1 / 50

# - PowerPoint PPT Presentation

More situations with Newton’s Second Law. Consider a problem that includes an air resistance term: F air resistance = bv 2 , where b is a coefficient that depends on the shape of the object and the density of the air. The direction of the air resistance term is opposite the motion.

Related searches for

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

## PowerPoint Slideshow about '' - moana

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
More situations with Newton’s Second Law

Consider a problem that includes an air resistance term: Fair resistance = bv2 , where b is a coefficient that depends on the shape of the object and the density of the air. The direction of the air resistance term is opposite the motion.

For medium speeds, (not real slow and not real fast), the coefficient for air resistance, b, can be expressed as:

b = (1/2)*C*Across-section*ρ

Where C is a constant that depends on the shape (for a sphere, C = ½), A is the cross-sectional area perpendicular to the velocity, and ρ is the density of the material through which the object is moving. For air at sea level, ρ = 1.2 kg/m3.

Consider an object dropped from a high place:

• we have Fgravity = W = mg directed down,

and with air resistance (AR)

• we have FAR = bv2 directed up.

Newton’s Second Law gives (for 1-D):

-mg + bv2 = ma , or -mg + bv2 = m(dv/dt)

which is a differential equation with v(t) being the solution.

-mg + bv2 = m (dv/dt)

This differential equation can be solved, but it is beyond the level of this course.

We can get an approximate solution, however, by using a numerical (discrete) technique rather than the differential equation (continuous function) technique.

-mg + bv2 = m (dv/dt)

The first step is to calculate the forces based on initial conditions. Let’s say we drop an object that has a mass of 2 kg and an air resistance coefficient 0.03 Nt-s2/m2 from a helicopter 1,000 meters above the ground.

Fgravity = mg = 2 kg * 9.8 m/s2 = 19.6 Nt.

FAR = bv2 = 0.03 Nt-s2/m2 * [0 m/s]2 = 0 Nt.

Using Newton’s Second Law, we can now calculate the initial acceleration:

F = ma gives: -19.6 Nt + 0 Nt = 2 kg * a,

or a = -9.8 m/s2; we now assume that the acceleration does not change much over the first second, and calculate the velocity at one second: v1 = vo + ao*(1 sec)

v1= 0 m/s + (-9.8 m/s2)*(1 sec) = -9.8 m/s.

x1 = xo + ½(vo+v1)*(1 sec)

x1 =1,000 m + ½ (0 m/s + -9.8 m/s)*(1 sec) =

995.1 m.

We now repeat the whole process with the new position and velocity providing new initial conditions:

Fgravity = mg = 2 kg * 9.8 m/s2 = 19.6 Nt.

(gravity remains the same)

FAR = bv2 = 0.03 Nt-s2/m2 * [-9.8 m/s]2 = 2.88 Nt.

(air resistance increased with increased speed)

F = ma  -19.6 Nt + 2.88 Nt = 2 kg * a,

or a = -8.36 m/s2; we now assume that the acceleration does not change much over the second second, and calculate the velocity at two seconds: v2 = v1 + a1*(1 sec)

v2 = -9.8 m/s + (-8.36 m/s2)*(1 sec) = -18.16 m/s.

x2 = x1 + ½ (v1+v2)*(1 sec)

x2= 995.1 m + ½(-9.8 m/s + -18.16 m/s)*(1 sec) = -981.12 m.

We simply keep this process up until x becomes zero.

Normally this would be a lot of steps, but we can use either a computer program to do this or a spreadsheet. We can then plot the graph of either v versus t or x versus t to see what the motion looks like. (See the Excel spreadsheet FallAR.xls which can be downloaded from my PHYS 150 web page.)

To see the effects of different air resistance coefficients, we simply change the value of b and run the program again or recalculate the spreadsheet.

To see the effects of different air resistance functions, such as F=-bv, simply change the F-air resistance function in cell E-5 and copy this change into all the following E cells.

This is the same iterative type process that is used in the Computer Homework Vol. 1 - #8 on Projectiles and in Vol. 1 - #9 on Gravity.

As we can see from the spreadsheet, the velocity does not continue to increase in a linear fashion (as it would without air resistance), but instead it approaches a terminal velocity.

This is because as the speed increases due to gravity, the air resistance also increases which acts against gravity.

As the air resistance approaches the strength of gravity, the net force approaches zero and so the acceleration also approaches zero, and the velocity levels off.

