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Ch.4 Forces And Newton’s Laws

Non Contact - Action at a distant force

Is the Force a Vector or a Scalar ? Units and Dimensions ? Mass – Measure of Mass is inertia

(How does it differ from Weight ?)

Higher mass – harder to change the motion, i.e. higher inertia What is a Force? A Push or a Pull

Newton’s First Law Of Motion

•

_{An object}

_{continues in a state of rest or in a}

state of motion

at a constant speed along a

straight line unless compelled to change that

state by

a net force.

•

_{What is the}

_{net force?}

If so, why will a ball rolling on a flat surface

stop?

•

_{Vector sum of all}

the forces acting on

the object is called

the net force.

( i.e. both magnitude

and direction needs

to be considered

)

What is the net force on the car ?

Inertial Reference Frame

Newton’s laws are valid in an inertial frame,

i.e. the acceleration of the frame should be

zero (

Constant Velocity

).

Ideally we do not have such a frame, but

earth is a very good approximation.

Newton’s Second Law

• First law: no net force no change in velocity

• _{Second law: What happens if there is an external}

force? (net)

Acceleration of the object Net Force

Acceleration inversely proportional to mass



m

1



Only two factors determine the

acceleration---

Force and Mass

Force Units Newton (N) = kg. m/s

2

What is 1 Newton

?

m

F

Newton’s Second Law Of Motion

When a net external force acts on an object of mass m, the acceleration a results is directly

proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force.



F





_

 or F ma

m F a

Units for Mass, Acceleration & Force

System Mass Acceleration Force

SI kilogram(kg) meter/second2(m/s2) newton(N) CGS gram(g) centimeter/sescond2

(cm/s2₎

dyne(dyn)

Example 1. Pushing A Car

Frictional force = 560 N

Push = 275 N+ 395 N = 670 N

Net force = 670 N – 560 N, N= 110 N F = ma , 110 N = 1850 kg*a

a = 110 N/1850 kg = 0.059 m/s2

Forces In 2-D

F_x = ma_x F_y = ma_y

Two forces are applied onto an object. 1.) 12N towards North, and 2.) 5N towards East, what is the direction of the motion of the object?

tan θ = 12/5 =2.4

θ = tan-1(12/5) = 67.38 0 12N

5N

θ

11

Example 2.

1300 kg raft

P= Force by the man

A= Force of wind

In 65 seconds,

where will the boat be if v_0x = 0.15

m/s ?

(a)

Resultant Force And Acceleration

Force X- component Y- Component

P 17 N 0 N

A 15cos67 N 15sin67 N

Resultant , 23 67 cos 15 17



F_x    N

, / 018 . 0 1300 23 ₂ s m kg N m F

a_x 



x  



F_y 15sin 67 0 14N

2 / 011 . 0 1300 14 s m kg N m F

Displacement Of The Boat In 65 sec

y

=

v

_0y

t + (½) a

_y

t

2

= 0

*

65 + (½)

*

0.011

*

(65)

2

= 23 m

For

x

: v

_0x

= 0.15 m/s, a

_x

= 0.018 m/s

2

t = 65 sec, x

= ?

x = v

_0x

t + (½)a

_x

t

2

= 0.15

*

65 +(1/2)

*

0.018

*

(65)

2

= 48 m

Which of these will cause the

acceleration to be doubled ?

a) All of the forces acting on the object

doubles

b) The net force acting on the object

doubles

c) Both the net force and the mass doubles

d) The mass of the object is reduced by a

factor of two.

Newton’s Third Law of Motion

•

_{Whenever one body exerts a force on a}

second body, the second body will exert

an oppositely directed force of equal

magnitude on the body.

•

_{(Action reaction law)}

M_S = 11000 kg m_A = 92 kg

Force by Astronaut on Space Ship = P Force on Astronaut by Space ship = P

What will be the acceleration of the two objects ?

