MSU Physics 231 Fall 2015 1

Physics 231Topic 4: Forces & Laws of Motion

Alex Brown September 23-28

2015

MSU Physics 231 Fall 2015 2

What’s up? (Wednesday Sept 23)

1) Homework set 03 due Tuesday Sept 29rd 10 pm

2) Learning Resource Center (LRC) in 1248 BPS

Monday 9 am to 1 pm and 3 pm to 5 pm

Tuesday 9 am to 9 pm

Friday 9 am to 1 pm

3) First midterm in here during class time on Wednesday Sept 30

Covers Chapters 1-4 and homeworks 01-03

MSU Physics 231 Fall 2015 3

• The 1st exam will be Wednesday September 30.

• The exam will take place right here in BPS 1410.

• About 15 multiple choice questions that you should be able to answer in 1 to 5 min each.

• We will have assigned seating, so you need to show up 10min early to guarantee your seat.

• If you are sick, you MUST have a note from a doctor on the doctor’s letterhead or prescription pad.

• You CANNOT use cell phones during the exam.

• You can (should!) bring a hand written equation sheet - 8.5x11 sheet of paper (both sides).

• Bring your calculator and several no. 2 pencils.

• Bring your MSU picture ID. 3

MSU Physics 231 Fall 2015 4

Recall from Last Chapter

Vectors and Scalars

Two Dimensional Motion Velocity in 2D

Acceleration in 2D

Projectile motion Throwing a ball or cannon fire

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Key Concepts: Forces & Laws of Nature

Force and Mass

Newton’s Laws of Motion Inertia, F=ma, Equal/Opposite Reactions

Application of Newton’s Laws Force diagrams

Component Forces

Friction and Drag Kinetic, static, rolling frictions

Covers chapter 4 in Rex & Wolfson

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Isaac Newton (1642-1727)

“In the beginning of 1665 I found the…rulefor reducing any dignity of binomial to a series. The same year in May I found the method of tangents and in November the method of fluxions and in the next year in January had the Theory of Colours and in May following I had the entrance into the inverse method of fluxions and in the same year I began to think of gravity extending to the orb of the moon…and…compared the force requisite to keep the Moon in her orb with the force of Gravity at the surface of the Earth.”

“Nature and Nature’s Laws lay hid in night:God said, Let Newton be! And all was light.”Alexander Pope (1688-1744)

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atvv f 0

221 attvx o

221 attvx f

tvtvvx f )( 021

)( 20

2

21 vvx fa

(A)

(B)

(C)

(D)

(E)

What causes theAcceleration?

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Newton’s LawsFirst Law: If the net force exerted on an object is zero the

object continues in its original state of motion; if it was at rest, it remains at rest.

If it is moving with a given velocity, it will keep on moving with the same velocity.

Second Law: The acceleration of an object is proportional to the net force acting on it, and inversely proportional to its mass: a = F/m or F=ma

Third Law: If two objects interact, the force exerted by the first object on the second is equal but opposite in direction to the force exerted by the second object on the first: F12=-F21

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move) toB (causes Bon A of Force am BB

ABF

move) A to (causesA on B of Force am A A

BAF

law 3rd Newtons - AB FFBA

AABBAB aFaF

AB mm

BA

B am

m-

Aa magnitudes 1

m

m

A

B B

A

a

a

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Identifying the forces in a system

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questionIn the absence of friction, to keep an object that istraveling with a certain velocity moving with exactly the same velocity over a level floor, one:

a) does not have to apply any force on the object

b) has to apply a constant force on the object

c) has to apply an increasingly strong force on the object

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Force, Mass and Weight

Forces are quantified in units of Newton (N).1 N = 1 kgˑm/s2

A force is a vector: it has direction.

The net force causes acceleration in the same direction as the net force.

am net

F

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Force due gravity

Fg = m g where g = 9.8 m/s2

For an object with m = 1 kg Fg = m g = 9.8 kgˑm/s2 = 9.8 Newton

F = ma = mg means that a = g the direction is down

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am net

F

If there is one force acting then

If there are many forces acting then

FF

net

..... 321net FFFF

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Two forces that act upon us nearly all the time.

