Chapter 4 Forces Maintaining Equilibrium or Changing Motion

Chapter 4 Forces: Maintaining Equilibrium or Changing Motion

Force » Force: a push or pull acting on a body that causes or tends to cause a change in the linear motion of the body (an acceleration of the body) – Characteristics of a force l l l magnitude represented with direction point of application. line of action sense (push or pull along the line of action) Vector an arrow

Classifying Forces l Internal Force: acts within the object or system whose motion is being investigated » action / reaction forces both act on different parts of the system – tensile-internal pulling forces when the structure is under tension – compressive- internal pushing (squeezing) forces act on the ends of an internal structure » do not accelerate the body, maintain integrity – Garfield slide

Classifying Forces l External Force: acts on object as a result of interaction with the environment surrounding it » non-contact - occur even if objects are not touching each other – gravity, magnetic » contact - occur between objects in contact – fluid (air & water resistance) – reaction forces with another body (ground, implement) l l vertical (normal) reaction force » acts perpendicular to bodies in contact shear reaction force » acts parallel to surfaces in contact (friction)

Friction l Component of a contact force that acts parallel to the surface in contact » » acts opposite to motion or motion tendency reflects interaction between molecules in contact reflects force “squeezing” surfaces together acts at the area of contact between two surfaces Static friction- when surfaces are not moving relative to each other l Dynamic friction- when surfaces move relative to each other l

Friction: create graph on board l Maximum static friction - maximum amount of friction that can be generated between two static surfaces » fixed value based on – nature of materials in contact – force holding the bodies together (normal contact force) l Dynamic friction - constant magnitude friction during motion » always less than maximum static friction

Reaction Force l Normal reaction force (normal contact force) » force acting perpendicular to two surfaces in contact. » magnitude intentionally altered to increase or decrease the amount of friction present in a particular situation – football coach on sled – push (pull) upward to slide object – pivot turn on ball of foot

Friction and Surface Area Friction force is proportional to the normal contact force l Friction is not affected by the size of the surface area in contact l » normal contact force distributed over the area in contact l Friction is affected by the nature of the materials in contact

Friction is FUN l Coefficient of friction (mu) - value that serves as an index of the interaction between two surfaces in contact. Examples of manipulating mu and Nc shoe design grips (gloves, tape, sprays, chalk) skiing: decrease for speed, increase for safety your examples? ? ?

Friction is FUN l Coefficient of friction (mu) - number that serves as an index of the interaction between two surfaces in contact. Homework for next day: perform the 4 (four) self-experiments on pages 53 to 56. Submit (next day) typed answers to the questions posed in the experiments. Ask for clarification of any experimental result that is confusing.

Addition of Forces (calculating the net [resultant] force) l Net force = vector sum of all external forces acting on the object (body) » account for magnitude and direction » ie. Add force of 100 N and 200 N

Addition of Forces (calculating the net [resultant] force) l Net force = vector sum of all external forces acting on the object (body) » account for magnitude and direction » ie. Add force of 100 N and 100 N – act in same direction? ? ? – Act in opposite direction? ? ? – Act orthogonal to each other? ? ? – Act at angles to each other? ? ?

Colinear forces Forces have the same line of action l May act in same or different directions l » ie tug of war teammates: 100 N, 200 N, 400 N – show force on rope graphically

Colinear forces Forces have the same line of action l May act in same or different directions l » ie tug of war teammates: 100 N, 200 N, 400 N » tug of war opponents: 200 N, 200 N – show force on rope graphically

Colinear forces Forces have the same line of action l May act in same or different directions l » ie tug of war teammates: 100 N, 200 N, 400 N » tug of war opponents: 200 N, 200 N » calculate resultant of the two teams – show force on rope graphically – calculate algebraically

Concurrent Forces do not act along same line, but do act through the same point l ie gymnast jumps up to grab bar. Coach stops swinging by applying force to front and back of torso. l » 20 N posterior directed push on front of torso » 30 N anterior directed push on back of torso » 550 N force from bar on gymnast’s hands » gymnast mass 50 kg

Concurrent Forces l gymnast hanging from grab bar. Coach applies force to front and back of torso to stop swing. » 20 N posterior directed push on front of torso » 30 N anterior directed push on back of torso » 550 N force from bar on gymnast’s hands » gymnast mass 50 kg