If P = 36 N, -36 N = 92 a_A , i.e. a_A = -0.39 m/s2

36 N = 11000 kg a_s , i.e. a_s = 0.0033 m/s2

Types Of Forces

Fundamental Forces

•

_{Gravitational force}

•

_{Strong Nuclear}

•

_{Electroweak Electromagnetic force}

All other forces are Non Fundamental

(can be explained by a fundamental force)

Maxwell, Unification theory

Gravitational Force

• _{Newton’s law of Universal gravitation}

G = 6.673 x 10 –11 N m2/kg2

Always attractive , Planetary motion, Satellites r center to center

2 2 1

r

m

m

G

Weight Of An Object

• _{Mass m on the surface of}

the earth

W = mg

M_e

Radius of earth = R_e

Distance ~ R_e (approx)

So:

= 9.799 m/s2

mg R mM G W e e _  ₂ 2 e e R M G g  2 6 24 2 2 11 ) 10 38 . 6 ( ) 10 98 . 5 )( / 10 67 . 6 ( m kg kg m N       2 2 1

r

m

m

G

F



Hubble Telescope

•

_{The mass of Hubble telescope is 11600 kg.}

(R

_e

= 6.38

*

10

6

m, M

_e

= 5.98

*

10

24

kg)

•

_{(i) What is the weight when resting on earth}

surface,

2 6

24 11

2 _{(6.38 10 )}

11600 10 98 . 5 10 67 . 6         r M M G

W e h

= 1.14*105 N

•

_{(ii) and in orbit 598 km above}

(r = R_e +598 km)

Check Your Understanding 2

m₂>m₁, and the net gravitational force acting on the third object is zero. Which of the drawings correctly represents the locations of the object?

The Normal Force & Newton’s 3

rd

Law

Net Force on the block: F_N – W

Table exert a force on the block: F_N

Normal Force: The force component that a surface exerts on the object (Normal to the contact surface)

How can the table exert a force ? (Consider the springs in a mattress)

What happens if I push or pull on the

block ?

What happens to the normal force if the rope has a pull = 15N?

Apparent Weight

• How do you feel when the elevator suddenly starts going up?

F_N – W = ma

F_N = Apparent weight W = True weight

Feel heavy?

Elevator Accelerating Up With “a”

Applying 2nd law vertically

upward

F_N – W = ma F_N = W+ma

Apparent weight is larger than the true weight (W = mg)

Elevator Accelerating Down With “a”

F_N – mg = ma

( if downward a = negative)

F_N = mg +ma

What if a = -g ?

Static And Kinetic Friction

Two surfaces touching each other

Contact surface is not smooth

In addition to the normal force, there is a force

parallel to the surface

Static Friction (f

_s

)

• As F increases, fs

increases from zero up to fs Max

• _{Just before motion}

f_sMax = μ_s F_N

Units and Dimensions of μ_s? (μ_s= Coefficient of static friction)



30

Force Needed To Start The Sled

The maximum force needed to just begin to move = need to

overcome the max frictional force i.e. F = f_sMax = μ

s FN = μs mg

If μ_s = 0.35, m = 38 kg f_sMax ₌ 0.35

* 38 * 9.8 = 130 N

What happens once it just starts moving ? Easier to move, i.e. needs force going down

i.e. f changes to f

Kinetic Friction

F

Kinetic Friction

• f_k = μ_kF_N

(μ_k Coefficient of kinetic friction)

μ_k is usually less than μ_s

Example 10: Sled riding : How far does the sled go before stopping ?

f

_k

= μ

_k

F

_N

= μ

_k

mg

Net force on the sled = kinetic frictional force

g

u

m

mg

u

m

f

a

k k _k

k













Acceleration m g u v v x k x

x _16.3

8 . 9 ) 05 . 0 ( 2 ) 0 . 4 ( 0 ) ( 2 2 2 0 2        

What is the meaning of the negative sign? Does the acceleration depend on the mass ? Stopping distance x? _v2 = v

Tension

Mass-less and non-stretch rope Force gets applied to the box undiminished.

Traction Of The Foot

• T1 and T2 keep the pulley on the foot at rest. i.e.

pulley is at equilibrium (Net Force = 0)



F

_y





T

₁

sin

35





T

₂

sin

35





0

0

35

cos

35

cos

₂

1

















F

_x

T

T

F

N

s

m

kg

)(

9

.

