Lecture 3. Newton s laws. Forces. Friction

Transcription

1 Lecture 3 Newton s laws Forces Friction

2 NEWTON S THIRD LAW If one object exerts a force on a second object, the second object exerts a force back on the first that is equal in magnitude and opposite in direction. For every action there is an equal and opposite reaction object F reaction of the table table F weight object Magnitude of force F weight object = F reaction table

3 Application: Newton s 2 nd Law F a External force? Define system System 1 (trolley, patient, and attendant): External force provided by the floor: reaction to the force the attendant exerts on the floor. System 2 (trolley with patient): External force provided by the attendant. define the system and determine the external forces

4 Application: system & external force Mass of patient = 50 kg Mass of trolley = 20 kg Mass of attendant=85 kg F a Mass System 1 = 155 kg Mass System 2 = 70 kg 1/ If the attendant exerts a force of 100 N on the floor, what is the magnitude of acceleration of the trolley and of the attendant? System 1: F=ma 100 N = (Mass System 1)a a = 100N/155kg = ms -2

5 Application: system & external force Mass System 2 = 70 kg F a 2/ What force does the attendant then exert on the trolley? System 2: F attn = (Mass System 2)*a = (70kg)(0.645ms -2 ) = 45.2 N 3/ Why is the force exerted by the attendant on the floor bigger than the force he exerts on the trolley? Because he also accelerates. Mass of attendant = 85 kg F=ma = 85kg.0.645ms -2 =54.8N

6 Adding Forces F is the Resultant force co-linear F 1 F=F 1 +F 2 F 2 Length of arrow represents magnitude of force perpendicular F 2 F q F 1 Magnitude F F F Direction tanq F F F 2 q tan F1 In general, we use calculation or a graphical method to determine the resultant force

7 Projection of a force Often need to split a force into two perpendicular components. Projection parallel to the plane a a W=mg mgcosa Projection perpendicular to the plane F 2 a F 1 w=mg Cosa = F 1 /mg Sina = F 2 /mg F 1 = mgcosa F 2 = mgsina

8 Example A gardener pushes a lawnmower with a force of 500N directed along its handle which makes an angle of 60º with the ground in order for it to move at a constant velocity. Determine (a). The component of this force which is directed horizontally H, in order to overcome the frictional forces and maintain a constant velocity, and, (b). the component of the force which is directed vertically downwards, V. Use method of components V 500N H = 500 cos60 0 = 250N H 60 0 V =500Sin60 0 = 433N

9 Classes of Forces WEIGHT (w=mg) due to gravity w=mg NORMAL FORCE (N) normal to a plane N=w=mg w = mg FRICTION (f ) parallel to a plane f F TENSION (T) is any force carried by a flexible string, rope etc. acts all along the string T= w w

10 Normal Force The normal force is the reaction by the plane to the perpendicular (or normal) component of force applied. N=mg W=mg a mgcosa a W=mg

11 Exercise: Two children pull in opposite directions on a toy wagon of mass 8.0kg. One exerts a force of 30N, the other a force of 45N. Both pull horizontally and friction is negligible. What is the acceleration of the wagon? System: the toy wagon; x positive to the right N=mg F 1 =30N F 2 =45N w=mg F ext, net =F 2 - F 1 = 45N - 30N = 15N (right) a = F ext, net /m = (15/8)ms -2 =1.875ms -2 to the right

12 Biting Force Force between bottom and top teeth Bite-force estimation for Tyrannosaurus rex Nature 382, (22 August 1996); 6,410 N to 13,400 N Muscles in the human jaw can provide a typical biting force of 600N muscle force pivot

13 Biting Force Average maximum human biting force 750 N Biting force varies depending on region of mouth Molar region N Premolar region N Cuspid region N Incisor region N force Energy of a bite is absorbed by Food Teeth Periodontal ligament bone

14 Applications of Newton s Laws Friction is a force that always acts to oppose the motion of one object sliding on another. Friction is a contact force resulting from surface roughness Surfaces slide over each other, rough bits catch each other and impede the motion. Surfaces not perfectly smooth. Schematic showing the nature of forces on a very small (atomic) scale. Surface roughness results in frictional forces

15 Applications of Newton s Laws The normal reaction force, N, is described by Newton s third law. The third law says that for every force there is an equal but opposite force. Friction and Normal reaction force are two types of contact forces If one object is sliding on another, the frictional force, F, always acts in the direction opposed to the motion, the normal force always acts in a direction perpendicular to the surface.

