of the object will decrease if the direction of the resultant force is opposite to the direction of the velocity of the object. The relationship between the resultant force and the acceleration can be explained by using Newton's second law, which states that:

If the resultant of forces working on an object is not zero, then the object will experience an acceleration in the same direction with the resultant force.

Mathematically, the Newton's second law can be written as follows.
𝞢 F = ma  or  F = ma   ..... (2)
where D is the magnitude of force, m is the object's mass, and a is the acceleration experienced by the object.

3. Newton's Third Law

If we press a wooden block with our finger, we will feel hurt. Why? The pain is actually caused by the force that is exerted by the wooden block. The magnitude of force exerted by the block to our finger is the same as the magnitude of force exerted by our finger to the block. The harder we press the block, the more hurt our finger will be. This phenomenon shows the existence of action-reaction force. Because the finger exerts a force on the block, the block will also exert a force on the finger with the same magnitude but in opposite direction. The force exerted by the finger is called an action, whereas the force exerted by the block is called reaction. This phenomenon is expressed in the Newton's third law, which states that:

If the first object exerts a force on the second object, then the second object will exert the same amount of force to the first object in the opposite direction.

Faction = - Freaction

The negative sign indicates that Faction and Freaction are in opposite direction. The forces of action and reaction are always the same, in opposite direction, at the same incident point, and working on two different objects.

In Figure 2.3 F₁ is the force exerted by the hand on the rope. And the reaction force F₁ is the force exerted by the rope on the hand. Vector F₂, shows the force exerted by the hand on the block, and the reaction F2, shows the force exerted by the block on the hand.

4. Analysis Of Object's motion in the absence of friction force

every object obeys the Newton's laws of motion. We will discuss several examples of the motions of objects in our daily lives. In this discussion, the friction that influences the motion of an object is neglected.

a. An object Moving on A Flat Plane

An object on a flat plane is given a horizontal force so that its acceleration becomes a= F/m

b. An Object Moving On an Inclined Plane

A block is moving on an inclined plane. Base on the Newton's second law, the acceleration of the block (a) can be determined by using the following equations.

∑F = ma  → w sin Ө = ma à mg sin Ө= ma

Therefore, the magnitude of acceleration of the object on an inclined plane with elevation angle of  Ө is

a = g sin  Ө

Where g is gravitational acceleration.

c. Two Objects on a Pulley

Block-1 (which mass is ) by a string through a pulley. In this case, the string's mass and the friction caused by the pulley. In this case, the string's mass and the friction caused by the pulley is neglected. Block-1 moves to the right with acceleration experienced by block-2 that moves downward. The force working is horizontal direction is the string's tension T. Whereas in a vertical direction, it works the string;s tension T and the weight force of block-2 (W₂).

Exercise 2.1

1. An object with mass of 50 kg is initially at rest. It is then accelerated to reach a velocity of 20 m/s within 5 seconds. What is the magnitude of force that accelerated the object?

2. Two object m₁ = 10 kg and m₂ = 15 kg, are set as shown in the picture. What is the acceleration of each object? (g = 10m/s²)

B ) Friction Force 

We way see some cleats at the bottom soccer shoes. They are attached for protecting the player who wears it from sliding. The cleats will increase the friction between the shoes and the surface of the soccer field. Meanwhile, the frictions among the engine parts of a car are always tried to be minimized by using lubricant oil. Therefore, some frictions are useful (advantageous), while some others are disadvantageous. The friction force is the force that occurs when two objects contact with each other. It is divided into two: static friction and kinetic friction forces.

1. Static Friction Force

Static friction force fs occurs between the contact surfaces of two objects when the external force that works on the system does not cause relative motion toward each other. A block with mass m is pulled by a force F on a rough plane. IF the block is not moving yet, it means that there is a force F. The magnitude of the force is equal to F but the direction is opposite to the direction of the force F. That force is what we called static friction force (fs).

If the force F keep on increasing, then the static friction force is also increasing until at a certain time it reaches its maximum value, that is Fsmax. The maximum value of the static friction force is proportional to the coefficient of static friction (μ_s) between the block and the plane surface and the normal force of the block (N). It can be written as (F_smax) = μ_s N. The magnitude of the static friction force ranges from zero to its maximum value so that it can be written. 

0 ≤ fs ≤μ_s N

The block is said to be about to move it the magnitude of the external force F is exactly the same as the maximum value of the static friction force, or it is expressed by F = F_smax.

2. Kinetic Friction Force

Kinetic friction force exits between the contact surfaces of two objects when there is relative motion toward each other. If the force F keeps on increasing so that it exceeds the magnitude of the maximum static friction force, then the block will move eventually. Because the moving block and the surface of the rough plane still contact with each other, then there must be friction force working on the system. However, that friction force is not static friction force, it is the kinetic friction force (f_k).

The magnitude of the kinetic friction force is proportional to the coefficient of kinetic  (μ_k) between the block and the surface of the rough plane and the normal force of the block (N). It is formulated

f_k = μ_k N