Concurrent Forces l gymnast hanging from grab bar. Coach applies force to front and back of torso to stop swing. » » l 20 N posterior directed push on front of torso 30 N anterior directed push on back of torso 550 N force from bar on gymnast’s hands gymnast mass 50 kg What is the resultant force? » Tip to tail method (fig 4. 8 & 4. 9 in text) » separate algebraic summation of horizontal and vertical – Pythagorean thereom to solve resultantl

Resolution of Forces l Forces are not colinear and not Hor & Vert? » ie figure 4. 1: forces on shot – 100 N from shot-putters hand – mass of shot = 4 kg » What is the resultant force on shot? – Draw Components of 100 N force l l solve graphically: tedious & imprecise trigonometric technique 100 N wt

Click here for a Quiz on the concepts of net force & friction

Trigonometry: SOH, CAH, TOA Click on Trigonometry above to go to a Penn State website on trig Key Equations: sin = opp/hyp l cos = adj/hyp l tan = opp/adj l l Pythagoras l a 2 + b 2 Pythagorean Humour = hyp (c) thereom c 2 opp adj Check this site.

Vector Composition Sample: l = 30 degrees l adj = 100 N l Find opp and hyp (c) opp adj

Vector Composition use l cos = adj/hyp l tan = opp/adj l hyp (c) opp adj

Vector Composition l l cos = adj/hyp cos 30 = 100/hyp = 100/. 866 hyp = 115. 47 N l l hyp (c) tan = opp/adj tan 30 = opp/100 opp =. 5774 x 100 opp = 57. 74 N opp adj

Vector Resolution Sample l = 35 degrees l hyp = 120 N l Find opp and adj hyp (c) opp adj

Trigonometric Calculations Use l sin = opp/hyp l cos = adj/hyp (c) opp adj

Vector Resolution l l sin = opp/hyp sin 35 = opp/120 opp = 120 X. 5736 opp = 68. 83 N l l hyp (c) cos = adj/hyp cos 35 = adj/120 adj = 120 X. 8192 adj = 98. 30 N opp adj

Free body diagram l Free body diagram - sketch that shows a defined system in isolation with all the force vectors acting on the system » defined system: the body of interest » vector: arrow to represent a force – length: size of the force – tip: indicates direction – location: point of application

Resolution of Forces l Forces are not colinear and not concurrent » ie figure 4. 1: forces on shot – 100 N from shot-putters hand – mass of shot = 4 kg = -40 N » What is the resultant force on shot? – Draw Components of 100 N force l l solve graphically: tedious & imprecise trigonometric technique 100 N 60 o wt -40 N

Recall l Concept of Net External Force

Recall Concept of Net External Force l Newton’s First Law of Motion l

Recall Concept of Net External Force l Newton’s First Law of Motion l » acceleration = 0 if net external force = 0 l Newton’s Second Law of Motion

Recall Concept of Net External Force l Newton’s First Law of Motion l » acceleration = 0 if net external force = 0 l Newton’s Second Law of Motion » a=F/m Static equilibrium: object is at rest (no motion) forces are in equilibrium (sum to 0 in all directions)

Free body diagram Me, at rest in front of class l sagittal plane view l What are the names of the forces? l How big are the forces? l What direction are the forces? l Where are the forces applied? l

Free body diagram ==>static analysis Me (75 kg): at rest (a = 0) in front of class l sagittal plane view F=ma F=0 l Wt & v. GRF Wt + v. GRF = 0 l Cof. M, at feet v. GRF = - Wt l 750 N, ? ? ? v. GRF = - (-750) l-&+ v. GRF = 750 N l

Free body diagram ==>static analysis Weightlifter (80 kg) l 100 kg bar overhead l at rest (a = 0) l sagittal plane view l What are the forces? l Where are the forces applied? l How big are the forces? l What direction are the forces? l

Free body diagram ==>static analysis Weightlifter (80 kg) l 100 kg bar overhead l at rest (a = 0) l sagittal plane view l Wt, bar & v. GRF l Cof. M, on hands, at feet l 800 N, 1000 N, ? ? ? l - , -, + l F=ma F=0 Wt + bar + v. GRF = 0 v. GRF = - Wt - bar v. GRF = - (-800) - (-1000) v. GRF = + 1800 N

Free body diagram ==>static analysis Child on swing (20 kg) l Mother’s force l » 40 N horizontal » 10 N upward at rest (a = 0) l force of swing on child? l

Free body diagram ==>static analysis Child on swing (20 kg) l Mother’s force l » 40 N horizontal » 10 N upward at rest (a = 0) l force of swing on child? l Fx = m a x Fy = m a y