Forces in action

1. Dynamics:

• Dynamics is the study of the motion of objects and the forces that cause motion. It is a fundamental part of physics and engineering, and is used to describe and predict the motion of everything from projectiles to planets.
• There are several key concepts in dynamics, including:
• – Motion: The change in position of an object over time.
• – Force: A push or pull that causes an object to change its motion.
• – Newton’s Laws: Three laws that describe the relationship between forces and motion.
• – Energy: The ability to do work, which can take various forms (kinetic, potential, thermal, etc.).
• – Momentum: The product of an object’s mass and velocity.
• a) [math]\text{Net force} = \text{mass} \times \text{acceleration}; \quad F = ma[/math]
• Newton’s Second Law, also known as the Law of Acceleration, relates the motion of an object to the force acting upon it. It states:
• “The acceleration of an object is directly proportional to the force applied and inversely proportional to its mass.”
• [math]\text{Net force} = \text{mass} \times \text{acceleration}[/math]
• [math]\vec{F} = m \vec{a}[/math]
• Where:
• [math]\vec{F} [/math]is the net force acting on an object, unit is N (newton)
• m is the mass of the object, unit is kg (kilogram)
• [math]\vec{a}[/math]is the acceleration of the object, unit is m/s (meter per second)
• 
  Figure 1 Newton’s second Law very clearly example with respect to two different masses, different acceleration, and different force
• This law means that:
• The more force applied to an object, the more it will accelerate ([math]\vec{a} \propto \vec{F}[/math] ) where the mass of an object will be constant.
• The heavier an object is (more massive), the less it will accelerate when a force is applied
• ([math]\vec{a} \propto \frac{1}{m}[/math] ).
• Some key aspects of Newton’s Second Law:
• Force and acceleration are vectors, so they have both magnitude and direction.
• Mass is a scalar quantity, so it has only magnitude (amount of matter).
• The law applies to all objects, big or small, and is a fundamental principle in understanding how objects move and respond to forces.

b) The newton as the unit of force:

• The newton (N) is the International System of Units (SI) derived unit of force. It is defined as:
• – 1 newton (N) = 1 kilogram-meter per second squared ([math]\text{kg} \cdot \frac{\text{m}}{\text{s}^2}[/math])
• – In other words, 1 N is the force required to accelerate a 1 kg mass by 1 m/s².
• The newton is named after Sir Isaac Newton, who laid the foundation for classical mechanics with his laws of motion.

c) Weight of an object; W = mg:

• The weight of an object is indeed calculated by multiplying its mass (m) by the acceleration due to gravity (g).
• [math]\text{Weight (W)} = \text{mass (m)} \times \text{acceleration due to gravity (g)}[/math]
• [math]W=mg[/math]
• Where:
• – W is the weight of the object (in newtons, N)
• – m is the mass of the object (in kilograms, kg)
• – g is the acceleration due to gravity (approximately 9.8 m/s² on Earth’s surface)
• This equation shows that weight is a force that acts on an object due to gravity. The unit of weight is typically measured in newtons (N)

d) The terms tension, normal contact force, upthrust and friction:

• These terms are all related to forces that act on an object:
• – Tension (T): The force transmitted through a string, cable, or wire when it’s stretched or pulled tight.
• 
• Figure 2 Tension forces are creating on the string
• – Normal contact force (N): The force exerted perpendicular to the surface of an object by another object that’s in contact with it (e.g., the force of the ground on your feet).
• – Upthrust (U): The upward force exerted on an object by a fluid (like water or air) when it’s partially or fully submerged.
• 
• Figure 3 Upthrust force apply on a boat
• – Friction (F): The force that opposes motion between two surfaces that are in contact (e.g., the force that makes it hard to slide a block of wood across a table).
• These forces can be summarized as follows:
• – Tension: Pulling force in a stretched string/cable/wire
• – Normal contact force: Perpendicular force between surfaces in contact
• – Upthrust: Upward force on an object in a fluid (buoyancy)
• – Friction: Force opposing motion between surfaces in contact

⇒ Free body diagram:

• A free body diagram is a graphical representation of an object and the forces acting upon it. It’s a crucial tool in physics and engineering to visualize and analyze the forces that affect an object’s motion.
• A free body diagram typically includes:
  1. The object: Represented by a box, circle, or other shape.
  2. Forces: Arrows that represent the forces acting on the object, labeled with their magnitude and direction.
  3. Axes: Coordinate axes (x, y, and sometimes z) that help to define the orientation of the forces.
     Types of forces that might be shown on a free body diagram:
     External forces:
     – Frictional force (f)
     – Normal force (N)
     – Applied force (F)
     – Gravity (g)
• Internal forces:
  – Tension (T)
  – Spring force (k)
  – Air resistance (D)
• Some benefits of using free body diagrams:
  – Help to clarify complex force systems
  – Simplify problem-solving
  – Enhance understanding of force interactions
  – Facilitate calculation of net forces and motion
  
• Figure 4 When external force applies on a moving object then it will change its motion

⇒ Examples:
(1)

Figure 5 The forces on a cyclist

• Figure 5 shows the forces on a cyclist accelerating along the road. The forces in the vertical direction balance, but the force pushing her along the road, F, is greater than the drag forces, D, acting on her. The mass of the cyclist and the bicycle is 100kg. Calculate her acceleration.

Given Data:
Drag Force = [math]\vec{D} = 180 N[/math]
Road force = [math]\vec{F} = 320N[/math]
The mass of the cyclist and the bicycle= 100kg
Find data:
Acceleration =?
Formula:

[math]\vec{F} – \vec{D} = m\vec{a}[/math]

Solution:
Her acceleration can be calculated as follows.

[math]\vec{F} – \vec{D} = m\vec{a}[/math]

Put values

[math]320 -180 =100 * \vec{a}[/math]
[math]\frac{140}{100} = \vec{a} [/math]
[math]\vec{a}=1.4m/s[/math]

(2)

• A passenger travels in a lift that is accelerating upwards at a rate of [math]1.5 \, \text{m/s}^2 [/math]. The passenger has a mass of 62kg. By drawing a free body diagram to show the forces acting on the passenger, calculate the reaction force that the lift exerts on her.

• Figure 6  A passenger in a lift

• Given data:
  Upward acceleration [math] = \vec{a} =1.5\,\text{m/s}^2[/math]
  Mass of the passenger = 62kg
• Find Data:
• Force applying downward by her weight = W =?
• Formula:
• [math]\vec{W} = m \vec{g}[/math]
• Solution:
• [math]\vec{W} = m \vec{g}[/math]
• [math]\vec{W} = (62)(10)[/math]
• [math]\vec{W} = 620 \, \text{N}[/math]
• Find Data:
  Reaction force (upward force) =[math]\vec{F_R}= ? [/math]
  Formula:

  [math]\vec{F_R} – \vec{W} = m\vec{a} [/math]
  [math]\vec{F_R} = \vec{W} + m\vec{a}[/math]

  Solution:

  [math]\vec{F_R} = \vec{W} + m\vec{a}[/math]

  Put values

  [math]\vec{F_R} = 620 + 62 \times 1.5 [/math]
  [math]\vec{F_R} = 713 \, \text{N}[/math]

Examples for free body diagram:

(1)

• After Galileo made some discoveries on friction, Newton expanded on them.
• Galileo saw balls tumbling over various curves.
  
• Figure 7 Both (starting and ending) points are equal without incline plane
• The ball will run down one side of the curve then up the other.
• Galileo noticed that if smooth surfaces were used, the ball got closer to its original height (its height at the starting point).
• Galileo reasoned that the ball would get to the original height if there were no friction.
• The ball would get to the original height if there were no outside force (unbalanced force).
  
• Figure 8 both points (ending and starting) are same but one end has an incline plane
• In all cases, the ball will return to its starting height regardless of the angles.
• The ball will go a larger distance but still not reach its initial height if the second curve’s inclination is smaller than the firsts because of friction, or the ball meeting resistance from the surface it is racing along.
• Even though the two ramps had different slopes, in the absence of friction, the ball would roll up the opposing slope to its original height.
  
• Figure 9 Second end is not equal to first starting point
• Galileo came to the conclusion that if the curve terminated without an inclination, the ball would continue indefinitely till friction finally stopped it.
• These findings led Newton to the conclusion that an item may remain in motion without the assistance of a force. The ball is really stopped from going any farther by an outside force.

