phy1012f dynamics

NEWTON’S LAWS FORCE and MOTION

PHY1012F

DYNAMICS

Gregor [email protected]

NEWTON’S LAWS FORCE and MOTIONPHY1012F

2

FORCE and MOTIONLearning outcomes:

At the end of this chapter you should be able to…Correctly identify specific forces acting on “the system of interest”.Draw free-body diagrams as part of a problem-solving strategy for dealing with dynamics problems. Apply Newton’s laws of motion usefully and consistently to dynamics problems – developing a “Newtonian intuition” of the connection between force and change in motion (acceleration).


3

FORCEThe force of one object on a second object is part of the mutual interaction between the two objects. The other part of this interaction is the equal and opposite force which the second object exerts on the first.It is more convenient to consider the forces acting on one object at a time. The other object is the necessary agent.The force of the agent has “some effect”, i.e. a push or a pull, on the other object.Force is a vector quantity. Whether the force is a push or a pull, graphically we always place the tail of its ray on the affected object.

F


4

THE FUNDAMENTAL FORCES OF NATUREOur current understanding is that there are only four fundamental forces in nature:

the gravitational force (between objects which have mass);the electromagnetic force (between objects which are charged);the strong nuclear force (between protons and neutrons);the weak nuclear force (which results in radioactive beta decay).(In mechanics problems the last two forces are not

significant, but the electromagnetic force is in fact responsible for macroscopic “contact” forces (qv).)


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SUPERPOSITION OF FORCESIf several forces act simultaneously on a body, we can determine what is variously called the

net force, resultant force, or total force,the sum of all the forces,

by applying the principle of superposition of forces and using vector addition:

net 1 21

N

i Ni

F F F F F

netF

resultantF

resF

totalF

F


6

FORCE and MOTIONAristotle :

Objects behave according to their “nature”. By nature, all objects tend to be at rest in their “proper” positions.

“Natural motion” is simply objects striving to return to where they naturally belong. Things of the earth (e.g. stones) fall; things of the air (e.g. smoke) rise. Heavier objects, being more of the earth, fall faster than light objects.

Motion imposed by an pushing or pulling agent is “violent motion” and persists only while the action is sustained .


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FORCE and MOTIONGalileo:

In the absence of air resistance, all objects fall at the same rate.

a

How about that!


8



a


9



a

In the absence of friction or other opposing forces, a hori-zontally moving object will continue moving indefinitely.The property of an object which causes it to tend to “continue doing what it is doing” is called inertia.


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2. The acceleration of a system is directly related to the net force acting on the system:

3. Forces are interactions between systems.

FORCE and MOTIONNewton:

1. Any system in mechanical equilibrium remains in mechanical equilibrium unless compelled to change that state by a non-zero net force acting on the system.

netFa m


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SYSTEMS and ENVIRONMENTSA “system” refers to the object (or collection of objects) on which the forces under consideration are acting.

Everything else is the environment.

We shall deal only with systems of constant mass.

In the particle model…the system is treated as a single point of mass;a system cannot exert forces on itself.


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NEWTON’S FIRST LAW“An object which is at rest will remain at rest, or an object which is moving will continue to move with constant velocity, if and only if the net force acting on the object is zero.”Key phrases:

at rest: the system is in static equilibriumconstant velocity: the system is in dynamic equilibriumconstant velocity: neither speed nor direction changesif and only if: no change no force; no force no changenet force: balanced forces may be present, but effect no

change


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NEWTON’S FIRST LAWstatic equilibriumdynamic equilibrium

Newton’s first law is also known as the law of inertia [Latin, inertia : slothfulness, sluggishness, laziness].

Inertia is the property of a body which causes it to resist acceleration.

mechanical equilibrium

A system in (mechanical) equilibrium has zero accel-eration and defines what we mean by zero total force.


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FORCES CATALOGgravitationalelectromagnetic

strong nuclearweak nuclear

long-range forces

short-range forces

act at a distance

contact forcesapply only while the specific agent is in contact with the system!


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FORCES CATALOG Weight, …

is the gravitational pull of the entire Earth on any system with mass (!), near the surface of the Earth;

is a vector quantity which points downwards (towards the centre of the Earth);

is a long-range force, and works equally on stationary or moving systems;

is sometimes used to refer to only the magnitude of the weight force, w, given by w = mg;

is sometimes used INCORRECTLY when referring to the mass of an object. (As is the verb “weigh”.)

w


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FORCES CATALOG Spring force, …

is applied by elastic objects, which have been temporarily deformed, as they attempt to regain their original shapes;

(Hooke’s Law)

where k, the spring constant, depends on the spring material.

spF

spF

sp sF k s

s

s

s

spF


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FORCES CATALOG Tension force, …

is applied by flexible materials, such as string, rope, wire, etc. (which are ideally modelled as massless and non-stretchable);

is applied in the direction of the string or rope (it is difficult to use a length of string to push something!);

has a scalar counterpart referred to simply as tension, which is equal in magnitude to the force applied by each end of the string – and therefore also equal to the force applied to each end of the string.