Gravitational Force (Fg)

Normal Force:elastic force acting perpendicular to the surface the object is resting on. Name: n

Fg = mg (referred to as weight)g = 9.81 m/s2

1)

2) Fg = mg = n (magnitudes)

3)

0 net nFF g

nFg

n

gF

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Clicker question

5 kg10 kg

5 kg

3 cases with different mass and size are standing on

a floor. which of the following is true:a) the normal forces acting on the crates is the

sameb) the normal force acting on crate b is largest

because its mass is largestc) the normal force on the green crate is largest

because its size is largestd) the gravitational force acting on each of the

crates is the same e) all of the above

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An object has a mass of 30.0 kg and three forces are acting on it as shown in the figure. What is the magnitude and direction of acceleration?

Decomposing Forces

70o

24o

300 N

690 N

mg

Force x(N) y(N)300 N -122 -274690 N 236 648Weight 0 -294Resultant 114 N 80 N

Net Force: F = (1142+80.02) = 139 NDirection: = tan-1(Fy/Fx) = 35.2o

Acceleration: a = F/m = 139/30.0 = 4.65 m/s235.2o

139 N

Net Force

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inclinded plane - angles to remember

90-

Choose your coordinate system in a clever way:Define one axis along the direction where you expectan object to start moving, the other axis perpendicularto it (these are not necessarily the horizontal and vertical direction).

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inclinded plane - angles to remember

Choose your coordinate system in a clever way:Define one axis along the direction where you expectan object to start moving, the other axis perpendicularto it (these are not necessarily the horizontal and vertical direction).

x

y

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A first example1) Draw all Forces acting on the red object2) If =40o and the mass of the red object

equals 1 kg, what is the resulting acceleration in the direction of motion.? (assume no friction).

Balance forces in directions where you expect no acceleration; whatever is left causes the object to accelerate!

Fgy = mg cos

Fg=mg

n

n = Fgy (magnitudes)ma = Fnet = Fgx = mg sin a = g sin a = 6.3 m/s2

Fgx = mg sin

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A ball is rolling from a slope at angle . It is repeated but after decreasing the angle . Compared to the first trial:

a) the normal force on the ball increasesb) the normal force gets closer to the gravitational forcec) the ball rolls slowerd) all of the above

questionClicker Question

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A ball is rolling from a ramp at angle = 10o. The length of the ramp is L=1.5 m. What is the time to reach the bottom (in s)

a) 2.71b) 1.33c) 0.55d) 0.20

questionQuestion

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TensionT

The magnitude of the force Tacting on the crate, is the sameas the tension in the rope.

Spring-scaleYou could measure the tension by insertinga spring-scale...

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Tension

wFg

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Tension

As long as it’s one rope, the tension is the same at both ends!

M

T2 = - T1 (vectors)

T2 = T1 (magnitudes)T2

T1

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Elevator An object of 1 kg is hanging from a spring scale in an elevator. What is the weight read from the scale if the elevator:a) moves with constant velocityb) accelerates upward with 3 m/s2 (start going up) c) accelerates downward with 3 m/s2 (start going down) d) decelerates with 3 m/s2 while moving up (end going up)

Before starting this:1) Imagine standing in an elevator yourself2) the scale reading is equal to the tension TF=ma so… T-mg=ma and thus…T=m(a+g)

T

a) a=0 T = 1*9.8 = 9.8 Nb) a=+3 m/s2 T = 1(3+9.8) = 12.8 N (heavier)c) a=-3 m/s2 T = 1(-3+9.8) = 6.8 N (lighter)d) a=-3 m/s2 T = 6.8 N

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Newton’s second law and tension

m1

m2

No friction.n

F1g

T1

T2

F2g

Object 1 (red): F1 = m1 a = T1

Object 2 (green): F2 = m2 a = F2g - T2 = m2 g – T1 = m2 g - m1a Solving for a

What is the acceleration ofthe objects?

a = m2 g /(m1+m2)

T2 = T1 (magnitudes)

F2g = m2g

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Problem

T

Draw the forces: what is positive & negative?

