applied mathematics: lecture 4 - duke...

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction

Newton’sThree Laws ofMotion

DifferentKinds ofForces andBasicExamples

MoreAdvancedProblems

Higher LevelProblems

Applied Mathematics: Lecture 4Exam Question 4: Newton’s Laws & Connected Particles

Brendan Williamson

June 2, 2015

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Introduction

• In this chapter, we will move away from traditional questionsinvolving acceleration, velocity and position. Instead we will lookat the effects impact, gravity and friction have on the accelerationof an object. For example, how will a block move when placed onan angled surface? What will a 10kg block do when it is hit witha 50g bullet? What happens when objects of different masses areput on each end of a pulley?

• Unlike the previous chapters, which required very little interactionwith physics concepts, the next couple of chapters will deal withthe ideas of force, friction and momentum. In particular, we willstart with looking at Newton’s three laws of motion.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Newton’s Laws of Motion

• First Law: A body will remain in a state of rest or at constantvelocity in a straight line, unless it is acted on by an external force.

• Meaning: Objects don’t start, or stop moving by themselves. Ifan object slows down, speeds up, or changes direction, some forceacted on it. This may have been an impact with another object,friction, gravity, or something else.

• Second Law: The change in momentum per unit time isproportional to the force applied, and takes place in the directionof the force.

• Meaning: Momentum is mass×velocity, so change in momentumper unit time is (mv −mu)/t = ma. So in mathematical terms,F ∝ ma, and since acceleration and force are vectors, they act inthe same direction. We can change the proportionality to equalityby noting that F = kma, and 1N=1kg m/s2 and so 1 = k(1)(1),so k = 1 and F = ma.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Newton’s Laws of Motion

• Third Law: To every action there is an equal and oppositereaction.

• Meaning: Nothing to do with good and bad karma. This is to dowith recoil from firing a gun or why you lean forward whenpushing something. When a force is applied in one direction, aforce of equal magnitude is applied in the opposite direction. Forexample, if you push someone on an ice rink, you will get pushedin the opposite direction with the same force.

• Notes: Collectively, the most important thing to take fromNewton’s Laws is that only force can cause acceleration, and itdoes so according to the equation F = ma. So when we want tocalculate the acceleration of something, we look at the forcesacting on it and go from there.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Motivating Examples

• One of the most important things in this chapter is figuring outwhere forces are coming from. For example, a car moving along astraight road will experience a force from the engine. There willalso be resisting forces from air friction, and maybe the brakes.

• Something hanging from a rope experiences a tensile force fromthe rope, pulling it upwards. It also experience a gravitationalforce pulling it down, and where possible, the tensile force willmatch the gravitational one, so that the object doesn’t move.Also, there is a tensile force of equal magnitude exerted on theother end of the rope (we will see more of this later).

• An object sitting on a horizontal table experiences what is called anormal reaction exerted on it by the table. The object is beingpulled down by gravity, but is not moving. This is because there isan equal force pushing it up, this is a force coming from thesurface of the table. You can also think of this as Newton’s ThirdLaw, the table is pushing the object in the opposite direction theobject is pushing down on the table, but we are only interested inthe object.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Motivating Examples

• An object falling from the sky into the ground will experience aresistance force when it hits the ground. If instead the object fallsinto water, it’s easier to imagine the deceleration of the objectonce it hits the water.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Examples

• A 1000kg car is travelling along a straight road. Its engineprovides a tractive force of 4,000N, and there is air resistance of1,000N. What is the acceleration of the car?

• If we consider the direction of the car to be the positive ~idirection, then the traction of the car is a force of 4, 000~iN andthe air resistance is a force −1, 000~iN, so that there is a netoverall force of 3, 000~iN acting on the car. ThusF = ma⇒ 3, 000 = 1000a⇒ a = 3m/s2.

• If a 100g rock, falling vertically downwards, hits soft earth at avelocity of 100m/s, and stops 0.2m into the ground, what is theresistance of the earth?