T

2T

1T

3T

(T1 = T2)


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FORCES CATALOG Normal force, …

is applied by any surface touching the system;

is applied at right angles (i.e. normal ) to the surface;

is the force which prevents a system falling as a consequence of its weight. (Bodies in free-fall are NOT weightless – they are simply normal-force-less!)

balances, on horizontal surfaces, the system’s weight…

More generally, however,

n

cosn w

n

wcosw

n

w


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FORCES CATALOG Friction, …

appears (it is a responsive force) when a system moves or tries to move, across a surface;

acts parallel to the contact surface;

is called static friction, , when it prevents motion (i.e. the system is stuck to the surface);

is called kinetic friction, , when it opposes motion

(i.e. the system is sliding over the surface);

The magnitude of … is proportional to the normal force on the system;depends on the nature of both contact materials.

f

sf

kf

f


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FORCES CATALOG Static friction, …

acts in equilibrium with an applied force parallel to the surface, adjusting its magnitude to balance the force – but only up to a maximum, . After this the system slips and starts to move;

(direction opposite attempted motion)

where s, the coefficient of static friction, is a dimensionless number which must be determined by experiment.

s maxf

s max sf n

sf

n

wsf F

0v


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FORCES CATALOG Kinetic friction, …

replaces static friction once the system has begun to slide;

has a magnitude (less than ) which is nearly constant, irrespective of the system’s speed relative to the surface;

(direction opposite motion)

where k, the coefficient of kinetic friction, is a dimensionless number which must be determined by experiment.

s maxf

k kf n

kf

kf F

vn

w


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FORCES CATALOG Rolling friction, …

results from “welds” formed between the molecules in a wheel and those of the surface over which it rolls – bonds which must be broken as the wheel rolls and its molecules lift off the surface;

(direction opposite forward motion)

where r, the coefficient of rolling friction, is very much lower than k or s.

r rf n

rf


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FORCES CATALOG Approximate values for some coefficients of friction:

materials static, s kinetic, k rolling, r

rubber / tarmac (dry)steel / steel (dry)steel / steel (lubricated)teflon / teflon or steelsynovial joints (human)

1.00

0.80

0.10

0.04

0.01

0.80

0.80

0.60

0.05

0.04

0.01

0.50

0.02

0.002

rubber / tarmac (wet)


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FORCES CATALOG Drag, …

is the resistive force experienced by a system moving relative to a fluid (gas or liquid);

increases with increasing relative speed;

for objects of moderate cross-sectional area, A, and moderate speed, v, moving through air at the surface of the Earth is approximated by

D

Note: Air resistance may be neglected in all problems unless a problem explicitly states that it must be taken into account.

214D Av


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FORCES CATALOG Thrust, …

is the reactive force which a surface (or a system’s own exhaust) exerts on the system as a result of the system exerting a force on the surface (or exhaust).

thrustF

Note: Even though the exhaust may be a gas, it nevertheless constitutes a tangible agent in contact with the system.


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FREE-BODY DIAGRAMS

1. Draw a picture of the situation.

2. Circle the system of interest on your picture.

3. Identify all the significant forces acting on the system and draw and label them. (Wherever the environment touches the system you will find contact forces. Include relevant long-range forces, such as gravity.)

4. Redraw the system as a particle with the lengths of the force vectors representing their magnitudes (if possible).

5. Draw convenient coordinate axes centred on the particle.

6. Resolve forces into components where necessary.

A free-body diagram represents a system as a particle and uses rays to show all the forces acting on the system.


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FREE-BODY DIAGRAMSA 5 kg box is being pulled up a 37° slope by a rope parallel to the slope. If k = 0.40 and the tension in the rope is 60 N, what is the acceleration of the block?

1. Draw a picture of the situation.

5 kg

37°


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FREE-BODY DIAGRAMS

2. Circle the system of interest on your picture.

37°

5 kg

A 5 kg box is being pulled up a 37° slope by a rope parallel to the slope. If k = 0.40 and the tension in the rope is 60 N, what is the acceleration of the block?