Moton with non-uniform acceleration:

• A motion with non-uniform acceleration is a type of motion where the acceleration of an object changes over time. In other words, the object’s speed and direction are not constant, and its acceleration is not uniform.
• Characteristics of motion with non-uniform acceleration include:
• – Changing acceleration: The acceleration of the object changes over time, either in magnitude or direction.
• – Non-constant speed: The object’s speed is not constant and changes over time.
• – Non-uniform velocity: The object’s velocity is not uniform and changes over time.
• – Curved or nonlinear motion: The object’s motion is not in a straight line and follows a curved or nonlinear path.
• a) Drag as the frictional force experienced by an object travelling through a fluid:
• Drag is the frictional force experienced by an object traveling through a fluid (such as air or water). It’s a type of resistance that opposes the motion of the object and slows it down.
• 
• Figure 10 Drag force apply on moving particle in fluid
• Drag is caused by the interaction between the object and the fluid it’s moving through. As the object moves, it creates a disturbance in the fluid, which creates a force that opposes the motion.
• b) The amount of drag depends on several factors, including:
• – Shape and size of the object
• – Velocity of the object
• – Density of the fluid
• – Viscosity of the fluid (thickness or stickiness)
• There are two main types of drag:
• – Frictional drag (or surface drag): caused by the friction between the object’s surface and the fluid
• – Pressure drags (or form drag): caused by the pressure difference between the front and back of the object
• Drag is an important consideration in many fields, such as:
• – Aerodynamics (study of air and flight)
• – Hydrodynamics (study of water and fluid flow)
• – Engineering (design of vehicles, aircraft, and buildings)
• – Sports (performance of athletes and equipment)
• 
• Figure 11  Aerodynamics and hydrodynamics

c) Motion of objects falling in a uniform gravitational field in the presence of drag:

• The motion of objects falling in a uniform gravitational field in the presence of drag is a classic problem in physics.
• When an object falls in a uniform gravitational field, it experiences a downward force due to gravity, which causes it to accelerate. However, as it falls, it also experiences an upward force due to drag, which opposes its motion.
• 
• Figure 12 Gravitational field apply on falling an object
• The drag force depends on the object’s velocity, shape, and size, as well as the density of the fluid (air or water) it’s falling through. As the object falls faster, the drag force increases, which slows down the object.
• There are two main regimes of motion:
• – Low-speed regime: Drag is proportional to velocity (Stokes’ law)
• – High-speed regime: Drag is proportional to velocity squared (Newton’s law)
• The motion of the object can be described by the following equations:
• – Without drag:
• [math]y = \frac{1}{2} g t^2 \, (\text{free fall})[/math]
• – With drag:
• [math]y = 1/2 g t^2 – \frac{b}{m} v^2 \, (\text{drag slows down the object})[/math]
• where:
• – y = position
• – g = gravitational acceleration
• – t = time
• – b = drag coefficient
• – m = mass of the object
• – v = velocity
• The terminal velocity is the maximum velocity an object can reach as it falls in a fluid, where the drag force equals the gravitational force.
• This topic has many practical applications, such as:
• – Parachuting
• – Skydiving
• – Aircraft design
• – Ballistics
• – Engineering design
• ⇒Terminal Velocity:
• Definition: The maximum velocity an object reaches as it falls through a fluid (air, water, etc.) when the force of gravity is balanced by the force of drag.
• – Formula:
• [math]v_t = \sqrt{\frac{2mg}{C_d \rho A}}[/math]
• – Factors affecting terminal velocity:
• – Mass (m)
• – Drag coefficient ([math]C_d[/math])
• – Fluid density (ρ)
• – Cross-sectional area (A)
• – Gravity (g)
• Techniques and Procedures to Determine Terminal Velocity in Fluids:
• Free Fall Method:
• – Measure the time it takes for an object to fall a known distance in a fluid.
• – Repeat the experiment with different objects or fluids.
• Velocity Measurement:
• – Use instruments like velocimeters or particle image velocimetry (PIV).
• – Measure the velocity of an object as it falls in a fluid.
• Drag Coefficient Method:
• – Measure the drag coefficient ( [math]C_d[/math]) using wind tunnel testing or computational fluid dynamics (CFD).
• – Calculate terminal velocity using the
• Terminal Velocity Apparatus:
• – Use a terminal velocity tube or a falling sphere viscometer.
• – Measure terminal velocity directly.
• High-Speed Imaging:
• – Use high-speed cameras to capture the motion of an object as it falls in a fluid.
• – Analyze the footage to determine terminal velocity.
• Computational Fluid Dynamics (CFD):
• – Use CFD simulations to model the motion of an object in a fluid.
• – Determine terminal velocity numerically.
• Experimental Data Analysis:
• – Analyze data from experiments (e.g., velocity vs. time graphs).
• – Determine terminal velocity from the data.
•  These techniques and procedures help researchers and engineers determine terminal velocity in various fluids, which is essential in understanding fluid dynamics and designing applications like parachutes, airbags, and submersibles.

