DEC 1013 ENGINEERING SCIENCEs

1. DEC 1013 ENGINEERING SCIENCEs 3. FORCE AND MOMENT NAZARIN B. NORDIN nazarin@icam.edu.my

2. What you will learn: • Definition: mass, force, moment, static equilibrium • Vectors: addition, subtraction, resultant • Polygon of forces: resultant/ equilibrium of many forces • Force: normal force, friction force (static and kinetic), coefficient of friction, µ • Friction torque (rotational motion) • Moment and momentum

3. Physical Quantities and Units • Measuring Things Physics is based on measurement. • We discover the physical world by measuring physical quantities like length, time and mass. • All physical quantities consist of a numerical magnitude and a unit.

4. Physical Quantities and Units • Physical quantities are quantities that can be measured. • Every physical quantity has a magnitude and a unit. • There are 7 fundamental or base units in physics. • For every unit, there must be a standard. • The S I (Systeme Internationale d’Unites) is based on the 7 fundamental units.

5. The 7 fundamental Units

6. Unit Conversions • When units are not consistent, you may need to convert to appropriate ones • Units can be treated like algebraic quantities that can “cancel” each other • See the inside of the front cover for an extensive list of conversion factors • Example: Appendix 5: Units of Mesaurement

7. Problem: A very boring-looking house is 50 feet long, 26 feet wide with 8.0 foot high ceilings. What is the volume of the house in cubic centimeters, cm3 and cubic meters, m3?? Appendix 5: Units of Mesaurement

8. Examples of various units measuring a quantity Appendix 5: Units of Mesaurement

9. Trigonometry Review Appendix 5: Units of Mesaurement

10. More Trigonometry • Pythagorean Theorem • To find an angle, you need the inverse trig function • for example, • Be sure your calculator is set appropriately for degrees or radians • Must beware of quadrant ambiguities Appendix 5: Units of Mesaurement

11. And more trigonometry.. • Pythagorean Theorem • To find an angle, you need the inverse trig function • for example, Appendix 5: Units of Mesaurement

12. Problem: A high fountain is located at the center of a circular pool as shown in figure P1.41 (page 21). The student measures the circumference of the fountain to be 15.0 m. Using a protractor he measures the angle of elevation of the fountain to be 55◦. How high is the fountain? Appendix 5: Units of Mesaurement

13. Concept Summary • Scientific notation allows us to express very large or very small numbers in a compact form. This reduces errors and simplifies multiplication and division. • Scientists use the metric system which is tied to atomic standards (except for the kilogram) so that any scientist in the world can get very high precision standards. Appendix 5: Units of Mesaurement

14. Mass vs. Weight • Weight measures the amount of force an object applies • The pound is a unit of weight, not mass • A mass of 1 kg produces a weight of 2.2046 lbs on the Earth • An astronaut that has a weight of 150 lbs on the Earth has a weight of 25 lbs on the Moon. However, regardless of where she is, her mass is 68 kg. Appendix 5: Units of Mesaurement

15. Mass • Definition: the amount of matter of a body measured in kilogram (kg)

16. Force • The unit of force is the Newton • 1 N = 0.2248 lbs AND 1 N = 0.1 kg • If I weigh 100 kg, the force of gravity on the surface of the Earth that is pulling on me (and the force that my feet exert on the Earth) is 981 N. (pf course, taking g = 9.81 m/s2). Appendix 5: Units of Mesaurement

17. What is a Force ? A force • is the agency of change. • causes a body to accelerate . • is a vector quantity. • is measured in Newton’s.

18. Newton’s 2nd Law “The acceleration of a body is directly proportional to the resultant force acting on the body and inversely proportional to the body’s mass” Mathematically: The resultant force is just the (vector) sum of all of the forces acting on the body. Often written as F.

19. Newton’s Third Law • Whenever an object A exerts a force on another object B, then B will also exert a force back on A. • These forces are ALWAYS equal in magnitude but they point in opposite directions. • Such forces are called “Newton’s third law force pairs”. • The forces are on differentbodies, so do not add to zero.

20. Scalar and vectors • A scalar quantity is completely described by a magnitude (a number). • A vector quantity is completely described by a magnitude and a direction. • Vectors are represented by a bold face type (e.g., A). • Alternatively, we can also write it as A

21. Scalars • A quantity that can be specified from its magnitude only with units • No direction needed • Examples are speed and distance • Represented by italics; v = 2.4 m/s

22. Vectors • A quantity with both magnitude and direction. • Examples are velocity and displacement. • Represented by boldface; v = 2.4 m/s to the north • 2 Ways to write out Vectors • Polar Notation; Rectangular Notation

23. Vectors: geometric representation A vector is geometrically represented as a line segment with an arrow indicating direction Length represents magnitude (F) Head Tail Direction of the arrow gives the direction of the vector Length of the arrow gives the magnitude of the vector

24. Vector Addition: The Parallelogram Method A C = A + B A + C B B • Draw A and B from the same origin • Draw a straight line parallel to A and another parallel to B so • that a parallelogram is formed • The diagonal will represent the resultant vector C • Addition of vectors is commutative i.e A + B = B + A