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FREE-BODY DIAGRAMS

3. Identify all the significant forces acting on the system and draw and label them.

37°

n

w

5 kgkf

T

netF


?!


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FREE-BODY DIAGRAMS

4. Redraw the system as a particle with the lengths of the force vectors representing their magnitudes.

n

w

T

kf

netF



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FREE-BODY DIAGRAMS

5. Draw convenient coordinate axes centred on the particle (one axis usually points in dir’n of motion).

n

T

y

x

w

kf

netF



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FREE-BODY DIAGRAMS

6. Resolve forces into components along the axes where necessary.

n

kf

T

y

x

w

37°

ˆcos37 jw

ˆsin 37 iw

netF



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NEWTON’S SECOND LAW“A system of mass m subjected to a number of forces whose vector sum is undergoes an acceleration (in the direction of ) which is directly proportional to and inversely proportional to the mass of the system.”Notes:

. is the vector sum of all the individual forces acting on the system.If , the system is in equilibrium. This does not mean that there are NO forces acting on the system, merely that the vector sum of all the forces is zero (i.e. the forces are all “balanced” – a term which may be used ONLY when referring to forces acting on a single system).

netF

netF

netF

netFa m

netF

net 0F


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NEWTON’S SECOND LAWMore notes:

m is the inertial mass of the system,which relates the response of the system to the total force on it. (Mass can thus be regarded as a numerical measure of inertia.)The left and right sides of this equation are not equivalent. On the left are the physical forces acting on the system; the right represents the system’s response to these forces.Units: [kg m/s2 newton, N]One newton is the force required to accelerate a 1 kg mass at 1 m/s2.

netF ma

netFa m


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n

kf

T

y

x

ˆcos37 jw

ˆsin 37 iw

netF

= 0.4m = 5 kgTx = 60 Nax = ?

y yF ma

cos37 0 since 0yn w a

ˆ ˆj cos 37 j 0n mg

ˆcos37 j 0n mg

cos37 39 Nn mg

x xF ma

ksin37 xT w f ma

kˆ ˆ ˆ ˆi sin 37 i i ixT mg n ma

kˆ ˆsin 37 i ix xT mg n ma

ksin 37x xT mg u n ma 23 m/s up slopexa


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A resourceful engineering student investigates the motion of a lift simply by standing on an bathroom scale in the lift and taking certain readings.

…on which floor?!

(The building has 3.63 m between floors.) 77 kg

54 kg

68 kg

68 kg

As the doors close on the ground floor, the scale reads 68 kg. The reading climbs to 77 kg for 3 s, returns to 68 kg for another 5 s, and then drops to 54 kg for a short while before settling once again on 68 kg as the doors open…

WHAT EXACTLY DOES THAT SCALE MEASURE?


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stop

start77 kg

54 kg

68 kg

68 kg

v

a

0a

a


38

a0y

y

0y0, v0y, t0

a1y= 0

y1, v1y, t1

y3, v3y, t3

a2y

y0 = v0y = t0 = 0a0y = ?y1 = ? v1y = v2y = ? t1 = 3 sa1y = 0y2 = ? t2 = 3 + 5 = 8 sa2y = ? y3 = ? v3y = 0 t3 = ?

y2, v2y, t2


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y

0

a1y= 0

a0y

a2y

w w

w

w

w

w

sp0F

0 0y yF ma

netF

netF

sp1F

sp 2F

spF

spF

spF sp 0 0

ˆ ˆ ˆj j jyF mg ma

077 9.8 68 9.8 68 ya 2

0 1.30 m/sya

2 2y yF ma

sp 2 2ˆ ˆ ˆj j jyF mg ma

254 9.8 68 9.8 68 ya 2

2 2.02 m/sya


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a0y

y

0y0, v0y, t0

a1y= 0

y1, v1y, t1

y3, v3y, t3

a2y

y0 = v0y = t0 = 0a0y = 1.30 m/s2

y1 = ? v1y = v2y = ? t1 = 3 sa1y = 0y2 = ? t2 = 3 + 5 = 8 sa2y = –2.02 m/s2

y3 = ? v3y = 0 t3 = ?

y2, v2y, t2

y1 = y0 + v0y(t1 – t0) + ½a0y(t1 – t0)2

y1 = 0 + 0 + ½ 1.3 32 = 5.84 m

v1y = v0y + a0y(t1 – t0)

v1y = 0 + 1.3 3 = 3.89 m/s


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a0y

y

0y0, v0y, t0

a1y= 0

y1, v1y, t1

y3, v3y, t3

a2y

y0 = v0y = t0 = 0a0y = 1.30 m/s2

y1 = 5.84 m v1y = v2y = 3.89 m/s t1 = 3 sa1y = 0y2 = ? t2 = 3 + 5 = 8 sa2y = –2.02 m/s2

y3 = ? v3y = 0 t3 = ?