• We need to find the deceleration of the rock, which is done usingthe equation v2 = u2 + 2as, giving us a = −25, 000m/s2. ThusF = 0.1× 25, 000 = 2, 500N.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Examples

• An object is hanging from a rope. The object is not moving, andthe tension in the rope is 49N. What is the mass of the object?

• There are two forces acting on the object, gravity and the tensionof the rope. They are acting in the opposite direction and so mustbe equal, since the object is not moving. Thus the force due togravity must also be 49N. So 49 = mg and so m = 5kg .

• Note: In diagrams tension forces are usually denoted with theletter T , or S if there is more than one string.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Different Types of Force

• So far we have seen a few different types of forces, namelytension, resistance and direct forces such as the engine of a car.In Q5 another type of force will be introduced, impact, which issynonymous with studying momentum, but for now we willconsider only two more forces, normal reactions and friction.

• Normal reactions are the type we talked about when consideringwhat force kept an object on a table stationary. They occur whenan object is pressing on another object, usually due to gravity.They are called normal reactions because they act perpendicularto the surface they originate from.

• For example, if an object resting on a table has a mass of 10kg, agravitational force of 98N is exerted on it, directed downwards.But it does not move, because the table exerts a reactionaryforce of 98N to counter this. If we were on the moon, the forcewould be a lot weaker, and the materials of the table would beunder a lot less strain!

• Normal reactions are usually denoted with the letter R indiagrams.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Different Types of Force

• Friction is closely related to normal reaction. If an object isheavier, and the surface it’s resting on is exerting a larger normalreaction to prop it up, the object will be more resistant tomovement, as it is pressed tighter against the surface and socausing more friction. So a larger normal reaction causes a largerfrictional force when movement is attempted.

• It happens that the frictional force is (approximately) proportionalto the normal reaction, and the constant of proportionality,usually denoted µ, is between 0 and 1. 0 would indicate aperfectly smooth and unrestricted movement, and 1 wouldindicate an incredibly rough surface. In practice, µ depends on thenature of the two surfaces in contact, but not their size or shape.Two blocks made of identical polished wood on the same marblesurface would have the same coefficient of friction, even if theywere of different sizes.

• The frictional force acts in the opposite direction to attemptedmovement, and its magnitude is only ever large enough to stopmovement. So if a frictional force of 30N can be applied but theforce attempting movement is 20N, then the frictional force willonly be 20N in the opposite direction.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Example

• A block of stone with a mass of 1,000kg is resting on horizontalground. The coefficient of friction between the stone and theground is µ = 3/4. A force of 10,000N is applied to the rockacting parallel to the ground (i.e. pushing the stone). Find theacceleration of the rock.

•In the above diagram we have all the forces acting on the stone:gravity, the normal reaction between the ground and the stone,the 10,000N force applied to the stone, and the frictional forceacting in the opposite direction to it.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Example

• Since there is no movement in the vertical direction,R = Mg = 9, 800N, and so F = µR = 7, 350N. Thus the overallforce acting on the stone is of 2,650N rightwards. Thus theacceleration of the stone is a = F/m = 2.65m/s2.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Example

• If it was 10 people that were pushing the stone, each with anon-competing force of 1,000N, how many of them would havebeen needed to ensure the stone moved?

• The frictional force provides a resisting force of 7,350N, so wewould need 8 people, providing a force of 8,000N to move thestone.

• The maximum amount of friction that can be put into play, inthis case 7,350N, is called the limiting friction.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





More Advanced Problems 1

• The remaining problems that we will see on this topic involve acombination of frictional surfaces (although we will sometimesignore friction) which are sometimes angled, along with pulleys,and string.

• Two particles are connected by a taut inelastic string resting on asmooth pulley above a table as shown below. The 4kg particle is1m above the table. What is the tension in the string? What isthe common acceleration of the particles? How long does it takefor the 4kg particle to hit the table? How much higher does the3kg mass travel after the 4kg mass hits the table?

