SPH4U: Lecture 8 Notes - The Burns Home Page

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SPH4U: Lecture 8 Notes - The Burns Home Page

SPH4U: Lecture 8Today’s Agenda• Friction Recap• Drag Forces‣ Terminal speed‣ A special very cool demo….• Dynamics of many-body systems‣ Atwood’s machine‣ General case of two attached blocks on inclined planes‣ Some interesting problemsFriction Review:• Surface friction is caused by the “microscopic” interactionsbetween the two surfaces:‣ See discussion in textSPH4U: Lecture 8, Pg 1SPH4U: Lecture 8, Pg 2Model for Surface Friction• The direction of the frictional force vector f F is perpendicular to thenormal force vector N, in the direction opposing relative motion ofthe two surfaces.Frictional force vs applied force• Graph of Frictional force vs Applied force:• Kinetic (sliding): The magnitude of the frictional force vector isproportional to the magnitude of the normal force N.f F = K NIt moves, but it heats up the surface it moves on!f F = S Nf F = K N• Static: The frictional force balances the net applied forces such thatthe object doesn’t move. The maximum possible static frictionalforce is proportional to N.f F S N and as long as this is true, then f F = f A in oppositedirectionIt doesn’t move!f Ff F = F AF ASPH4U: Lecture 8, Pg 3SPH4U: Lecture 8, Pg 4Page 1


Kinetic Friction:draw a free body diagram, and follow the rules!• K is the coefficient of kinetic friction.i : F K N = maj : N = mgso F K mg = maStatic Friction:• The “coefficient of static friction,” S , determines maximumstatic frictional force, S N, that the contact between the objectscan provide.• S is discovered by increasing F until the object starts to slide:F MAX - S N = 0 N = mg (in this case)F MAX = S mg S F MAX / mgNjNjFmaiF MAXi K Nmg S NmgSPH4U: Lecture 8, Pg 5SPH4U: Lecture 8, Pg 6Lecture 8, Act 1friction dynamics• A block of mass m, when placed on a rough inclined plane( > 0) and given a brief push, keeps moving down the plane withconstant speed.‣ If a similar block (same ) of mass 2m were placed on the sameincline and given a brief push, it would:m(a)stop(b) accelerate(c) move with constant speedLecture 8, Act 1Solution• Since the velocity is constant, it’s just broken free from S. F.‣ Net force down ramp is essentially zero• Draw FBD and find the total force in the x-directionF NET,X = mg sin q K mg cos qji K NNqmgmg sinq= ma = 0 (first case)qDoubling the mass will simplydouble both terms…net forcewill still be zero!Speed will still be constant!Increase the friction and thedownhill force by the samefactor nothing changes!SPH4U: Lecture 8, Pg 7SPH4U: Lecture 8, Pg 8Page 2


Friction in Fluids: Drag Forces• When an object moves through a viscous medium, like air orwater, the medium exerts a “drag” or “retarding” force thatopposes the motion of the object relative to the medium.Many-body Dynamics• Systems made up of more than one object• Objects are typically connected:F DRAG‣ By ropes & pulleystoday‣ By rods, springs, etc.later onvF g = mgjSPH4U: Lecture 8, Pg 9SPH4U: Lecture 8, Pg 10Atwood’s Machine:Masses m 1 and m 2 are attached to an ideal massless stringand hung as shown around an ideal massless pulley.Atwood’s Machine...• Draw free body diagrams for each object• Applying Newton’s Second Law: ( j -components)• What are the tensions in thestring T 1 and T 2 ?• Find the accelerations, a 1 anda 2 , of the masses.• Use FBD• Solve for motion• You’re a hero!Fixed Pulleyjma 11m a 22‣ T 1 - m 1 g = m 1 a 1‣ T 2 - m 2 g = m 2 a 2Yikes! 2 eqn, but 4 unk???But T 1 = T 2 = Tsince pulley is idealand a 1 = -a 2 = -a.since the masses areconnected by the stringFree Body DiagramsT 1 T 2aj1 a 2T 1T 2SPH4U: Lecture 8, Pg 12m 1 gm 2 gSPH4U: Lecture 8, Pg 11Page 3


