BB101 ENGINEERING SCIENCE CHAPTER 3: FORCES

LEARNING OUTCOME CLO 1 Identify the basic concept of force. (C1) CLO 3 Apply concept of force in real basic engineering problems. (C2, A1)

UNDERSTANDING OF FORCE Apply the concept of force Define force and its units. Differentiate between weight and mass Define Newton’s Second Law. Define forces in equilibrium. Apply the concept of force in solving problems. Calculate Resultant Force. Understand the concept of Moment Force Define Moment Force Describe principle of moment Apply the concept and formula of moment force in solving the related problems

Resultant force using resolution force CONCEPTUAL MAP FORCE Definition   Effect of force Type of force Weight and mass INTRODUCTION Force in equilibrium Resultant force using resolution force Newton Second Law MOMENT Formula of moment Definition   Concept Force Moment Resultant Moment

Application Force in our life… http://lmspsp.cidos.edu.my/course/view.php?id=333

INTRODUCTION Definition of Force and its units. Force is the action of pushing or pulling on an objects. The SI unit of force is Newton (N) or kgms-2. The effect of force on an object: A stationary object to move. A moving object to change its speed. A moving object to change its direction of motion. An object to change its size and shape. Application of force Using in mechanical system Transferred using mechanical instrument, eg: gear, pulley, screw and piston. How to measure the force? using a spring balancing together with Newton scale (N).

THE TYPES OF FORCES The balance force is: When two or more external forces acting on a body produce no net force, Pulling forces = pushing forces The following are some of the situation where forces are balanced on a body : a pile of book resting on a hard surface a car moving at constant velocity along a straight road an airplane is flying horizontally at a constant height with a constant velocity 10 N

THE TYPE OF FORCES Unbalanced force is : When two or more forces acting on a body are not balanced, there must be a net force on it. This net force is known as the unbalanced force or the resultant force. Pulling forces ≠ pushing forces The effects of unbalanced forces acting on an object are shown in the following examples Golfer hits a stationary golf ball A footballer kicks a fast moving ball towards him When the engine of a moving car is shut down 8 N 12 N

Weight (N) = mass (kg) x gravity (ms-2) WEIGHT & MASS The weight of an object is defined as the gravitational force acting on the object. Weight (N) = mass (kg) x gravity (ms-2) W = mg where, W = weight, m = mass, g = gravitational field (9.81ms-2) The SI unit for weight is Newton (N) and it is a vector quantity The mass of an object is a measure of its inertia Mass is constant quantity and it is a scalar quantity. It is the same irrespective of where the object is. Mass (kg) ≠ weight (N)

WEIGHT & MASS Weight Mass The difference between weight and mass is summarized in below table Weight Mass Dependent on the acceleration due to gravity Is a constant quantity Is a vector quantity Is a scalar quantity Is measured in Newton (N) Is measured in kilogram (kg)

Please think of …….. Why we say, we weigh our body, while the given unit is in ‘kg’. Remember “kg” is unit of mass, not the unit of weight. Therefore, why we didn’t say we scale the mass. So, please think of ….. Unit in ‘kg’

NEWTON’S LAW Definition : When net force acting on an object is not zero, the object will accelerate at the direction of exerted force F  a F = ma Where, F = Force m = mass a = acceleration (9.81ms-2)

GROUP ACTIVITY Gallery walk

Exercise 1 1) Determine the direction of situations below and give the explanation. c) 10 N 5 N b) 8 N 4 N a) 6 N 15 N 2) A certain force is applied to a 2.0 kg mass. The mass is accelerated at 1.5 ms-2. if the same force is applied to a 5.0 kg mass, what is the acceleration og the mass? 3) A car of mass 700 kg accelerates from rest to 105 km h-1 in 10 s. what is the accelerating force developed by the car engine? 4) A toy car of mass 800 g is pulled along a level runway with a constant speed by a force of 2 N. a) what is the friction on the toy car? b) when the force is increased to 6 N, what is i) the unbalanced force acting on it? ii) the acceleration of the toy car?