80

/

)

cos

35

35

2

.

2

(

2

2















2

T

cos

35

2

mg

cos

35

F





2

T

cos

35

F

T

₁

=T

₂

( Same rope)



F

_y





T

₁

sin

35





T

₂

sin

35





0













F

_x

F

T

₁

cos

35

T

₂

cos

35







F

2

T

cos

35

(Say T₁=T₂ = T)

T = 2.2 g (why?) F = 35 N



F

_x



0

F



2

T

cos

35



Force x Component y Component T₁ -T₁ sin 10.00 +T

1 cos 10.00

T₂ +T₂ sin 80.00 -T

2 cos 80.00

W 0 -W

0

0

.

80

sin

0

.

10

sin

₂

1















F

_x

T

T

0

0

.

80

cos

0

.

10

cos

₂

1















2 1

0

.

10

sin

0

.

80

sin

T

T

_

















0

0

.

80

cos

0

.

10

cos

0

.

10

sin

0

.

80

sin

2

2



























W

T

T

            0 . 80 cos 0 . 10 cos 0 . 10 sin 0 . 80 sin 2 W T

Setting W=3150N, T₂=582N T₁=3.30*103N

Force x Component y Component W -W sin 30.00 -W cos 30.00

L 0 +L

T +T 0

R -R 0

0

0

.

30

sin















F

_x

W

T

R



F

_y





W

cos

30

.

0





L



0

N

N

W

Check Your Understanding 4

Which of the following could lead to equilibrium? a) Three forces act on the object. The forces all point

along the same line but may have different directions.

b) Two perpendicular forces act on the object. c) A single force acts on the object.

d) In none of the situations described in (a), (b) and (c) could the object possibly be in equilibrium.

Non-equilibrium Applications

•

_{Non zero net force}

•

_{Net force – x component, y component}

•

_{Acceleration in x direction and in y}

Towing A Super Tanker

D=75.0*103N, m=1.50*108N, R=40.0*103N

Force x Component y Component T₁ +T₁cos 30.00 _+T

1sin 30.00

T₂ +T₂cos 30.00 _-T

2sin 30.00

D +D 0

R -R 0



F

_y





T

₁

sin

30

.

0





T

₂

sin

30

.

0





0

a_x=2.0*10-3     0 . 30 cos 2 D R ma T x



















0

.

30

cos

2

10

0

.

75

10

0

.

40

)

/

10

00

.

2

)(

10

50

.

1

(

8

kg

3

m

s

2 3

N

3

N



F

_x



T



m

₂

a

_x



(

27000

kg

)(

0

.

78

m

/

s

2

)



21000

N

x

x

D

T

m

a

F













₁

N

s

m

kg

T

a

m

D



_{1 x}







(

8500

)(

0

.

78

/

2

)



21000

= 28000N

If the drawbar has no mass

T=T`

The Motion Of A Water Skier

(a) The skier is floating motionless in the water.

The skier is floating

motionless in the water, so her velocity and acceleration are both zero. Therefore, the net force acting on her is zero, and she is in equilibrium.

(b) The skier is being pulled out of the water and up onto the skis.

As the skier is being pulled up and out of the water, her velocity is increasing. Thus she is accelerating, and the net force acting on her is not zero. The skier is not in equilibrium. The direction of the net force is shown in the figure above.

(c) The skier is moving at a constant speed along a straight line.

The skier is now moving at a constant speed along a straight line, so her velocity is constant. Since

her velocity is constant, her acceleration is zero. Thus, the net force acting on her is zero, and she is again in equilibrium, even though she is moving.

(d) The skier has let go of the tow rope and is slowing down.

After the skier lets go of the tow rope, her speed decreases, so she is decelerating. Thus, the net

force acting on her is not zero, and she is not in equilibrium. The direction of the net force is shown in the figure.

MAX N

s

x

mg

F

ma

F













sin

10

.

0



m

F

mg

a

MAX









s N



sin

10

.

0

0

0

.

10

cos













F

y

mg

F

N





mg

cos

10

.