Fg

Fg

T

What is the tension in the string and what will be the acceleration of thetwo masses?

Object 1 (left): T – m1g = m1aObject 2 (right): – T + m2g = m2a

two equations and two unknownsm1

m2

a = g (m2- m1)/(m1+ m2) = 2.45 m/s2

T = m1(a+g) = 36.8 N

T1

T2

T = T1 = T2

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Walking on the moon. The acceleration on the moon is 1/6 thatof the earth. If m2=140 kg what does m1 have to be to make theacceleration of m2 the same as that on the moon?

A) 80B) 100C) 120D) 140E) 160

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A 1000-kg car is pulling a 300 kg trailer. Their accelerationis 2.15 m/s2. Ignoring friction, find:a) the net force on car (c)b) the net force on the trailer (t)c) the net force exerted by the trailer on the car

1000 300Fengine

Ftc= Fct

Example

Fe

FctFtc

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A 1000-kg car is pulling a 300 kg trailer. Their accelerationis 2.15 m/s2. Ignoring friction, find:a) the net force on car (c)b) the net force on the trailer (t)c) the net force exerted by the trailer on the car

1000 300Fengine

Ftc= Fct

Force provided by engine Fe = (mc+mt ) a = 2795 NFct = mt a = 645 N, so Ftc = 645 N

)a Fc = Fe - Ftc = (2795 – 645) N = 2150 N )b Ft = Fct = 645 N

c) Ftc = 645 N

Example

Fe

FctFtc

ct

MSU Physics 231 Fall 2015 32

M

A block of mass M is hangingfrom a string. What is thetension in the string?

a) T=0 b) T=Mg with g=9.81 m/s2

c) T=0 near the ceiling and T=Mg near the blockd) T=M e) one cannot tell

ceiling

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10 Kg

A homogeneous block of 10 kg is hanging with 2 ropes from a ceiling. The tension in each of the ropes is:a) 0 Nb) 49 Nc) 98 Nd) 196 Ne) don’t know

Question

MSU Physics 231 Fall 2015 34

900

1kg

T T

A mass of 1 kg is hanging from a rope as shown in the figure.If the angle between the 2 supporting wires is 90 degrees,what is the tension in each rope?

Tx

450450

Tx

T

y

T

y

Fg

x yT left rope T sin(45) T cos(45)T right rope -T sin(45) T cos(45)gravity 0 -19.81sum: 0 2 T cos(45) - 9.81

The object is stationary, so:2 T cos(45) - 9.81 = 0 so, T=6.9 N

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puck on ice

After being hit by a hockey-player, a puck is moving over a sheet of ice (frictionless). The forces on thepuck are:

a) No force whatsoeverb) the normal force onlyc) the normal force and the gravitational forced) the force that keeps the puck movinge) the normal force, the gravitational force and the force that keeps the puck moving

MSU Physics 231 Fall 2015 36

Friction

Friction are the forces acting on an object due to interaction

with the surroundings (air-friction, ground-friction etc).Two variants:

• Static Friction: as long as an external force (F) trying to make an object move is smaller than fs,max, the static friction fs equals F but is pointing in the opposite direction: - so they cancel and there is no movement!

fs,max = sn s = coefficient of static friction

n = magnitude of the normal force

MSU Physics 231 Fall 2015 37

Friction

Friction are the forces acting on an object due to interaction

with the surroundings (air-friction, ground-friction etc).Two variants:

• Kinetic Friction: After F has surpassed fs,max, the object starts moving but there is still friction. However, the friction will be less than fs,max

fk = kn k = coefficient of kinetic friction n = magnitude of normal force

fk points in the opposite direction to the motion

MSU Physics 231 Fall 2015 38PHY 231 38

MSU Physics 231 Fall 2015 39

MSU Physics 231 Fall 2015 40

Friction

20 kgA person wants to drag a crate of 20 kg over the floor. If he pulls the crate with a force of 98 N, the crate starts to move. What is s?

pull

Answer: fsmax = sn = sMg = 209.8s = 196s

Just start to move: Fpull-fsmax = 0 fsmax=Fpull 98=196s

so: s=0.5

Fg

n

fs

After the crate starts moving, the person continues to pull with the same force. Given k=0.4, what is the acceleration of the crate?

