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Chapter 6

Applications of Newton’s

Laws

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Units of Chapter 6

• Frictional Forces

• Strings and Springs

• Translational Equilibrium

• Connected Objects

• Circular Motion

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6-3 Translational Equilibrium

When an object is in translational equilibrium,

the net force on it is zero:

(6-5)

This allows the calculation of unknown forces.

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6-3 Translational Equilibrium

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6-4 Connected Objects

When forces are exerted on connected objects,

their accelerations are the same.

If there are two objects connected by a string,

and we know the force and the masses, we can

find the acceleration and the tension:

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6-4 Connected Objects

We treat each box as a separate system:

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6-4 Connected Objects

If there is a pulley, it is easiest to have the

coordinate system follow the string:

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T=m1a

M2g-T=m2a

The object was doing circular motion, if the

string is suddenly broken, which path will the

object follow?

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6-5 Circular Motion

An object moving in a circle must have a force

acting on it; otherwise it would move in a straight

line. The direction of the

force is towards the

center of the circle.

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Circular motion

planet, swing, and my key chain..

We know Fnet ≠ 0, because v is not constant,

a ≠ 0. a =???, Is a constant?

Vi

Vf

DV=aavet

Vi

Vf

DV=at

In circular motion, V direction keeps changing

Acceleration direction also keeps changing. 10

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6-5 Circular Motion Centripetal acceleration:

Direction: always pointing to the center, in the

radius direction.

The acceleration needed for stay on circular orbit.

acp = v2/r ( r is radius of circle )

The fast the velocity is, the larger acp is needed to

change velocity direction to stay in circular orbit.

The smaller the radius of the orbit is, the more acp

is needed .

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6-5 Circular Motion

Therefore the force, needed to keep an object of

mass m moving in a circle of radius r.

The magnitude of the force is given by:

(6-15)

When the velocity is faster, a larger Fcp is needed to

change the velocity direction so that it stay in

circular orbit. (Please SLOW DOWN when you

make turns). The smaller the radius of the orbit is,

the more acp is needed. (It’s hard to make sharp

turns with small radius. ) 12

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To do circular motion object Needs centripetal acp

Total force needed in radius direction = m acp = m

This total force needed in radius direction has a

fancy name: Centripetal force.

SF= ma

SFr = m acp =Fnet r = Fcp

Fnet r = m v2 /r

r

v2

Centripetal force is not a new force.

It is simply the total force in radius direction.

Fnet r or Fcp is the required

(needed) net force (force

sum) along the radius

direction of circular motion,

it is not another new force. 13

6-5 Circular Motion

This total force in the radius direction, Fcp , may

be provided by the tension in a string, the

normal force, or friction, or gravity…or

combination of any of these real forces, which

keep the object in a circular orbit, or part of a

circular arc.

You should find the existing

forces along the radius

direction which provide acp ,

then set the sum to be

Fnet r = Fcp =m v2 /r

You should never add Fcp to

existing forces.

SFr = fs = m v2 /r along a flat track. 14

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When the radius of the orbit is horizontal.

It needs an horizontal direction total force

along the radius direction to turn the

direction along the circle.

If there is no friction or friction is not enough,

we need N to provide some horizontal Fcp.

If friction =0, SFr = N sinq = m v2 /r

6-5 Circular Motion

On roller coaster:

Fcp depends on positions….

At point B, Fcp = N

Because only N is along the r

direction in this case.

At point A, Fcp = N – mg

Because along the radius directon

N is pointing to the center,

and mg is opposite.

At point C, Fcp = N + mg

Both N and Mg are pointing to the center. 16

N

N

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SFr = T sin q = m v2 /r

Problem solving stratege:

1. Identify all forces and

force components in the

radius direction.

2. Sum F in radius direction

and make it = m v2 /r

If you know v, r, you can solve force

If you know forces and r, you can solve v.

You should use v at that point and F at that point

r is radius of ORBIT! 17

Demo: 2, A ball attached to a string rotate

horizontally on a table.

Ask for N

at round

hill top.

Fnet in r direction point to center = mg –N = m v2 /r

N = m g – m v2 /r = 9810 – 1000 *102/20 =4800 (N)

N and mg are not balanced now. ay≠ 0 . On the hill

top, ar is downward. mg has to win to give the

downward ar, so that the car will follow the hill and

goes down but not fly out horizontally. 18

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m v2 /r , is the needed net force toward center,

Faster V , Fnet r needed ↑ easier to fly off orbit

Larger r , Fnet r needed ↓ easier to stay in orbit

(more gradual turn)

When existing forces can not provide any force

along radius direction ( pointing to the center),

objects can not stay in circular orbit, it will fly away

along tangent direction!!! (inertia) The broken key train…

When actual forces inward in r

direction is more than the needed

m v2 /r ,

Objects can not be kept running in

the circular object. r will decrease,

It will fall inward. Enough speed at

point C, is important.

SFr > m v2 /r

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When actual forces do not provide

enough m v2 /r ,

That means

Objects can not be kept running in

the circular object. r will increase,

It feels like being pushed outward

( that’s how centrifuge works), It’s

important for bio.

SFr < m v2 /r

Actually there is not such a centrifugal force, just

the effect of your inertia. Your dogs know that to

shake water off.

That’s also how Spin dryers works.

Larger v, no enough F in radius direction to hold

the water droplet. 20

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Example 2: orbit around earth

To let an object fly around the earth close to earth

surface, you need to launch it with what velocity?

Fnet along radius = mg = m acp= m

= g , so v2 = g.r so, v =

Close to earth surface. g=9.81 m/s2

Earth radius r = 6400km = 6.4 x 106m

V= 7.9 x 103(m/s) ~ 5 mile/s

Time to circle the world :

T= 2 p r / v = 2 p 6.4 x 106 / 7.9 x103

= 5.1 x 103 (s) = 85 (min)

r

v2

r

v2

rg.

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6-5 Circular Motion

An object may be changing its speed as it

moves in a circle; in that case, there is a

tangential acceleration as well:

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Summary of Chapter 6

• Friction is due to microscopic roughness.

• Kinetic friction:

• Static friction:

• Tension: the force transmitted through a

string.

• Force exerted by an ideal spring:

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Summary of Chapter 6

• An object is in translational equilibrium if the

net force acting on it is zero.

• Connected objects have the same acceleration.

• The force required to move an object of mass m

in a circle of radius r is:

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Extra notes for circular motion:

Circular motion : v keeps changing, maybe both speed and

direction are changing. At least v direction is changing.

Hence a≠0.

Acceleration NEEDED to stay on circular orbit:

acp=v2/r , pointing to the center along radius direction.

Total force NEEDED in the radius direction to stay

and maintain circular motion:

Fneeded= macp=mv2/r , pointing to center

1. When total force in r direction pointing to center equals to

mv2/r , it has what it needs to stay in orbit. ΣFr = mv2/r ,

it stays in circular orbit. Neither flies away, nor falls into the

center.

This is how you solve related forces. Velocities or radius, etc.

1. Find ALL REAL Forces and force components,

2. ADD ALL in radius direction ΣFr, toward center.

3. Make ΣFr At that position equals to mv2/r at that position

4. Solve Equation ΣFr =mv2/r