m 1 g - T = m 1 a (a)T - m 2 g = m 2 a (b)• Two equations & two unknowns‣ we can solve for both unknowns (T and a).• Add (b) + (a):‣ g(m 1 - m 2 ) = a(m 1 + m 2 )(m1- m2)‣ a =g(m1+ m2)• Subract (b) - (a):2(m1- m2)‣ 2T - g(m 1 + m 2 ) = -a(m 1 - m 2 ) = - gm1+ m2‣ Use polynomial arithmetic…‣ T = 2gm 1 m 2 / (m 1 + m 2 )Atwood’s Machine...aFree Body DiagramsTm 1 gTam 2 g• So we find:m ma ( 1 2 )( m m )g1 22 m1m2T = g(m + m )12Atwood’s Machine...aTm1Tm2jaSPH4U: Lecture 8, Pg 13SPH4U: Lecture 8, Pg 14Is the result reasonable?Check limiting cases!m ma ( 1 2 )( m m )g1 22 m m1 2T g( m m )12A related situation:Attached bodies on two inclined planessmooth peg• Special cases:i.) m 1 = m 2 = m a = 0 and T = mg. OK!ii.) m 2 or m 1 = 0 |a| = g and T= 0. OK!m 1m 2• Atwood’s machine can be used to determine g (bymeasuring the acceleration a for given masses).(m2 m1)g =+ a(m2- m1)If m 1 is almost m 2 , then acceleration will be small.You can measure motion for a long time. More accurate….SPH4U: Lecture 8, Pg 15q 1 q 2all surfaces frictionlesspeg is frictionlessSPH4U: Lecture 8, Pg 16Page 4


How will the bodies move?From the free body diagrams for each body, and the chosencoordinate system for each block, we can apply Newton’sSecond Law:Taking “x” components:1) T 1 - m 1 g sin q 1 = m 1a 1X2) T 2 - m 2 g sin q 2 = m 2 a 2XBut T 1 = T 2 = Tand -a 1X = a 2X = a(constraints)yNm 1m 1 gq 1xT 1T 2xm 2 gyNm 2Solving the equationsUsing the constraints, we get 2 eqn and 2 unks,solve the equations.T - m 1 gsin q 1 = -m 1 a(a)T - m 2 gsin q 2 = m 2 a(b)Subtracting (a) from (b) gives:m 1 gsin q 1 - m 2 gsin q 2 = (m 1 +m 2 )aSo:m ma 1 sin q1 2 sin q2gm1 m2q 2SPH4U: Lecture 8, Pg 18SPH4U: Lecture 8, Pg 17m 1m 2q 1 q 2m ma 1 sin q1 2 sin q2gm1 m2m m 1 2m ma 1 sin q1 2 sin q2gm1 m2q 1 q 2Special Case 1:Special Case 2:BoringTTAtwood’s Machinem 1m 2If q 1 = 0 and q 2 = 0, a = 0.m1m2If q 1 = 90 and q 2 = 90,m ma ( 1 2 )( m m )g1 2SPH4U: Lecture 8, Pg 19SPH4U: Lecture 8, Pg 20Page 5


m m 1 2m ma 1 sin q1 2 sin q2gm1 m2q 1 q 2Special Case 3:m 1Lab configurationm 2Lecture 8, Act 2Two-body dynamics• In which case does block m experience a larger acceleration?In (1) there is a 10 kg mass hanging from a rope. In (2) ahand is providing a constant downward force of 98.1 N. Inboth cases the ropes and pulleys are massless.mma10kgaF = 98.1 NIf q 1 = 0 and q 2 = 90,ma 2( m m g1 2 )Case (1) Case (2)(a) Case (1) (b) Case (2) (c) sameSPH4U: Lecture 8, Pg 21SPH4U: Lecture 8, Pg 22aLecture 8, Act 2Two-body dynamics• In which case does block m experience a larger acceleration?In (1) there is a 10 kg mass hanging from a rope. In (2) ahand is providing a constant downward force of 98.1 N. Inboth cases the ropes and pulleys are massless.10kgm W =10kgm(a) Case (1) (b) Case (2) (c) sameamF = 98.1 NCase (1) Case (2)• Add (a) and (b):m W g = (m + m W )amWga m m• Note:WLecture 8, Act 2Solution• For case (1) draw FBD and write F NET = ma for each block:T = ma (a)(a)m W g -T = m W a(b)mT mWg m mWa10kg(b)mm W =10kgSPH4U: Lecture 8, Pg 23SPH4U: Lecture 8, Pg 24Page 6