F=ma F = 98-0.4Mg = 98 - 0.4209.8 = 98-78.4 = 19.6 N19.6 = 20a a = 0.98 m/s2

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General strategy

• If not given, make a drawing of the problem.• Put all the relevant forces in the drawing, object by object.

• Think about the axis• Think about the signs

• Decompose the forces in direction parallel to the motion (x) and perpendicular (y) to it.

• Write down Newton’s law for forces in the parallel direction and perpendicular direction.

• Solve for the unknowns.• Use these solutions in further equations if necessary• Check whether your answer makes sense.

MSU Physics 231 Fall 2015 42

A) If s=1.0, what is the angle for which the block just starts to slide?

B) The block starts moving. Given that k=0.5, what is theacceleration of the block?

A) Parallel direction (x): m g sin - s n = 0 (F=ma) Perpendicular direction (y): m g cos - n = 0 so n = m g cos Combine: m g sin - s m g cos = 0 s = sin/cos = tan = 1 so =45oB) Parallel direction: m g sin(45o) - km g cos(45o) = ma (F=ma) a = 3.47

Example ProblemForce due to friction: fs or fk

Fgx = mgsinFgy = mg cos

Fg=mg

n = -Fgy

MSU Physics 231 Fall 2015 43

Fg

Tn

fk

F2g

T2

If a=3.30 m/s2 (the 12 kg blockis moving downward), what isthe value of k?

25.0

cos

sin)(

)(sin

cos

mg

MgmgamM

amMMgmgf

mgf

k

k

kk

Solve for k

All the Forces Come Together

fk = kmgcos

MSU Physics 231 Fall 2015 44

0.5 kg

A20o

Is there a value for the staticfriction of surface A for whichthese masses do not slide?If so, what is it?

1 kgnfsmax

Fg

T

T

Fg

Example

No sliding: same as previous problem with a=0 and angle=-20 -3.35 - 4.9 + 9.2s=0

s = 0.9

MSU Physics 231 Fall 2015 45

A 2000 kg sailboat is pushed by the tide of the sea with a force of 3000 N to the East. Because of the wind in its sail it feels a force of 6000 N toward to North-West direction. What is the magnitude and direction of the acceleration?

Problem

E

S

W

N6000N

3000N

Horizontal VerticalDue to tide: 3000 N 0 NDue to wind: - 6000 cos(45)=- 4243 6000 sin(45)= 4243Sum: -1243 N 4243 N

Magnitude of resulting force:Fsum = [(-1243)2+(4243)2] = 4421 N

Direction: angle with respect to north = tan-1(1243/4243) =16.30

F=ma so a = F/m = 4421/2000 = 2.21 m/s2

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Rotational motion and forces

centripetal acceleration must caused by a net force acting

Fnet = m ac

MSU Physics 231 Fall 2015 47

Example from Homework set 02

“weight” is due to the normal force of the wall on the person

n= Fnet=mv2/R

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Ball on string –

T= Fnet=mv2/R

T= Mg

Fnet provided by tension in the rope

MSU Physics 231 Fall 2015 49

Car on curve without friction –

n

mg

mg

Fnet=mv2/R

Fnet provided by normal force

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Car on curve with friction –

Fnet=mv2/r = fs

mg

n

Fnet provided by friction

MSU Physics 231 Fall 2015 51

Moon in orbit around the earth –

Fnet provided by gravitational force between the moon and the earth