We can calculate the terminal velocity without having to set up the spreadsheet by recognizing that there will be no acceleration and hence no increase in speed when the air resistance equals the gravity: mg = bv2; solving for the terminal speed gives: vterminal = [mg/b]1/2.

Using the numbers we used earlier,

vterminal = [2 kg * 9.8 m/s2 / .03 Nt-s2/m2]1/2= 25.56 m/s = 57 mph.

b = (1/2)*C*Across-section*ρ

For a sphere, C = ½; for air, ρ = 1.2 kg/m3;

for a tennis ball, r = 3.25 cm, m = 56 grams;

for a golf ball, r = 2.1 cm, m = 45 grams;

for a baseball, circumference = 23 cm, m = 145 grams.

From these values, we should be able to get the b value and hence the terminal speed for these three balls: golf ball: 73 mph; tennis ball: 53 mph; baseball: 75 mph.

Note that these values do NOT take into account any aerodynamic lift due to the seams or dimples and spin.

We use ramps and pulleys to make it easier to do certain things.

We’ll first look at ramps, and then pulleys.

Why is it easier to push something up a ramp than it is to lift it?

Let’s look at a picture of the situation:

Fc

P

W=mg

In the static case with no friction, Newton’s 2nd law gives:

F// = P - mg sin() = ma//

F = Fc - mg cos() = ma = 0

Fc

P

W=mg

In the absence of friction, to balance gravity we only need the Pull, P to be

Pbalance = mg sin(). Since sin() is always less than or equal to 1, Pbalance will always be less than or equal to mg: Pbalance mg

Fc

P

W=mg

θ

Pbalance = mg sin()

The smaller the angle, the less P is required. But what price do we pay for decreasing the pull? The price we pay is for an increased distance, L, to pull the object: h/L = sin(),or L = h/sin().

Fc

P

L

h

W=mg

θ

If we now include friction, Newton’s 2nd law gives: F// = P - mg sin() - Ff = ma//

F = Fc - mg cos() = ma = 0

Fc

P

Ff =Fc

W=mg

θ

From the perpendicular equation, we see that Fc = mg cos(), and since Ff = mFc we have Ff = mmg cos(q), and so the parallel equations, with a// = 0(just balancing weight and friction) gives: P = mg sin() +  mg cos()

Fc

P

Ff =Fc

W=mg

θ

P = mg sin() +  mg cos() or

P = mg [sin() +  cos()]

As long as  and  are such that

sin() +  cos() < 1, the pulling force, P, will be less than the weight, which means that the ramp will make it easier to raise the object up.

Let’s let: f() = sin() +  cos()

(the fraction of mg that the pull needs to be).

To find the minimum (or maximum) of this function with respect to angle, we solve:

df()/d = 0 = cos() -  sin()

Which leads to the condition for minimum (or maximum): tan() = 1/.

The condition for minimum (or maximum) pull is: tan() = 1/.

Let’s look at an example: Suppose m = 0.3.

Then qm = arctan (1/.3) = 73.3o. The fraction, f, with

• = 73.3o and m = 0.3 gives a value:

f(73.3o) = sin(73.3o) + 0.3 cos(73.3o)= 1.044 .

If we use a different angle, say 45o, we get

f(45o) = sin(45o) + 0.3 cos(45o)= 0.919 .

Thus the angle 73.3o gives a maximum! Of course, with a value greater than 1 at the critical angle of 73.3o, we realize it would be easier to just lift it up than slide it at this angle!

Pulleys can do two different things for us:

1. Pulleys can change the direction of a force - sometimes it is easier to pull down than to lift up.

2. Pulleys can “add ropes” to an object to reduce the tension in the rope, and hence reduce the pull we need to apply.

Consider the single pulley below:

All the pulley does is change the direction of the rope so we can pull down on the rope instead of lifting up on it.

One of the effects of the

pulley is that there are

essentially two ropes coming

from it - one to the weight and

one to us. This will cause it’s

attachment to the ceiling to have

twice the force on it!

P

W = mg

P = Tension = W

Consider the single pulley in this diagram:

In this case, the single pulley essentially adds a second rope to the weight. We still have to pull up, but now the weight is split between the two ropes, so we only have to pull with

half the weight!

P

T

T

P=T, 2T = W; so P=W/2

W

We now consider putting two pulleys together as in the diagram below:

The left pulley essentially adds a second rope to the weight, reducing the tension in the rope by half. The right pulley simply redirects the tensionso the pull is down instead of up.

