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MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

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Page 1: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 1

Physics 231Topic 4: Forces & Laws of Motion

Alex Brown September 23-28 2015

Page 2: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 2

Page 3: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 3

What’s up? (Wednesday Sept 23)

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

2) Learning Resource Center (LRC)  in PBS 1248   note expanded hours

Monday 9 am to 8 pm

Tuesday 9 am to 9 pm

Friday 9 am to 4 pm

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

Covers Chapters 1‐4 and homeworks 01‐03

Page 4: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 4

• 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. 4

Page 5: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 5

Recall from Last Chapter

Vectors and Scalars

Two Dimensional Motion Velocity in 2D

Acceleration in 2D

Projectile motion Throwing a ball or cannon fire

Page 6: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 6

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

Page 7: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 7PHY 231 7

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)

Page 8: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 8

atvv f 0

221 attvx o

221 attvx f

tvtvvx f )( 021

)( 20

221 vvx fa

(A)

(B)

(C)

(D)

(E)

What causes theAcceleration?

Page 9: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 9PHY 231 9

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

Page 10: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 10

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 amm

-

Aa magnitudes 1mm

A

B B

A

aa

Page 11: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 11PHY 231 11

Identifying the forces in a system

Page 12: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 12PHY 231 12

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

Page 13: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 13PHY 231

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

Page 14: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 14PHY 231

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

Page 15: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 15

am net

F

If there is one force acting then

If there are many forces acting then 

FF

net

..... 321net FFFF

Page 16: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 16PHY 231 16

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

Page 17: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 17PHY 231 17

Clicker question

5 kg 10 kg5 kg

3 cases with different mass and size are standing ona 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

Page 18: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 18

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) = 35.2oAcceleration:  a = F/m = 139/30.0 = 4.65 m/s235.2o

139 N

Net Force

Page 19: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 19

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).

Page 20: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 20

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

Page 21: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 21

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 sina = g sina = 6.3 m/s2

Fgx = mg sin

Page 22: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 22

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

Page 23: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 23

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

Page 24: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 24

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...

Page 25: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 25

Free-body diagram for net force

wFg

Page 26: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 26

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

Page 27: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 27PHY 231 27

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

Page 28: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 28

Newton’s second law and tension

m1

m2

No friction.n

F1g

T1

T2

F2g

Object 1 (red):         F1= m1 a = T1Object 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

Page 29: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 29

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 unknowns

m1

m2

a = g (m2‐m1)/(m1+ m2) = 2.45 m/s2

T = m1(a+g) = 36.8 N    

T1T2

T = T1 = T2

Page 30: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 30

Clicker question: 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

m2

xx xx

m2 g – m1 g = (m1 + m2) a a = g/6

solve for m1

m1 = m2 (g - a)/(a + g) = = m2 (5/7) = 100

Page 31: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 31

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 300FengineFtc= Fct

Example

Fe

FctFtc

Page 32: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 32

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 300FengineFtc= 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 Nc) Ftc = 645 N

Example

Fe

FctFtc

ct

Page 33: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 33

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

Page 34: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 34

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

Clicker question

Page 35: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 35

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

TyTy

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

Page 36: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 36PHY 231 36

Clicker question: 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

Page 37: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 37

Friction

Friction are the forces acting on an object due to interactionwith 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

Page 38: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 38

Friction

Friction are the forces acting on an object due to interactionwith 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 frictionn = magnitude of normal force

fk points in the opposite direction to the motion

Page 39: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 39PHY 231 39

Page 40: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 40

Page 41: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 41

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=196sso:  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

Page 42: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 42

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.

Page 43: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 43

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 cosCombine:  m g sin ‐ s m g cos = 0

s = sin/cos = tan = 1  so =45o

B) 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 = mg sinFgy = mg cos

Fg=mg

n = -Fgy

Page 44: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 44

Clicker question: 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

m2

xx xx

m2 g – m1 g = (m1 + m2) a a = g/6

solve for m1

m1 = m2 (g - a)/(a + g) = = m2 (5/7) = 100

Page 45: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 45

The 12 kg block is moving downward, what is the value of a?

Similar problem

m1 =

m2 =

-m2 g sin + m1 g

= (m1 + m2) a

solve for a

xx

0.4)(

sin

21

21

mmgmgma

Page 46: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 46

F2g

T2n

F1g

T1

All the Forces Come Together

m1 =

m2 =

Page 47: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 47

F2g

T2

n

fk

F1g

T1

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

25.0cos

sin)(

2

1221

gmgmgmamm

k

All the Forces Come Together

-fk -m2 g sin + m1 g = (m1 + m2) a

with fk = k m2 g cos 

solve for k 

Page 48: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 48

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

F2g

T

T

F1g

Example

-fs + m2 g sin + m1 g = (m1 + m2) a = 0

with fsmax = sm2 g cos 

solve for s

90.0cos

sin

2

12

gmgmgm

s

Page 49: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 49

Rotational motion and forces

centripetal acceleration must caused by a net force acting

Fc = m ac

Page 50: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 50

Example from Homework set 02

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

n= Fc=mv2/R = m R = m g g/R)1/2 = 0.61 radians/s

Page 51: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 51

Ball on string –

T = Fc = mv2/R

T= Mg

Fc provided by tension in the rope

Page 52: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 52

Car on curve without friction –

n

mg

Fc = mv2/R

Fc provided by normal force

Page 53: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 53

Car on curve without friction –

n

mg

mg

Fc = mv2/R

Fc provided by normal force

Rgv2

tan

Page 54: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 54

Car on curve with friction –

Fc = mv2/R = fs

mg

n

Fc provided by friction

Rgv

s

2

R

R

Page 55: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 55

Moon in orbit around the earth –

Fnet provided by gravitational force between the moon and the earth

Page 56: Physics 231 - Michigan State Universitybrown/231/Ch4b.pdf · MSU Physics 231 Fall 2015 1 Physics 231 Topic 4: Forces & Laws of Motion Alex Brown September 23-28 2015

MSU Physics 231 Fall 2015 56

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