0

F

_N











g

sin

10

.

0

g

cos

10

.

0

a

MAX



_s











(

9

.

80

m

/

s

2

)

sin

10

.

0

(

0

.

350

)(

9

.

80

m

/

s

2

)

cos

10

.

0

=1.68m/s2

m

₁

=8.00 kg,

m

₂

=22.0 kg,

T=T

’

T=86.3 N, a=5.89 m/s

2

a

m

T

W

F

_x





₁

sin

30

.

0







₁



For mass m₁

What is the mass m₁ in y direction?

For mass m₂

)

(

2

2

m

a

W

T

F

_y











Hoisting A Scaffold

m=155 kg, T=540 N



F

_y





T



T



T



W



ma

_y

2

/ 65 . 0 155

1520 )

540 (

3 3

s m kg

N N

m W T

Check Your Understanding 5

Two boxes have masses m₁ and m₂, and m₂ is greater than m₁. The boxes are being pushed across a frictionless

horizontal surface. As the drawing shows, there are two

possible arrangements, and the pushing force is the same in each. In which arrangement does the force that the left box applies to the right box have a greater magnitude, or is the magnitude the same in both cases?

Velocity, Acceleration And

Newton’s Second Law Of Motion

0

3 2

1











F

_x

F

F

F

N

N

N

F

F

F

₃





(

₁



₂

)





(

3000



5000

)





8000

v=850m/s

If the craft is moving with constant velocity, are there other forces?

The Importance Of Mass

?

g_moon=1.60m/s2, T=24N, μ

m

F T

m F

a x k N

x     



m

mg

T

a

k moon

x









earth earth

mg

W



kg

s

m

N

g

W

m

earth

earth

₉

_.

₀

/

80

.

9

88

2







2 2 / 3 . 2 0 . 9 ) / 60 . 1 )( 0 . 9 )( 20 . 0 ( 24 s m kg s m kg N    m mg T

a k moon

x



  

Conceptual Questions 1.

Why do you lunge forward when a car suddenly stops ? Pressed backward when the car suddenly accelerates ?

REASONING AND SOLUTION When the car comes to a sudden halt, the upper part of the body continues forward (as predicted by Newton's first law) if the force exerted by the lower back

muscles is not great enough to give the upper body the same deceleration as the car. The lower portion of the body is held in place by the force of friction exerted by the car seat and the floor. When the car rapidly accelerates, the upper part of the body tries to remain at a constant velocity (again as predicted by Newton's first law). If the force provided by the lower back muscles is not great enough to give the upper body the same acceleration as the car, the upper body appears to be pressed backward against the seat as the car moves forward.

Conceptual Question 6.

Father and daughter on ice pushing each other, who has the

higher acceleration ?

REASONING AND SOLUTION

Since the father and the daughter are standing on ice skates, there is virtually no friction between their bodies and the ground. We can assume, therefore, that the only horizontal force that acts on the daughter is due to the father, and

similarly, the only horizontal force that acts on the father is due to the daughter. a.) According to Newton's third law, when they push off against each other, the force exerted on the father by the daughter must be equal in magnitude and opposite in direction to the force exerted on the daughter by the father. In other words, both the father and the daughter experience pushing forces of equal magnitude.

b.) According to Newton's second law, Therefore, . The magnitude of the net force on the father is the same as the magnitude of the net force on the daughter, so we can conclude that, since the daughter has the smaller mass, she will acquire the larger acceleration.

F

Problem

4) A 5.0 kg projectile accelerates from rest to 4.0*103 m/s . The net

force on the projectile is 4.9 * 105 N. What is the time for the

projectile to come to the speed?

F=4.9*105N, M=5.0kg, v₀=0, v_f=4.0*103

sec

10

08

.

4

10

8

.

9

10

0

.

4

₂ 4 3 0 















a

v

v

t

v = v₀ + at ,

Problem

8)An arrow starting from rest leaves the bow with a speed of 25.0 m/s. If the average force on the arrow is doubled what will be the speed?