EXERCISE 2 An astronaut has mass of 70 kg. what is his weight if (a) he is on the surface of the Earth where the gravitational field strength is 9.8 N kg-1? (b) he is on the surface of the moon where the gravitational field strength is 1/6 of that on the surface of the Earth? A spacecraft of mass 800 kg is orbiting above the Earth’s surface at a height where its gravitational field strength is 2.4 N kg-1. (a) what is meant by gravitational field strength at a point in the gravitational field? (b) Calculate the gravitational force experienced by the spacecraft A rock has a mass of 20.0 kg and weight of 90.0 N on the surface of a planet. (a) What is the gravitational field strength on the surface of the planet? (b) what are the mass and the weight of the rock on the surface of the Earth where its gravitational field strength is 9.8 N kg-1?

PROPERTIES OF VECTOR Showed by symbol and arrow Write as vector AB magnitude direction Write as vector AB A B

VECTOR OF FORCES Showed by arrow. The length of this diagonal represents the magnitude resultant force, F and its direction. magnitude direction F = 15 N Force at 30 ° 30° F = 30 N Vertical vector F

FORCES IN EQUILIBRIUM An object at rest is in equilibrium. This is because the forces acting on it are balanced and the resultant force is zero. Fx 2 Fy 1 Fy 2 Fx 1 Horizontal force Fx1 = Fx2 Therefore Fx1 – Fx2 = 0 Vertical force Fy1 = Fy2 Therefore Fy1 – Fy2 = 0 Since the resultant force on an object in equilibrium is zero, if the forces are resolved into horizontal and vertical components, then The sum of all the horizontal components of the forces = 0 The sum of all the vertical components of the forces = 0

ADDITION FORCES : RESULTANT FORCES A resultant force is a single force that represents the combined effect of two or more forces with magnitude and direction Vertical force = The forces acting at y-axial Horizontal force = The forces acting at x-axial The effects of force are depends on: The magnitude – the value of forces in Newton’s unit The directions – left, right, up and down A B C AB + BC = AC AC is resultant force

ADDITION FORCES: RESOLUTION FORCE A resolution forces is a single force that can be resolved into two perpendicular components F θ Fy Fx Figure shows a force F is resolved into two perpendicular components Fx and Fy. With that, Fx = F cos  Fy = F sin  Fx is the Vertical component of force whereas Fy is the Horizontal component of force

PROBLEM SOLVING Example 1: The table being pulled by two forces with the magnitude of 6N and 8N respectively. The angle between the two forces is 60° Solution Method 1 : can be determined by using the parallelogram of forces Method 2 : can be resolved into two perpendicular components (using formula)

Method 1 : Parallelogram Steps: Choose the scale . Eg: 1 cm = 1m. Using the graph paper and set the point Draw the forces F1 and F2 from a point with an angle of  with each other. Draw another two lines to complete the parallelogram Draw the diagonal of the parallelogram. The length of this diagonal represents the magnitude resultant force, F and its direction, α can be determined by measuring the angle between the diagonal with either one side of the parallelogram  F1 F2 F  F1 F2 F

Method 1 : Parallelogram Steps: Fx = ______ N Fy = ______ N 6.0 N 60° 8.0 N Fx and Fy are the vertical and horizontal components of the force : Magnitude of Fx = 6.0 cos 60° = 3.0 N Magnitude of Fy = 8.0 sin 60° = 7.0 N

METHOD 2 : FORMULA Using formula : R = √ FY 2 + FX 2 = √ 62 + 82 = √ 62 + 82 = √ 36 + 64 = √ 100 = 10 m

Exercise 3 45 N 15 N 20 N Who will win? Calculate the total net force between the following interaction: 5 N 10 N 25 N a) b) 30 N 55 N 8 N 32 N c) d) 45 N

THE TOTAL OF FORCE 1. The Total of force between two or more interaction Example 1: F1 F2 Total net force F = F1 + F2 F2 F3 F1 Total net force F = (F2 + F3 ) - F1

THE TOTAL OF FORCE 1. The Total of force between two or more interaction Example 2: Total net force = ( 5 + 10 ) – 15 N = 0 (equilibrium state) 5 N 10 N 15 N Total net force = 28 – 24 N = 4 N (move to right side) 28 N 24 N

