Chapter 6 Applications of Newton’s Laws

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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=m

₁

a

M

₂

g-T=m

₂

a

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 F

_net

≠ 0, because v is not constant, a ≠ 0. a =???, Is a constant?

V

_i

V

_f

DV=a_ave

t

V

_i

V

_f DV=at

In circular motion, V direction keeps changing

Acceleration direction also keeps changing.

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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.

a

_cp

= v

²

/r ( r is radius of circle )

The fast the velocity is, the larger a

_cp

is needed to change velocity direction to stay in circular orbit.

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

_cp

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 F

_cp

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 a

_cp

is needed. (It’s hard to make sharp

turns with small radius. )

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To do circular motion object Needs centripetal a

_cp

Total force needed in radius direction = m a

_cp

= m This total force needed in radius direction has a fancy name: Centripetal force.

SF= ma

SF_r

= m a

_cp

=F

_{net r}

= F

_cp

F

_{net r}

= m v

²

/r

r v²

Centripetal force is not a new force.

It is simply the total force in radius direction.

F

_{net r}

or F

_cp

is the required (needed) net force (force sum) along the radius

direction of circular motion,

it is not another new force.

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6-5 Circular Motion

This total force in the radius direction, F

_cp

, 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 a

_cp

, then set the sum to be

F

_{net r}

= F

_cp

=m v

²

/r

You should never add F

_cp

to existing forces.

SFr = f_s = m v²

/r along a flat track.

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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, SF_r = N sinq = m v² /r

6-5 Circular Motion On roller coaster:

F

_cp

depends on positions….

At point B, F

_cp

= N

Because only N is along the r direction in this case.

At point A, F

_cp

= N – mg

Because along the radius directon N is pointing to the center,

and mg is opposite.

At point C, F

_cp

= N + mg

Both N and Mg are pointing to the center.

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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!

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Demo: 2, A ball attached to a string rotate horizontally on a table.

Ask for N at round hill top.

F

net in r direction point to center

= mg –N = m v

²

/r

N = m g – m v

²

**/r = 9810 – 1000 *10**

²

/20 =4800 (N) N and mg are not balanced now. a

_y

≠ 0 . On the hill top, a

_r

is downward. mg has to win to give the downward a

_r

, so that the car will follow the hill and goes down but not fly out horizontally.

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m v

²

/r , is the needed net force toward center, Faster V , F

_{net r}

needed ↑ easier to fly off orbit Larger r , F

_{net 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 v

²

/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 v

²

/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.

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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?

F

_net

along radius = mg = m a

_cp

= m = g , so v

²

= g.r so, v =

Close to earth surface. g=9.81 m/s

²

Earth radius r = 6400km = 6.4 x 10

⁶

m V= 7.9 x 10

³

(m/s) ~ 5 mile/s

Time to circle the world :

T= 2 p r / v = 2 p 6.4 x 10

⁶

/ 7.9 x10

³

= 5.1 x 10

³

(s) = 85 (min)

r v²

r

v

² _g._r

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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:

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

Total force NEEDED in the radius direction to stay and maintain circular motion:

F_needed= ma_cp=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. ΣF_r = 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 ΣF_r, toward center.

3. Make ΣF_r At that position equals to mv²/r at that position 4. Solve Equation ΣF_r=mv2/r