2 1 2

2 2 1

a

a

v

v



(Say mass m, travels a distance x before leaving the bow) initial velocity v₀=0, find velocity v₁

v₁2 = v

o2 + 2a1x (first case)

v₂2 = v

If the net force is doubled in case 2, i.e. 2 1 2 1

a

a

v

v



1 1 2 1

2

a

a

v

v



2 1

2

v

v



s

m

v

₂



25

2



35

.

4

/

F = ma a₂ = 2a₁

F=ma₁ 2F=ma₂ 2a₁=a₂

F/m=a₁

Problem

15)A duck of mass 2.5 kg has a force of 0.1 N due east. Water exerts a force 0.2 N 52o south of east . Velocity of the duck

is 0.11 m/s due east. Find the displacement of the duck in 3.0s while the forces are active.

x component of force = 0.1 + 0.2 cos 52 = R_x=0.2231N y component of force = - 0.2 sin 52 = R_y= -0.1576 N v₀₌ 0.11 m/s

0.1 N 52o

0.2 N

R_x = 2.5 a_x a_x = R_x /2.5 = 0.08924m/s2

Start

End x = ?

y = ?

x = v_ox t + ½ a_x t2 = 0.11

* 3 + ½ a_x 32 =0.73158m

y = v_oy t + ½ a_y t2

= 0 + ½ a_y * 9= -0.28368m

_Displaceme

_nt

_x

2

_y

2

₀

_.

₇₈₄₇

_m









3878 .

0 tan  

x y

 tan1 21.19

x y 

Problem

20)REASONING AND SOLUTION The forces that act on the rock are shown at the right. Newton's second law (with the direction of motion as positive) is

mg R



F



mg

–

R



ma

Solving for the acceleration a gives







2







₂ / 2 . 4 45 250 / 8 . 9 45 s m kg N s m kg  

Problem

4) A 5.0 kg projectile accelerates from rest to 4.0*103 m/s . The net force on the projectile is 4.9 * 105 N. What is the time for the

projectile to come to the speed?

F=4.9*105N, M=5.0kg, v₀=0, v_f=4.0*103

sec

10

08

.

4

10

0

.

4

₂

4 3

0 













v

v

t

v

=

v

₀

+ at ,

Problem

8)

An arrow starting from rest leaves the bow with a

speed of 25.0 m/s. If the average force on the arrow

is doubled what will be the speed?

1 2

2 1

a

a

v

v



(Say mass m, travels a distance x before leaving the

bow) initial velocity v

₀

=0, find velocity v

₁

v

₁2

= v

o2

+ 2a

1

x (first case)

v

₂2

= v

If the net force is doubled in case 2, i.e.

2 1 2

1

a

a

v

v



1 1 2

1

2

a

a

v

v



2

1

2

v

v



F = ma a

₂

= 2a

₁

F=ma₁ 2F=ma₂ 2a₁=a₂

F/m=a

₁

Problem

15)

A duck of mass 2.5 kg has a force of 0.1 N due east. Water

exerts a force 0.2 N 52

o

south of east . Velocity of the duck

is 0.11 m/s due east. Find the displacement of the duck in

3.0s while the forces are active.

x component of force = 0.1 + 0.2 cos 52 = R_x=0.2231N y component of force = - 0.2 sin 52 = R_y= -0.1576 N v₀₌ 0.11 m/s

0.1 N 52o

0.2 N

R_x = 2.5 a_x a_x = R_x /2.5 = 0.08924m/s2

Start

End x = ?

y = ?

x

=

v

_ox

t + ½ a

_x

t

2

= 0.11

*

3 + ½ a

_x

3

2

=0.73158m

y

=

v

_oy

t + ½ a

_y

t

2

= 0 + ½ a

_y *

9= -0.28368m

_Displaceme

_nt

_x

2

_y

2

₀

_.

₇₈₄₇

_m









Problem

20)

REASONING AND SOLUTION

The forces that act

on the rock are shown at the right. Newton's second

law (with the direction of motion as positive) is

mg R



F



mg

–

R



ma

Solving for the acceleration a gives







2







₂

/

2

.

4

45

250

/

8

.

9

45

s

m

kg

N

s

m

kg



