Lecture 7

ForcesImportant note:

First Exam is next Tuesday, Feb. 6, 8:15-9:45 pm(see link on Canvas for locations)

Today’s Topics:

• Forces– The gravitational force– The normal force– Frictional Forces

• Next lecture…. Lots of examples

Newton’s Law of GravitationEvery particle exerts an attractive force on every other particle.

A particle is a piece of matter, small enough in size to be regarded as a mathematical point.

The force that each exerts on the other is directed along the line joining the particles.

221

rmmGF = 2211 kgmN10673.6 ×´= -G

Defining weightThe weight of an object on or above the earth is the gravitational force that the earth exerts on the object.

The weight always acts downwards, toward the center of the earth.

On or above another astronomical body, the weight is the gravitational force exerted on the object by that body.

2rmM

GW E=

mgW =

2rMGg E=

At the earth’s surface:

( )( )( )

2

26

242211

2

sm 80.9m 106.38kg 1098.5kgmN1067.6

=

´

´×´=

=

-

E

E

RMGg

a) Fg is greater on the featherb) Fg is greater on the stonec) Fg is zero on both due to vacuumd) Fg is equal on both alwayse) Fg is zero on both always

What can you say about

the force of gravity Fg

acting on a stone and a

feather on the moon?

The force of gravity (weight) depends on the mass of the object!! The stone has more mass, and therefore more weight.

ACT: Stone and feather I

a) it is greater on the featherb) it is greater on the stonec) it is zero on both due to vacuumd) it is equal on both alwayse) it is zero on both always

What can you say

about the acceleration

of gravity acting on the

stone and the feather?

The acceleration is given by F/m so here the mass divides out. Because we know that the force of gravity (weight) is mg, then we end up with acceleration g for both objects.

Follow-up: Which one hits the bottom first?

ACT: Stone and feather II

Defining the Normal forceThe normal force is one component of the force that a surfaceexerts on an object with which it is in contact – namely, thecomponent that is perpendicular to the surface.

Block on two tablesA block is balanced in the space between two tables as shown below. What forces are acting on the block?

Wby Earth

Nby right tableNby left table

DEMO: Nail bed

If the block rests on 100 mini-tables, each table exerts a relatively small force:

=each table 100WN

!F

net= 0

⇓

NR= N

L=W

2(½ with enough symmetry)

The acceleration of both masses together is the same in either case. But the contact force is the only force that accelerates m1

in case A (or m2 in case B). Because m1 is the larger mass, it requires the larger contact force to achieve the same acceleration.

If you push with force F on either the heavy box (m1) or the light box (m2), in which of the two cases below is the contact force between the two boxes larger?

a) case A

b) case B

c) same in both cases

F m2

m1

A

F m2

m1

B

ACT: Contact force

In case 1, the force F is pushing down (in

addition to mg), so the normal force

needs to be larger. In case 2, the force F

is pulling up, against gravity, so the

normal force is lessened.

Case 1

Case 2

Below you see two cases: a man pulling or pushing a sled with a force F that is applied at an angle 𝜃. In which case is the normal force greater?

a) case 1

b) case 2

c) it’s the same for both

d) depends on the magnitude of the force F

5) depends on the ice surface

ACT: Normal force

Apparent Weight

mamgFF Ny =-+=åmamgFN +=

apparent weight

trueweight

Inclined plane (no friction)

θgmW !!

=

NF!

mamgmgFN

==-

qq

sin0cos

qq

sincos

gamgFN

==

θmg

qsinmg

qcosmg

NF

a) case Ab) case Bc) both the same (N = mg)d) both the same (0 < N < mg)e) both the same (N = 0)

Consider two identical blocks, one resting on a flat surface (case A) and the other resting on an incline (case B). For which case is the normal force greater?

N

WWy

x

y

f

𝜃

𝜃

In case A, N = W.

In case B, due to the angle of

the incline, N < W. In fact,

N = W cos 𝜃.

ACT: Incline

Frictional Forces

When an object is in contact with a surface, the component of the contact force that is parallel to the surface and opposes motion is called the frictional force.

Static FrictionWhen the two surfaces are not sliding across one anotherthe friction is called static friction.

MAXss ff £

NsMAXs Ff µ=

Kinetic FrictionOnce there is motion:

Kinetic friction opposes the relative sliding motion motions that actually does occur.

fk = µkFNµk < µs

1

2

In case 1, the force F is pushing down(in addition to mg), so the normal force is increased. In case 2, the force F is pulling up, against gravity, so the normal force is lessened. The frictional force is proportional to the normal force.

a) pushing her from behindb) pulling her from the frontc) both are equivalent

Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this?

ACT: Going sledding

Inclined plane with friction

θgmW !!

=

NF!

θmg

qsinmg

qcosmg

NF

0sin

0cos

=-

=-MAXS

N

fmgmgF

qq

Increasing 𝜃, when will the block start to slide down the plane?

What is the block’s acceleration?

θgmW !!

=

NF!

θmg

qsinmg

qcosmg

NF

mafmgmgF

k

N

=-

=-

qq

sin

0cos

a) component of the gravity force parallel to the plane increased

b) coefficient of static friction decreased

c) normal force exerted by the board decreased

d) both a) and c)e) all of a), b), and c)

A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why?

As the angle increases, the component of weight parallel to the plane increases and the component perpendicular to the plane decreases (and so does the normal force). Because friction depends on normal force, we see that the friction force gets smallerand the force pulling the box down the plane gets bigger.

Static frictionNormal

Weight

ACT: Sliding box

ExampleAn ice skater is gliding across the ice with an initial velocity of +6.3 m/s. The coefficient of kinetic friction between the ice and the skate blades is 0.081, and air resistance is negligible. How much time elapses before his velocity is reduced to +2.8 m/s?

v0 = +6.3 m/s v = +2.8 m/s Δx = ?a = ?Δt = ?

fk = -µkFN = -µkmg

From Newton’s Second Law:

ma = −µkmga = −µkg

v = v0 + aΔtΔt = (v −v0 )/( − µkg)

=(2.8 m/s − 6.3 m/s)/(– (0.081)(9.80 m/s2))

Δt = 4.4 s