Newton’s First Law •  If there is no net force, the velocity of a mass remains

constant (neither the magnitude nor the direction of the velocity changes). –  Objects at rest feel no net force. –  Objects in motion with a constant velocity feel no

net force. movie

clicker question •  The diagram below shows two

successive positions of a particle; it's a segment of a full motion diagram. Which of the following vectors best represents the acceleration between

a)

b) c) d)

!vi and

!vf !vi

aΔt

recall that

!vf =!vi +!aΔt

When a ball on the end of a string is swung in a horizontal circle: What is the direction of the acceleration of the ball?

A.  Tangent to the circle, in the direction of the ball’s motion B.  Toward the center of the circle

Checking Understanding

Slide 6-15

Forces and uniform circular motion

•  Consider the mass undergoing horizontal circular motion at the right. At any point, the instantaneous velocity is tangential to the circle. Because the direction of the tangent changes, however, the direction of the velocity changes.

vt

ac

Δ!v ="aΔt = !v f −

!v i ≠ 0

rvv

tv

tva ttc

2

00 limlim ==ΔΔ=

ΔΔ= →Δ→Δ ω

θ

rv=ω

•  This means there is an acceleration pointing inward towards the center of the circle. If we look at the component of the acceleration in this direction:

•  Here we have used:

•  The accelerations ac is caused by the tension force supplied by the string and points along the string towards the center of the circle.

acΔt =

Quiz

•  The boat moves at a fixed velocity relative to the water. On which route(s) do you go the least distance as you cross the river? –  a) B –  b) H –  c) F –  d) K –  e) C

Reading Quiz 1.  A “net force” is

A.  the sum of the magnitudes of all the forces acting on an object. B.  the difference between two forces that are acting on an object. C.  the vector sum of all the forces acting on an object. D.  the force with the largest magnitude acting on an object.

Slide 4-7

Newton’s second Law

•  Newton’s 2nd law: –  is the acceleration. –  m is the mass of the object. It is proportional to the number of

nucleons (neutrons and protons) in an object. –  is the net (total) force acting on the object. It is the vector sum

of all forces acting on the object. •  Both and are vectors with a length and a direction. •  m is a scaler. It relates the x and y components of the force and

acceleration vectors and also their lengths The direction of the force is the same as that of the acceleration.

F!

amF !!=

F!

a!

yy

xx

maFmaF

==

)tan()tan(

)()( 2222

ax

y

x

y

x

yF

yxyx

aa

mama

FF

maaammamaF

θθ ====

=+=+=

Here, θF and θa are angles with respect to the x axis.

The angles are equal because the acceleration is in the direction of the net force.

a!

Reading Quiz 1.  Which of the following statements about mass and weight is correct?

A.  Your mass is a measure of the force gravity exerts on you. B.  Your mass is the same everywhere in the universe. C.  Your weight is the same everywhere in the universe. D.  Your weight is a measure of your resistance of being

accelerated.

Slide 5-5

Inertial mass

•  The unit of mass is kg. •  The inertial mass m is given by the

number of neutrons and protons (nucleons) in the object. It governs how hard it is to accelerate an object.

kgNM nucleons-271.66x10 ; =≈ µµ

Example

•  A bicycle has a mass of 13.1 kg and its rider has a mass of 81.7 kg. The rider is pumping hard so that a horizontal net force of 9.78 N accelerates them in the westward direction. What is the acceleration? –  a) 0.103 m/s2 west –  b) 0.103 m/s2 east –  c) 0.120 m/s2 west –  d) 0.120 m/s2 east –  e) don’t know.

2/103.07.811.13

78.9 smkgkg

NmmFa

riderbike

=+

=

+=mbike 13.1 kg

mrider 81.7 kg

F 9.79 N

Example

•  A catapult on an aircraft is capable of accelerating a plane from 0 to 56 m/s in a distance of 80 m. Find the magnitude of the average net force that the catapult exerts on a 13300 kg jet.

!F = m!a

mplane 13300 kg

v0 0

v 56 m/s

!x 80 m

2 20v v 2a x− = Δ

2 2 20v v va

2 x 2 x−

= =Δ Δ

2 20v vF m

2 x−

=Δ

( )( )

2556m / s13,300 2.6x10 N

2 80m= =

quiz

•  A constant force is exerted for a short time interval on a cart that is initially at rest on an air track. This force gives the cart a certain final speed. Suppose we repeat the experiment but, instead of starting from rest, the cart is already moving with constant speed in the direction of the force at the moment we begin to apply the force. After we exert the same constant force for the same short time interval, the increase in the cart’s speed –  a) is equal to two times its initial speed. –  b) is equal to the square of its initial speed. –  c) is equal to four times its initial speed. –  d) is the same as when it started from rest. –  e) cannot be determined from the information provided.

0 0Fv v at v tm

= + = +

Fv tm

⇒ Δ =

Example

•  A 36.5 kg sphere is acted upon by a 133 N force directed along the positive x axis. What is the magnitude and direction of a second force, such that the sphere experiences an acceleration of 3.96 m/s2 directed at 210 above the positive x axis.

F1 133 N

F2 ?

m 36.5 kg

!2 ?

1F!

netF!

2F!

[ ] 2netF 36.5kg 3.96m/ s 144.5N⎡ ⎤= =⎣ ⎦21o ( )onet,xF 144.5Ncos 21 134.9N= =

( )onet,yF 144.5Nsin 21 51.8N= =2,x net,x 1,xF F F 134.9N 133N 1.9N= − = − =

2,y net,y 1,yF F F 51.8N= − =

( ) ( )2 22F 1.9N 51.8N 51.8N= + =1 o

251.8Ntan 881.9N

θ − ⎛ ⎞= =⎜ ⎟⎝ ⎠

Reading Quiz 3.  Which of these is not a force discussed in this chapter?

A.  The tension force. B.  The normal force. C.  The orthogonal force. D.  The friction force.

Slide 4-11

5 kg

T

Examples of Forces

5 kg

W=mg

5 kg

T

T

•  Gravitational force: –  Objects feel a force from gravity equal to their weights. The

weight W of an object equals its mass times the gravitation acceleration g. The weight of an object is less on the moon, but the mass remains the same.

–  The direction of the gravitation force is towards the center of the Earth.

•  String tension: –  If a string pulls on an object, we call its force the string tension.

It is always in the direction of the string. Strings cannot push. –  A pulley can redirect the tension force.

•  The net or total force is the vector sum of all forces on an object. •  The acceleration is proportional to the net force.

Example

•  A 10 kg mass hangs from a string. The string pulls it so that it is accelerating upwards with an acceleration of 1 m/s2. What is the magnitude of the tension in the string. –  a) 10 N –  b) 88 N –  c) 98 N –  d) 108 N

10 kg

T

a 1 m/s2

m 10 kg T mg ma− =T ma mg m(a g)= + = +

2T 10kg(10.8m / s ) 108N= =

mg