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• Since there is an upwards force of T (tension on the string) and adownward force of Mg , resulting in an overall force of F = Mawhere a is the acceleration of the connected particles, we havethat

T − 3g = 3a

4g − T = 4a

⇒ 4g − (3a + 3g) = 4a

⇒ g

7= a

⇒ T =24g

7.

The original two equations come from the forces acting on eachparticle, namely tension and gravity, and the fact that theresulting overall force is equal to Ma, where a is acting upwardson the 3kg particle, and downwards on the 4kg particle. To seehow long it takes the 4kg particle to reach the ground, we haveu = 0, s = 1, a = g/7, and so t =

√7/2g .

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• To calculate how high the 3kg particle travels after the 4kgparticle hits the table, note first that there is no tension acting onit; the string is slack. So the only force acting on it is gravity. Sowe set a = −g , v = 0. For our third piece of information, wecalculate the velocity of the 4kg particle at the time when it hitthe table. Calculating that v =

√g/14, and so letting

u =√

g/14, a = −g , v = 0 we can calculate that s = 1/28m.

• It’s worth noting that only the ratio between the masses of theparticles is important in calculating a. You can see this byreplacing 3 and 4 with 3M and 4M respectively, or also bynoticing that the distance, velocity and acceleration vectors areindependent of the kg measurement.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• An object of mass 2M is placed on a table. It is connected to asecond particle of mass M by a taut inelastic string passing over asmooth pulley hanging over the edge of the table, as in thediagram below.

• Assume first that the table is smooth. What is the commonacceleration of the particles? What is the common acceleration ofthe particles if the coefficient of friction between the particle ofmass M and the table is µ = 0.2? What is the smallest value of µs.t. the acceleration of the particles is zero?

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• The following are the forces acting on each object:

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• The vertical forces on the 2M object cancel out and are of littleconcern. From the horizontal force we get a resultant force of Tto the left, and so T = 2Ma. From the M object we deduce thatMg − T = Ma, and so a = g/3, T = 2Mg/3.

• If µ = 0.2, then on the 2M object we would have an extra force ofmagnitude F = µR = µ2Mg = 0.4Mg , and so we have

T − 0.4Mg = 2Ma

Mg − T = Ma

g/5 = a.

• To see the smallest µ to guarantee inertia, after similarcalculations we will end up with the equations

T − 2µMg = 2Ma

Mg − T = Ma

(1− 2µ)g/3 = a.

Setting µ ≥ 1/2 will guarantee a = 0.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• This time consider a third object, attached to the object on thetable and hanging over the other side of the table, as in thefollowing diagram:

• What is the common acceleration of the objects (i) if the table issmooth? (ii) if the table and 2M particle have a coefficient offriction of µ = 0.5?

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• (i) Like before, we construct a set of equations based on theforces at play. This time to have two strings, with tensions S andT . It should be clear that acceleration will take place in the”clockwise” direction.

S −Mg = Ma

T − S = 2Ma

3Mg − T = 3Ma.

By solving these equations, we get that a = g/3.

• (ii) In this case, we need to look more indepth at the forces actingon the 2M mass:

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• In this case, since there is no vertical movement, R = 2Mg and soµR = Mg . Our equations now look like

S −Mg = Ma

T − S −Mg = 2Ma

3Mg − T = 3Ma.

Solving these equations will now give a = g/6.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• So far all the problems we have seen involved pulleys, strings andflat surfaces. Now consider a particle on an angled surface, as inthe following diagram:

• What is its acceleration? Assume that the wedge is immovable. Ifthe slope is smooth, then the forces acting on the wedge are asfollows:

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• In order to solve the problem, we need to resolve the forces alongperpendicular axes. This is best done along the angled slope, sothat the Mg force is resolved as follows:

• Since there is no movement in the direction perpendicular to theslope, we have that R =

√3Mg/2. The object moves down the

slope with a force of Mg/2, and so to solve for the acceleration ofthe object, we set Mg/2 = Ma⇒ g/2 = a.