25. Resolving a Vector into components y A x

26. ^ r UNIT VECTORS A unit vector r is a vector having length 1 and no units. It is used to specify the direction of a vector: • So we can write A = Ar • The unit vectors i, j, kpoint in the x, y and z axes • respectively. A j In terms of unit vectors, we can express a 2-D vector as follows: y x i z k

27. Scalar (Dot) product There are two types of vectormultiplications: • Dot product (results in a scalar) • Cross product (results in a vector) Dot product of two vectors A and B is defined as A q B

28. Properties of Dot products Dot products of unit vectors The scalar product of any two vectors Let Then A•B = Ax Bx + Ay By + Az Bz

29. Two special cases • When angle between two vectors is zero • When angle between two vectors is 90o

30. Cross Product Vector or cross product A x B - vector result is C = A x B Magnitude of C = |C| =|A x B| = AB sinq C has DIRECTION to A and B Right hand rule

31. Properties of Cross-products • A x B = - B x A • The vector product of any vector with itself is zero i. e A x A =0 So i x i= 0, jxj= 0, k xk = 0 3. For 2 perpendicular vectors A andB, |A x B| = AB sin900 = AB

32. y x j k z y x j z j 4. i x j = k j x k = i k x i = j 5. j x i = -k k x j = -i i x k = -j i i k -j

33. Adding Vectors • Vectors can be added graphically. • When adding two or more vectors, the answer is called the resultant. • Vectors can be moved parallel to themselves in a diagram as long as they don’t change direction or length. • Draw vectors using head to tail method.

34. Friction is a result of irregularities in the surfaces of objects. • The force required to overcome friction is called the static frictionforce. • The force needed to keep a constant speed is called the kinetic frictionforce.

35. Imagine your car broke down and you have to push it. Which takes more force, to get it started rolling or to keep it rolling? • To get it started • Static friction is greater than kinetic friction.

36. Friction depends on 3 things: • The friction force depends on whether or not the surfaces are moving. • The friction force depends on the materials of which the surfaces are made of. • The friction force depends on how hard the surfaces are pressed together. This is called the Normal Forceand depends on mass and gravity also known as Weight

37. Coefficients of Friction Equations • Static force=Coefficient of static x Normal Force (Weight) • Fs = us x N • Kinetic Force=Coefficient of kinetic x Normal Force (Weight) • Fk = uk x N

38. fFRICTION Friction • What does it do? • It opposes relative motion of two objects that touch! • How do we characterize this in terms we have learned (forces)? • Friction results in a force in the direction opposite to the direction of relative motion (kinetic friction, static – impending mot) j N FAPPLIED i ma some roughness here mg

39. Surface Friction... • Friction is caused by the “microscopic” interactions between the two surfaces:

40. Surface Friction... • Force of friction acts to oppose relative motion: • Parallel to surface. • Perpendicular to Normal force. j N F i ma fF mg

41. Model for Sliding (kinetic) Friction • The direction of the frictional force vector is perpendicular to the normal force vector N. • The magnitude of the frictional force vector |fF| is proportional to the magnitude of the normal force |N |. • |fF| = K| N | ( =K|mg | in the previous example) • The “heavier” something is, the greater the friction will be...makes sense! • The constant K is called the “coefficient of kinetic friction.” These relations are all useful APPROXIMATIONS to messy reality.

42. Model... • Dynamics: i :F KN = ma j :N = mg so FKmg = ma (this works as long as F is bigger than friction, i.e. the left hand side is positive) j N F i ma K mg mg

43. Forces and Motion • A box of mass m1 = 1.5 kg is being pulled by a horizontal string having tension T = 90 N. It slides with friction (mk= 0.51) on top of a second box having mass m2 = 3 kg, which in turn slides on a frictionless floor. (T is bigger than Ffriction, too.) • What is the acceleration of the second box ? (a) a = 0 m/s2 (b) a = 2.5 m/s2 (c) a = 3.0 m/s2 Hint: draw FBDs of both blocks – that’s 2 diagrams slides with friction (mk=0.51) T m1 a = ? m2 slides without friction

44. Act 1 Solution • First draw FBD (free body diagram) of the top box: N1 m1 f = mKN1 = mKm1g T m1g

45. Act 1 Solution • Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2. • As we just saw, this force is due to friction: m1 f1,2 = mKm1g f2,1 m2

46. Solution • Now consider the FBD of box 2: N2 (contact from…) (friction from…) f2,1 = mkm1g m2 (contact from…) (gravity from…) m1g m2g

47. mKm1g = m2a a = 2.5 m/s2 Act 1 Solution • Finally, solve F = ma in the horizontal direction: f2,1 = mKm1g m2

48. Inclined Plane with Friction: • Draw free-body diagram: ma KN j N  mg  i

49. i mg sinKN=ma j N =mg cos  mg sinKmgcos =ma Inclined plane... • Consider i and j components of FNET=ma : KN ma j N  a / g= sin Kcos   mg mg cos  i mg sin 

50. Static Friction... • So far we have considered friction acting when the two surfaces move relative to each other- I.e. when they slide.. • We also know that it acts in when they move together:the ‘static” case. • In these cases, the force provided by friction will depend on the OTHER forces on the parts of the system. j N F i fF mg