P

Consider this combination of pulleys:

The rightmost pulley just redirects the tension so we pull down.

The other four pulleys combine

to provide five ropes to

hold the weight, thus

reducing the tension in

the rope by a factor of 5.

P

Another Example of Newton’s 2nd Law:Car going around a turn

Consider first a car making a right turn on a level road. To make the turn, the car must go in a circle (for a 90o turn, the car must go in a circle for 1/4 of the complete circle). This means that there will be an acceleration towards the center of the circle, which is to the right for a right turn.

From the circular motion relations, we know:

a = w2r, and v=wr; or a = v2/r. That says that we have a bigger acceleration the faster we go and the shorter the radius (sharper the turn).

What are the forces that cause this circular acceleration?

Gravity (weight) acts down

Contact Force acts up (perpendicular to the surface)

Friction - which way

does it act (left or right)?

Fc

a=v2/r

W=mg

Car is heading into the screen.

Without friction the car will NOT make the turn, it will continue straight - into the left ditch! Therefore, friction by the road must push on the car to the right. (Friction bythe caron the road will be opposite - to the left.)

The fastest the car can go around the turn without sliding is when the friction is maximum: Ff = mFc.

Fc

Ff mFc

a=v2/r

W=mg

We now apply Newton’s Second Law - in rectangular components:

SFx = Ff = ma = mv2/r

To make the car go around the turn fastest, we need the maximum force of friction: Ff = mFc

SFy = Fc - W = 0

which says Fc = mg, so

mmg = mv2/r, or vmax = [mgr]

Fc

Ff= mFc

a=v2/r

W=mg

vmax = [mgr]

Note that the maximum speed (without slipping) around a turn depends on the coefficient of friction, the amount of gravity (not usually under our control), and the sharpness of the turn (radius).

If we go at a slower speed around the turn, friction will be less than the maximum: Ff < mFc.

There is one other thing we can do to go faster around the turn - bank the road! How does this work?

By banking the road, we have not added any forces, but we have changed the directions of both the contact force and the friction force!

Have we changed the direction of the acceleration? No - the car is still travelling in a horizontal circle.

Fc

a=v2/r

Ff

q

W=mg

Since the acceleration is still in the x direction, we will again use x and y components (rather than // and )

SFx = Fc sin(q) + Ff cos(q) = mv2/r

SFy = Fc cos(q) - Ff sin(q) - mg = 0

If we are looking for the max speed, we will need the max friction: Ff = mFc .

This gives 3 equations for

3 unknowns: Ff, Fc, and v.

Fc

a=v2/r

Ff

q

W=mg

SFx = Fc sin(q) + Ff cos(q) = mv2/r

SFy = Fc cos(q) - Ff sin(q) - mg = 0

Ff = mFc . Using the third equation, we can eliminate Ff in the first two:

Fc sin(q) + mFc cos(q) = mv2/r

Fc cos(q) - mFc sin(q) - mg = 0

We can now use the second equation to find Fc:

Fc = mg / [cos(q) - m sin(q)], and use this in the first equation to get:

v = [gr {sin(q)+m cos(q)} / {cos(q) - m sin(q)} ]1/2

v = [g r {sin(q)+m cos(q)} / {cos(q) - m sin(q)} ]1/2

Notice that the mass cancels out. This means that the mass of the car does not matter! (Big heavy trucks slip on slippery streets just like small cars. When going fast, big heavy trucks flip over rather than slide off the road; little cars don’t flip over like big trucks. But flipping over is not the same as slipping! We’ll look at flipping in Part 4 of the course.)

Note also that when q = 0, the above expression reduces to the one we had for a flat road:

vmax = [mgr]1/2 .

v = [g r {sin(q)+m cos(q)} / {cos(q) - m sin(q)} ]1/2

Note that as m sin(q) approaches cos(q), the denominator approaches zero, so the maximum speed approaches infinity!

What force really supports such large speeds (and so large accelerations)? As the angle increases, the contact force begins to act more and more to cause the acceleration. And as the contact force increases, so does friction.

Actually, there is a limit on the maximum speed because there is a limit to the contact force.

Banked turn - Minimum Speed

Is there a minimum speed for going around a banked turn? Consider the case where the coefficient of friction is small and the angle of bank is large. In that case the car, if going too slow, will tend to slide down (to the right) so friction should act to the left.

Can you get an equation for the minimum speed necessary?

What changes in what we did for max speed?