Lecture 8, Act 2Solution• For case (2) T = 98.1 N = maa98.1Na m 10kg10kgCase (1)m98.1Na m• The answer is (b) Case (2) In this case the block experiences alarger acceleratioina98.1Na mmF = 98.1 NCase (2)SPH4U: Lecture 8, Pg 25Problem: Two strings & Two Masses onhorizontal frictionless floor:• Given T 1 , m 1 and m 2 , what are a and T 2 ?T 1 - T 2 = m 1 a (a)T 2 = m 2 a (b)‣ Add (a) + (b):T1T 1 = (m 1 + m 2 )a a =m1+ m‣ Plugging solution into (b):m2T2= T1m + m12m 2m 1T 2 T 12aSPH4U: Lecture 8, Pg 26iLecture 8, Act 3Two-body dynamics• Three blocks of mass 3m, 2m, and m are connected bystrings and pulled with constant acceleration a. What is therelationship between the tension in each of the strings?• Draw free body diagrams!!T 3 = 3maLecture 8, Act 3Solution3mT 3a3mT 3 T 2 T2m1mT 2 - T 3 = 2maT 3 T 2 2mT 2 = 2ma +T 3 > T 3T 2 T 1(a) T 1 > T 2 > T 3 (b) T 3 > T 2 > T 1 (c) T 1 = T 2 = T 3T 1 - T 2 = mamT 1 = ma + T 2 > T 2T 1 > T 2 > T 3SPH4U: Lecture 8, Pg 27SPH4U: Lecture 8, Pg 28Page 7


• Alternative solution:Consider T 1 to be pullingall the boxesT 2 is pulling only theboxes of mass 3mand 2mT 3 is pulling only thebox of mass 3mLecture 8, Act 3Solution3m3m3maT 3 T 2 T2m1maT 3 T 2 T2m1maT 3 T 2 T2m1miProblem: Accelerometer• A weight of mass m is hung from the ceiling of a car with amassless string. The car travels on a horizontal road, andhas an acceleration a in the x direction. The string makesan angle q with respect to the vertical (y) axis. Solve for qin terms of a and g.aqT 1 > T 2 > T 3SPH4U: Lecture 8, Pg 29SPH4U: Lecture 8, Pg 30Accelerometer...Accelerometer...• Draw a free body diagram for the mass:‣ What are all of the forces acting?qT (string tension)• Using components (recommended):i: F X = T X = T sin q = maj: F Y = T Y mgq= T cos q mg = 0 Tqjimmg (gravitational force)mammgiSPH4U: Lecture 8, Pg 31SPH4U: Lecture 8, Pg 32Page 8


Accelerometer...Accelerometer...• Using components :i: T sin q = ma• Alternative solution using vectors (elegant but not assystematic):• Find the total vector force F NET :j: T cos q- mg = 0• Eliminate T :T sin q= ma tan q aT cos q= mg gTmaqmmgijmgqTF TOTqT (string tension)mmg (gravitational force)SPH4U: Lecture 8, Pg 33SPH4U: Lecture 8, Pg 34Accelerometer...tan q a gAccelerometer...• Alternative solution using vectors (elegant but not assystematic):• Find the total vector force F NET :• Recall that F NET = ma:T (string tension)qT• Somgqmama atan q mg gmmg (gravitational force)tan q a g• Let’s put in some numbers:• Say the car goes from 0 to 60 mph in 10 seconds:‣ 60 mph = 60 x 0.45 m/s = 27 m/s.‣ Acceleration a = Δv/Δt = 2.7 m/s 2 .‣ So a/g = 2.7 / 9.8 = 0.28 .‣ q = arctan (a/g) = 15.6 degaqCart w/accelerometerSPH4U: Lecture 8, Pg 35SPH4U: Lecture 8, Pg 36Page 9

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