THE TOTAL OF FORCE 2. The total of force acting at on angle  Method 2: Scale drawing : Using paper graph to get accurate reading / value Scale 1cm = 1N Use the protractor to measure the angle of 30° 10 N 30° FX 10 cm Fx = _________ cm = _________ N 30° Fx

THE TOTAL OF FORCE 2. The total of force acting at on angle  Example 3: Method 1 : Analysis FX = 10 cos θ = 10 cos 30° = 8.67 N 10 N 30° FX

THE TOTAL OF FORCE 2. The total of force acting at on angle  Example 4 Method 1: Analysis Fx = FX1 + FX2 FX1 = 10 cos 35° = 8.19 N Fx2 = 12 cos 40° = 9.19 N Fx = 8.19 + 9.19 = 17.38 N 10 N 35° FX 40° 12 N

THE TOTAL OF FORCE 2. The total of force acting at on angle  Method 2: Scale drawing Using paper graph to get accurate reading/value Scale 1cm = 1N Use the protractor to measure the angle of 35° 10 N 35° 40° FX 40° 12 N 10 cm 12 cm Fx = _________ cm = _________ N 35° Fx

MOMENT The moment of a force can be worked out using the formula: moment = force applied × perpendicular distance from the pivot. If the magnitude of the force is F newtons and the perpendicular distance is d metres then: MOMENT = Force X Perpendicular distance (arm) = F X d = (Newton) x (meter)  SI unit  = Nm Force (F) Distance (d)

MOMENT OF FORCES For an object to be in static equilibrium, the sum of the forces must be zero, but also the sum of the torques (moments) about any point. For a two-dimensional situation with horizontal and vertical forces, the sum of the forces requirement is two equations: The total of anti-clockwise moment = the total of clockwise moment 2. The total of normal force = the total of interaction force F1 F2 RF d1 m d2 m F1 d1 = F2 d2 Nm F1 + F2 = RF N

MOMENT OF FORCES For an object to be in static equilibrium, The centre of gravity can be determine using moment resultant method Resultant moment = the total of force moment Given, the centre of gravity at x is A Then, label the reference point of moment at A Resultant moment = ( F1 + F2 + F3 ) x The total of force moment = F1 (0) + F2 (x1) + F3 (x1 + x2) So, ( F1 + F2 + F3 ) x = F1 (0) + F2 (x1) + F3 (x1 + x2) F3 A B F1 x2 x F2 x1 F1 (0) + F2 (X1) + F3 (X1 + X2) ( F1 + F2 + F3 ) X =

MOMENT OF FORCES Example 5: Determine the centre of gravity for force action, so that the bar remains in horizontal equilibrium A B 20 N 50 N 4 m x

MOMENT OF FORCES Example 5 : Solution Method 2: 20 ( 0 ) + 50 ( 4 ) 20 ( 0 ) + 50 ( 4 ) ( 20 + 50 ) 200 70 = 2.86 m = x Method 1: Given the centre of gravity at x form A is, The total of anti-clockwise moment = the total of clockwise moment 20 ( x ) = 50 ( 4 - x ) 20 ( x ) = 200 – 50 x 20 x + 50 x = 200 x 70 x = 200 x = 2.86 m

MOMENT OF FORCES Example 6: Determine the centre of gravity for force action, so that the bar remains in horizontal equilibrium P Q 50 N 100 N 4 m x 25 N 1 m

MOMENT OF FORCES Example 5 : Solution Method 2: 50 ( 0 ) + 25 ( 1 ) + 100 ( 5 ) 50 + 25 + 100 525 175 = 3.00 m x = Method 1: Given the centre of gravity at x form P is, The total of anti-clockwise moment = the total of clockwise moment 50 + 25 ( - 1 ) = 100 ( 5 - ) 50 + 25 - 25 = 500 – 100 175 = 525

EXERCISE 4 1. Determine the centre of gravity for force action, so that the bar remains in horizontal equilibrium 5 N 8 N 15 N 2 m 8 m 12 N 10 N 1 m 4 m 55 N 6 m 16 N 12 m 45° 25 N a) b) c) d)

EXERCISE 5 1. Find the interaction for both of following points, RA and RB : a) b) 0.5m 0.5m 1m RA RB 2N 3N 1kg 1kg 1m 1m 1m 1m RA RB 2kg

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