Fc

Ff

a=v2/r

q

W=mg

Multiple Objects:Newton’s 2nd and 3rd Laws

Consider a dish on a tablecloth on a table. If we pull on the tablecloth (P), can we keep the dish on the table or will it be pulled off?

Note that while we pull (P) to the leftonthe tablecloth, the pull by the tableclothonus is to the right (not shown).

Dish

P

Tablecloth

Table

Multiple Objects:Newton’s 2nd and 3rd Laws

Consider a dish on a tablecloth on a table. If we pull on the tablecloth (P), can we keep the dish on the table or will it be pulled off?

We must be very careful to consider what forces are acting on and what forces are acting by each object in the problem.

On the diagrams in the next slides, all forces onthe tablecloth will be colored in blue. All forces onthe dish will be colored in orange.

Let’s look first at the weight. We recognize that both the tablecloth and dish will have a mass, mc and md, and so both will also have a weight:

Wc = mcg and Wd = mdg , both directed down.

The earth is doing the pulling, so by Newton’s 3rd Law, the tablecloth and dish will also pull on the Earth in the opposite direction (up) with the same magnitude! However, the mass of the earth will make the acceleration of the earth negligibly small.

Dish

P

mdg

Tablecloth

mcg

Table

We now recognize that the dish is pushing on the tablecloth (due to the dish’s weight). This is actually a contact force (Fcd) downonthe tablecloth by the dish.

But by Newton’s 3rd Law, there must be an equal and opposite force (Fcd) uponthe dishby the tablecloth.

Fcd

Dish

P

mdg

Tablecloth

mcg

Fcd

Table

We now recognize that the table is pushing on the tablecloth. This is actually a contact force (Fct) up onthe tablecloth bythe table.

But by Newton’s 3rd Law, there must be an equal and opposite force (down) on the table: Fct(not shown). However, we will assume the table is solid and will not fall down.

Fcd

Dish

Fct

P

mdg

Tablecloth

mcg

Fcd

Table

We now recognize that there is friction between the tablecloth and the dish. The dish is resisting the pull to the left, so the direction of the force (Ff-cd) onthe tablecloth will be to the right.

But by Newton’s 3rd Law, there must be an equal and opposite force on the dish: Ff-cd. Note that this will be directed to the left (same direction as P) trying to move the dish off the table.

Fcd

Ff-cd

Dish

Fct

P

Ff-cd

mdg

Tablecloth

Fcd

mcg

Table

We now recognize that there is friction between the tablecloth and the table. The table is resisting the pull, so the direction of the force (Ff-ct) onthe tablecloth will be to the right.

But by Newton’s 3rd Law, there must be an equal and opposite force by the tableclothonthe table (not shown). But since the table is bolted down, we assume it will not move.

Fcd

Ff-cd

Ff-ct

Dish

Fct

P

Ff-cd

mdg

Tablecloth

Fcd

mcg

Table

Now we can use Newton’s 2nd Law:

DISH: Fx = Ff-cd = mdad

Fy = Fcd - mdg = 0

Tablecloth: Fx = P - Ff-cd - Ff-ct = mcac

Fy = Fct - Fcd - mcg = 0

Fcd

Ff-cd

Ff-ct

Dish

Fct

P

Ff-cd

mdg

Tablecloth

Fcd

mcg

Table

DISH: Fx = Ff-cd = mdad, or ad = Ff-cd / md

Fy = Fcd - mdg = 0, or Fcd = mdg

Tablecloth: Fx = P - Ff-cd - Ff-ct = mcac, or

P = mcac + Ff-cd + Ff-ct

Fy = Fct - Fcd - mcg = 0, or

Fct = Fcd+mcg = (md+mc)g

Ff-cd = cdFcd = cd mdg, or

ad = cd mdg / md = cd g

Ff-ct = ctFct = ct (md+mc)g

ad = cd g

This says that there will always be a constant acceleration on the dish, but this will last only as long as the tablecloth is in contact with the dish. To minimize the speed, we need to minimize the time the acceleration lasts. It will also help to reduce the speed of the dish if we reduce the friction between the dish and the tablecloth.

P = mcac + Ff-cd + Ff-ct

= mcac +cd mdg+ ct (md+mc)g

This says that the Pull we need will depend on the masses of the dish and tablecloth, on both coefficients of friction, and upon the amount of acceleration of the tablecloth we want. We would like to have as large an acceleration as possible in order to keep the time down.