• In this case solving for R was unnecessary, but what if there is nowfriction? Set µ = 0.2, the coefficient of friction between the blockand the slope. Then we have a force acting up the slope of sizeµR =

√3Mg/10, and so the overall force acting down the slope is

Mg(1/2−√

3/10) and so in this case a = (1/2−√

3/10)g .

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• What if the object in Problem 4 was attached by a string toanother object hanging off the vertical side of the wedge?

• Like before, first assume the slope is smooth. We have adownwards force of Mg and an upwards force of T acting on theblock hanging vertically, so that we have the equationMg −T = Ma, where a is the common acceleration of the blocks.From the block on the slope, we have a force of T acting up theslope, and a force of Mg/2 acting down, just like before. So wehave

Mg − T = Ma

T − Mg

2= Ma

⇒ Mg

4= a.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Angle of Friction

• Before looking at more complicated problems involving force, weneed to introduce one more definition. When a surface andparticle have coefficient of friction µ, the angle of friction isdefined as the value λ such that tanλ = µ. While this may seemarbitrary, in fact the value of λ is the minor angle in theright-angled triangle made by the normal reaction and thefrictional force, as shown below:

• To see another intuitive description of λ, look at problem 9(c) onthe worksheet, where it can be shown that in the setup as inProblem 4 of these notes, the angle of inclination must be lessthan λ to ensure movement of the block.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





More Advanced Problems 6: Higher Level Only

• What if we adjusted the setup of Problem 5 so that the verticalside was now angled? Consider the following diagram:

• If we assume all surfaces are smooth, the forces acting on theobjects are as follows:

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





More Advanced Problems 6: Higher Level Only

• As before, we individually resolve the forces into componentsalong and perpendicular to the slopes in the following way:

• Assuming that the block of mass M is the one that falls, this willyield

T − 2Mg sin 30 = 2Ma

Mg sin 45− T = Ma

⇒ 1−√

2

3√

2= a.

Introducing friction into the system is dealt with the same way asbefore.

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Higher Level Problems, Relative Acceleration:Compound Pulleys

• Consider the following system:

• Particles B, D and E have mass 3M, M and 4M respectively.Pulley C is weightless. Particle B and the pulley C are connectedby a taut string passing over pulley A. Particles D and E aresimilarly connected by a taut string passing over pulley C . Findthe common acceleration of B and C , and the commonacceleration of D and E relative to B and C .

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Higher Level Problems, Relative Acceleration:Compound Pulleys

• The forces acting on the particles are as follows:

• S and T are the tensile forces, a is the acceleration of B and C , bis the acceleration of D and E relative to C . So we have thefollowing equations:

S − 3Mg = 3Ma

S − 2T = 0

T −Mg = M(b − a)

4Mg − T = 4M(a + b).

Solving these equations will give a = g/31, b = 18/31g .

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction





Higher Level Problems, Relative Acceleration: MovableWedges

• When first dealing with particles on wedges, we assumed thewedge was immovable. Now we will remove that assumption, andin doing that we will need to give it a mass.

• The movement here is based on normal reactions between theobjects and between the wedge and the floor, and gravity. Allsurfaces are assumed to be smooth, and this will always be thecase for problems of this type. The forces on the objects are asfollows:

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






•• R is the normal reaction between the block and the wedge, and S

is the normal reaction between the wedge and the surface it restson. a is the acceleration of the block relative to the wedge, and bis the acceleration of the wedge.

• Resolving the forces acting on the block and the accelerations intodirections perpendicular and parallel to the slope of the block, wehave the following:

AppliedMathematics:

Lecture 4

BrendanWilliamson

Introduction






• Thus we have two equations:

√3Mg

2− R =

Mb

2

Mg

2= M

(a−√

3b

2

).

• Resolve the R force acting on the wedge into its vertical andhorizontal components, we get another pair of equations:

√3R

2+ 3Mg = S

R

2= 3Mb.

The third equation is superfluous but using the other 3, we canshow that b =

√3g/13, and therefore that a = 8g/13.

applied mathematics: lecture 4 - duke...

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