16 Applications of Newton s Laws Imagine an object lying on a frictional surface and a very small horizontal force, A, acts on it. The force diagram is drawn as follows. N F A W The forces acting on the object are its weight w, the normal force N, the frictional force F and the applied force A.

17 Applications of Newton s Laws To a good approximation the frictional force (F) of an object on a surface is proportional to the normal force of the surface on the object F N F = µn where the constant of proportionality is known as the Coefficient of friction µ. The most familiar thing about friction is that the heavier an object the greater the friction. Large force required to move heavy object N=m h g Small force required to move light object N=m l g W f =m h g W e =m l g

18 Kinetic Friction Kinetic friction is the force that opposes motion due to physical contact between substances, and is parallel to the contact surfaces. It is proportional to the normal force: F= k N k is the coefficient of kinetic friction (between moving substances) N=mg F= k N A w=mg Const. speed: (A=F) Acceleration: (A>F) Deceleration: (A<F)

19 Static Friction Static friction is the force that prevents motion due to physical contact between substances, and which is parallel to the contact surface. F = A A It behaves like the normal force: F=A up to a maximum of s N s is the coefficient of static friction (between immobile substances) If A< s N: static friction prevents motion If A> s N: static friction isn t strong enough motion kinetic friction. Generally k < s

20 Applications of Newton s Laws Friction Graphical representation Frictional force F s,max = s N Static F = k N Kinetic Applied Force Values of coefficients of friction s and k depend on the nature of the surfaces Typically 0< k < s <1 Typical values for coefficient of static friction s of some substances Rubber on concrete s = 1 Synovial joints s 0.016

21 EXAMPLE You press a book flat against a vertical wall. In what direction is the frictional force exerted by the wall on the book? (a)downward (b) upward (c) into the wall (d) out of the wall w EXAMPLE A crate is sitting in the centre of a flatbed truck. The truck accelerates eastwards, and the crate moves with it (not sliding on the bed of the truck). In what direction is the frictional force exerted by the bed of the truck on the crate? (a)to the west (b) to the east (c) there is no frictional force because the crate does not slid H F N

22 Friction Dental considerations Abrasion Abrasion or wear due to surface roughness Dental restorations and appliances should have smooth surfaces Abrasive resistance of material depends on Hardness Surface roughness Example Excessive wear of tooth enamel by opposing ceramic crown caused by High biting force Rough ceramic surface Remedy- broader contact areas---reduces stresses polish ceramic to reduce abrading surface

23 Friction Lubrication in the mouth is provided by saliva Saliva greatly reduces the frictional force and helps prevent excessive wear of teeth Saliva reduces friction force by up to a factor of 20. Lubricating properties of saliva are maintained over the range of jaw biting forces (0 600N) muscle force pivot Reduced friction reduces wear ensures good contact between opposite teeth is not impaired and surfaces used for chewing are maintained. Saliva, also protects the tongue and other soft tissues inside the mouth

24 EXAMPLE A crate slides at constant velocity down a plane inclined at an angle a to the horizontal. If the coefficient of kinetic friction between the surfaces of the crate and the plane is 0.6, determine the angle a. a mgcosa a W=mg F N mgcosa k F N mgsina k mgcosa mgsina k k Constant velocity k mgsina mgcosa a tan tan 0.6 k Tana

25 EXAMPLE You need to move an object across a horizontal surface by pushing it. Its mass is 50kg and the coefficients of friction are µ s =0.5 and µ k =0.4. How much force A do you have apply horizontally to a) get it moving? b) To keep it moving? When the object is stationary the friction that impedes motion is determined by µ s. The forces on the box are outlined below N = mg F max = s N A w=mg The object will start to move when the applied force A becomes very slightly greater than the maximum frictional force F max =µ s N. For our purposes we will say this happens when A=F max

26 In the y direction At all times there is no acceleration in the y direction, so the sum of the forces in the y direction must be zero. ΣF y = 0 Taking up as the positive direction N w = 0 or N = w But w = mg so N = mg In the x direction The object will begin to move when A = F max but F max =µ s N so it moves when A=µ s N as N = mg this becomes A=µ s mg

27 A=µ s mg Therefore the applied force to get the object moving must be A = (0.5)(50kg)(9.8m/s 2 ) = 245 N When the object starts moving the friction becomes controlled by µ k. Thus the equation above becomes A=µ k mg The force needed to keep the object moving is then A = (0.4)(50kg)(9.8m/s 2 ) = 196 N Note: larger force required to get it moving than to keep it moving.