y2, v2y, t2

y2 = y1 + v1y(t2 – t1) + ½a1y(t2 – t1)2

y2 = 5.84 + 3.89 5 + 0 = 25.3 m

v3y = v2y + a2y(t3 – t2)

0 = 3.89 + (–2.02)(t3 – 8) t3 = 9.93 s


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a0y

y

0y0, v0y, t0

a1y= 0

y1, v1y, t1

y3, v3y, t3

a2y

y0 = v0y = t0 = 0a0y = 1.30 m/s2

y1 = 5.84 m v1y = v2y = 3.89 m/s t1 = 3 sa1y = 0y2 = 25.3 m t2 = 3 + 5 = 8 sa2y = –2.02 m/s2

y3 = ? v3y = 0 t3 = 9.93 s

y2, v2y, t2

y3 = y2 + v2y(t3 – t2) + ½a2y(t3 – t2)2

y3 = 25.3 + 3.89 1.93 + ½ (–2.02) 1.932

y3 = 29.0 m

She stops on the 8th floor. 29.03.63


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smgm

y

x

Determine the maximum starting acceleration of a F1 car.

In Formula 1 the thrust may be provided only by friction between the tyres and the track…

thrust s maxF f

an

wthrustF

n

w

thrustF

x xF ma

sˆ ˆi ixn ma

y yF ma

0 since 0yn w a

s max xf ma

ˆ ˆj j 0n mg n mg s

xna m

29.8 m/ssg


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30°

y

x

A cricket pitch roller has a mass of 350 kg. The coefficient of rolling friction between the roller and the pitch is 0.06. If the handle of the roller makes an angle of 30° with the horizontal, explain (quantitatively) why it is easier for the groundsman to pull the roller at a constant speed rather than to push it.

0a

rf

An

w

w

n

AF

rf

w

n

BF

rAf AF

A sin 30F

A cos30F

y

x

0a Bn

w

30°rBf

BF

B sin 30F

B cos30F


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0x xF ma

0y yF ma

A sin 30 0n F mg

y

x

0a

An

w

rAf AF

A sin 30F

A cos30F

y

x

0a

Bn

w

rBf

B sin 30F

B cos30F

A3430 0.5n F

A rAcos30 0F f

A rcos30 0F n

A A0.866 0.06 3430 0.5 0F F

A 230 NF

0x xF ma

0y yF ma

B sin 30 0n F mg

B3430 0.5n F

B rBcos30 0F f

B rcos30 0F n

B B0.866 0.06 3430 0.5 0F F

B 246 NF


46

NEWTON’S THIRD LAW“If system A exerts a force on another system B, then B exerts a force of the same magnitude on A but in the opposite direction.”

Notes: and are known as an action/reaction pair.Remember that and act on different systems and can therefore never be described as “balanced”.Systems connected by massless strings passing over massless, frictionless pulleys act as if they interact via an action/reaction pair. String merely “transmits” the force.

A on B B on AF F

A on BF

B on AF

A on BF

B on AF


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y

Two blocks are connected by a light, inextensible string which passes over a frictionless pulley as shown. The coefficient of friction for the 4 kg block and the 37° slope is k = 0,15. Determine the tension in the string.

4 kg3 kg

37°

4a

3a

a

n

kf

4w

3w

3T

37°

x

y

4 cos37w 4 sin 37w

We have assumed that the system accelerates in the direction shown. 4T

Constraints:a4x = –a3y

T4 = T3 = T

4a

3a


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x4 kg

3 kg37°

a

a

ayn

kf T

4 cos37w 4 sin 37w 3w

T

y

4 4 4y yF m a

4 cos37 0n w

31.3 Nn

4 4 4x xF m a

4 k 4 4sin37 xT w u n m a

4 328.34x y

Ta a

3 3 3y yF m a

3 3 3 yT w m a

329.4 3 yT a

28.329.4 3 4TT

28.9 NT


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FORCE and MOTIONLearning outcomes:

At the end of this chapter you should be able to…Correctly identify specific forces acting on “the system of interest”.Draw free-body diagrams as part of a problem-solving strategy for dealing with dynamics problems. Apply Newton’s laws of motion usefully and consistently to dynamics problems – developing a “Newtonian intuition” of the connection between force and change in motion (acceleration).

phy1012f dynamics

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