Equilibrium:

• Equilibrium is a state where a system is in balance and there is no net change or movement. In physics, equilibrium occurs when the forces acting on an object are balanced, resulting in no acceleration or net force.
• 
• Figure 13 Equilibrium position maintain
• Types of Equilibrium:
• – Static Equilibrium: The object is at rest and there is no net force or moment.
• – Dynamic Equilibrium: The object is moving, but the forces are balanced, resulting in no acceleration.
• Conditions for Equilibrium:
• – Force Balance: The sum of all forces acting on an object must be zero.
• – Moment Balance: The sum of all moments (torques) acting on an object must be zero.
• Types of Equilibrium in Mechanics:
• – Stable Equilibrium: The object returns to its original position after a disturbance.
• – Unstable Equilibrium: The object moves away from its original position after a disturbance.
• – Neutral Equilibrium: The object remains in its new position after a disturbance.
• a) Moment of force:
• “The moment of force, also known as torque, is a measure of the rotational effect of a force around a pivot point or axis”.
• Formula:
• [math]\text{Moment of force } (\tau) = \text{Force } (F) \times \text{Distance } (d)[/math]
• From the pivot point to the point where the force is applied.
• Unit: Nm (Newton-meter)
• 
• Figure 14 Force is applying on a fixed point

• b) Couple; torque of a couple:
• “A couple is a pair of forces that are equal in magnitude and opposite in direction, which creates a rotational effect or torque”.
• Formula:
• [math]\text{Torque of a couple } (\tau) = \text{Force } (F) \times \text{Distance } (d)[/math]
• Between the two forces.
• The torque of a couple is independent of the pivot point.
• c) The principle of moments:
• “The sum of the moments (torques) acting on an object is equal to the moment of the resultant force”.
• Formula:
• [math]\Sigma \tau = 0 \, (\text{for an object in equilibrium})[/math]
• This principle is used to solve problems involving multiple forces and torques.
• d) Centre of mass; centre of gravity; experimental determination of centre of gravity:
• Centre of mass: The point where the total mass of an object can be considered to be concentrated.
• Centre of gravity: The point where the weight of an object can be considered to act.
• Experimental determination:
• – Suspension method: Suspend the object from different points and draw a line through each suspension point. The intersection of these lines is the centre of gravity.
• – Balancing method: Place the object on a fulcrum and adjust the position until it balances. The centre of gravity is at the balancing point.
• e) Equilibrium of an object under the action of forces and torques:
• An object is in equilibrium when the net force and net torque acting on it are zero.
• Conditions:
• [math]\Sigma F = 0 \, (\text{net force equals zero})[/math]
• [math]\Sigma \tau = 0 \, (\text{net torque equals zero})[/math]
• f) Condition for equilibrium of three coplanar forces; triangle of forces:
• The three forces must be concurrent (meet at a point) and the vector sum must be zero.
• Triangle of forces method:
• – Draw the three forces as vectors.
• – Construct a triangle with the vectors as sides.
• – The resultant force is the closing side of the triangle.
• If the triangle is closed, the forces are in equilibrium.

4) Density and pressure:

• Density and pressure are two important physical quantities that are related but distinct.
• ⇒ Density:
• Density is the mass per unit volume of a substance.
• – Formula:
• [math]\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}[/math]
• – Units:[math]\text{kg/m}^3[/math]
• Examples:
• (1)
• Find the density of a cube with a mass of 50 kg and a volume of 0.25 m³.
• Given data:
• Mass of a cube = m = 50kg
• Volume = V = 0.25 m³
• Find data:
• Density =[math]\rho [/math] =?
• Formula:
• [math]\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}[/math]
• Solution:
• [math]\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}[/math]
• [math]\text{Density } (\rho) = \frac{50}{0.25}[/math]
• [math]\text{Density } (\rho) = 200 \, \text{kg/m}^3[/math]
• (2)
• A rectangular block has a length of 5 cm, a width of 3 cm, and a height of 2 cm. If it has a mass of 120 g, what is its density?
• Given data:
• Volume of a rectangular block =[math]V = \text{length} \times \text{width} \times \text{height}[/math]
• [math]V = 5 \, \text{m} \times 3 \, \text{m} \times 2 \, \text{m} = 30 \, \text{m}^3[/math]
• Mass = m = 120 g = 0.12kg
• Find data:
• Density =[math]/rho[/math]=?
• Formula:
• [math]\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}[/math]
• Solution:
• [math]\text{Density } (\rho) = \frac{\text{Mass } (m)}{\text{Volume } (V)}[/math]
• [math]\text{Density } (\rho) = \frac{0.12}{30}[/math]
• [math]\text{Density } (\rho) = 0.0004 \, \text{kg/m}^3[/math]
• Pressure:
• Pressure is the force per unit area applied to an object or surface.
• – Formula:
• [math]\text{Pressure } (P) = \frac{\text{Force } (F)}{\text{Area } (A)}[/math]
• – Units: Pa (Pascals)
• Relationship between density and pressure:
• – In a fluid (liquid or gas), density and pressure are related through the equation of state:[math]P = \rho g h[/math] , where g is the acceleration due to gravity and h is the height of the fluid column.
• – In a gas, density is directly proportional to pressure at constant temperature
• [math]\text{Ideal Gas Law: } PV = nRT[/math]
• ⇒ Upthrust on an object in a fluid:
• The upthrust, also known as buoyancy, is the upward force exerted on an object by a fluid (liquid or gas) when the object is partially or fully submerged.
• [math]p = h \rho g[/math]
• This equation represents the hydrostatic pressure (p) at a depth (h) in a fluid of density (ρ) under the influence of gravity (g).
• The magnitude of the upthrust depends on the density of the fluid, the volume of the displaced fluid, and the acceleration due to gravity.

• Examples:
• (1)
• A person stands on a scale that reads 500 N. If the person’s feet cover an area of 0.1 m², what is the pressure exerted on the scale?
• Given data:
• Force exerted on the scale by a person = 500 N
• Area covers his feet = 0.1 m²
• Find data:
• Pressure on the scale = P =?
• Formula:
• [math]\text{Pressure } (P) = \frac{\text{Force } (F)}{\text{Area } (A)}[/math]
• Solution:
• [math]\text{Pressure } (P) = \frac{\text{Force } (F)}{\text{Area } (A)}[/math]
• [math]\text{Pressure } (P) = \frac{500}{0.1}[/math]
• [math]\text{Pressure } (P) = 5,000 \, \text{Pa} \, (\text{pascals})[/math]
• (2)
• A swimmer dives to a depth of 10 m in a pool. If the density of water is 1,000 kg/m³, what is the pressure at that depth?
• Given data:
• Density of water =[math]\rho = 1000 \, \text{kg/m}^3[/math]
• Depth of water = h =10 m
• Gravitational acceleration = g =[math]g = 9.8 \, \text{m/s}^2[/math]
• Find data:
• Pressure applied on the depth of that pool = P =?
• Formula:
• [math]p = h \rho g[/math]
• Solution:
• [math]p = h \rho g[/math]
  
• [math]P = (10)(1000)(9.8)[/math]
• [math]P = 98,000 \, \text{Pa} \, (\text{pascals})[/math]
• Archimedes’ Principle:
• 
• Archimedes was well known for his inventions and scientific discoveries. The most famous of these were the Archimedes’ Screw (a device for raising water that is still used in crop irrigation and sewage treatment plants today) and Archimedes’ principle of buoyancy.
• Archimedes’ principle states that a body immersed in a fluid is subjected to an upwards force equal to the weight of the displaced fluid.
• This is a first condition of equilibrium. We consider that the above force, called force of buoyancy, is located in the centre of the submerged hull that we call centre of buoyancy.
• 
• Figure 15 Archimedes’ principle is applying on different objects
• Mathematically, this is expressed as:
• [math]\text{Buoyancy Force} = \rho V g[/math]
• Where ρ is the fluid density, V is the displaced volume, and g is the acceleration due to gravity.
• Archimedes’ Principle helps us understand why objects float or sink in fluids and is crucial in designing ships, submarines, and other underwater vehicles.