25 Open query

#$&*

course PHY 201

11/20/2011 2:28 PM

025. `query 25*********************************************

Question: `qprinciples of physics and gen phy

4.26 free-body diagram of baseball at moment

hit, flying toward outfield

gen phy list the forces on the ball while in

contact with the bat, and describe the

directions of these forces

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Gravity pulling the ball down.

The x and y components of the moment of impact.

confidence rating #$&*:

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Given Solution:

`a** Gravity exerts a downward force equal to

the weight of the ball.

While in contact with the ball, and only while

i contact, the bat exerts a normal force, which

pushes outward along a line originating from

the central axis of the bat. This force is

perpendicular to the surface of the bat at the

point of contact.

Unless the direction of the ball is directly

toward the center of the bat, which will not be

the case if the ball is hit at an upward angle

by a nearly level swing, there will also be a

frictional force between bat and ball. This

frictional force will be parallel to the

surface of the bat and will act on the ball in

the 'forward' direction.

COMMON STUDENT ERROR: The gravitational force

and the force exerted by the ball on the bat

are equal and opposite.

The force of the bat on the ball and the

gravitational force are not equal and opposite,

since this is not an equilibrium situation--the

ball is definitely being accelerated by the net

force, so the net force is not zero. **

COMMON STUDENT ERROR: Confusing motion in a

direction with force in that direction.

There is no force associated with the motion of

the ball. The velocity of the ball in will

remain unchanged if there is no net force on

the ball. Furthermore, if the is net force has

zero component in the x direction, the x

velocity remains unchanged; the analogous

statement holds for the y direction.

STUDENT QUESTION

I got confused about the motion in the

direction with the force in that direction. I

think I understand.

INSTRUCTOR RESPONSE

The force tells you the direction of the

acceleration, not the direction of the

velocity. From the force you can therefore tell

the direction of the change in velocity, not

the direction of the velocity itself.

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Question: `qgen phy list the forces on the ball

while flying toward the outfield, and describe

the directions of these forces

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Gravity is still pulling down on the ball.

Air resistance pushing against the path of the

ball.

confidence rating #$&*:

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Given Solution:

`a**After impact the forces are gravity, which

is constant and in the y direction, and air

resistance. The direction of the force of air

resistance is opposite to the direction of

motion. The direction of motion is of course

constantly changing, and the magnitude of the

force of air resistance depends on the speed of

the ball with respect to the air, which is also

changing. **

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Self-critique (if necessary):

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Question: `qgen phy give the source of each

force you have described

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Gravity is the attraction of the ball to the

Earth.

Air resistance is the friction between the

surface of the ball and the air around it.

confidence rating #$&*:

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Given Solution:

`a** The gravitational force is the result of

the gravitational attraction between the ball

and the Earth.

The normal force is the result of the elastic

compression of bat and ball.

The frictional force is due to a variety of

phenomena related to the tendency of the

surfaces to interlock (electromagnetic forces

are involved) and to encounter small 'bumps' in

the surfaces. **

ERRONEOUS STUDENT ANSWER:

the air, the pitcher, the bat/ batter.

friction. gravity

INSTRUCTOR RESPONSE: All these are sources of

force in one or both situations (bat striking

ball, ball flying toward outfield) except the

pitcher. The pitcher exerted a force

previously, and that force was instrumental in

delivering the ball to the batter, but that

force ended well before any of these events

occurred.

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Question: `qgen phy what is the direction of

the net force on the ball while in contact with

the bat?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

?

confidence rating #$&*:

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Given Solution:

`a** We assume that the y axis is directed

vertically upward, and the x axis is

horizontal.

The normal force will vary from 0 at the

instant of first contact, to a maximum at the

instant of greatest compression, and back to 0

at the instant contact ceases. So there is no

single normal force. However we can represent

'the' normal force as the average normal force.

The gravitational force will remain constant.

The frictional force will vary with the

changing normal force, and we will speak here

of the average frictional force.

The average normal force will be the greatest

of these forces, much greater than friction or

gravity. The frictional force will likely also

exceed the gravitational force.

The y component of the normal force will

overwhelm the y components of the frictional

force and the gravitational force, both of

which are downward, giving us a net y component

slightly less than the y component of the

normal force.

The x component of the normal force will be

reinforced by the x component of the frictional

force, making the x component of the net force

a bit greater than the x component of the

normal force.

This will result in a net force that is

'tilted' forward and slightly down from the

normal force (see the figure at the end for a

vector diagram showing normal, frictional,

gravitational and resultant forces).

Note that the frictional and gravitational

forces will tend to 'spin' the baseball as well

as contributing to its translational

acceleration. The spinning effect is a topic

for a later chapter. **

IMPORTANT NOTE: It is essential that you

sketch a diagram showing these forces. You are

very unlikely to understand the explanation

given here without a picture. Even with a

picture this might be challenging. If you are

not sure you understand, you should submit a

copy of this question and solution, along your

questions and/or commentary (mark insertions

with ****).

STUDENT COMMENT: Not sure about the frictional

force. Why is it down? How do we calculate

it?

INSTRUCTOR RESPONSE: The frictional force

exerted on the ball by the bat is perpendicular

to the normal force, so the frictional force is

exerted in the plane tangent to both the ball

and the bat (imagine a flat piece of cardboard

sandwiched between the ball and the bat; it

lies in this tangent plane. If you have a line

segment connecting the middle of the ball with

the middle of the bat, it is perpendicular to

the tangent plane (this line segment would cut

through the piece of carboard at a right

angle). Note that the direction of the normal

force on the ball is along this line.).

It is clear that the x component of the

frictional force on the ball is in the

'forward' direction of motion. It is also clear

that the in the tangent plane, the 'forward'

direction is also downward. So the frictional

force has a positive x component, and a

negative y component.

Assuming the ball does not 'slip' in contact

with the bat, the frictional force is the force

of static friction. The force of static

friction cannot exceed the product of the

coefficient of friction and the normal force:

f_static < = mu * N, where f_static is the

force of static friction, mu is the coefficient

of static friction and N is the normal force.

STUDENT QUESTION

So the net force is tilted slightly, what does

this mean, is it parallel to the ball and then

slightly

perpendicular at the same time so it tends to

curve??????????????????????????????????????????

???????????

Is there a ready made sketch in our notes so

that I can see if what I drew is correct?

INSTRUCTOR RESPONSE

The ball is represented in the figure below by

the light pink circle, the bat by the green

circle. The ball has arrived from the left,

the bat is being swung to the right. The

normal force acts perpendicular to the surface

where the ball and bat make contact; the normal

force is represented by the vector pointing

toward upper right. The frictional force acts

parallel to the surface of contact, and is

represented by the shorter vector pointing down

and to the left. (The ball and bat actually

compress significantly, the ball more than the

bat, and that compression is the source of the

normal force. However the compression is not

depicted in the figure.)

The figure does not represent the gravitational

force on the ball, which would be depicted as a

downward force acting at the center of the

ball. In a typical 'hit', the gravitational

force would be much less than either the

frictional or the normal force.

The three forces are shown in the figure below,

head-to-tail, along with the resultant force

(the resultant is in red; the gravitational

force is in the downward vertical direction and

would likely be much less than depicted here).

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Question: `qgen phy what is the net force on

the ball while flying toward the outfield?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Fnet will be the sum of gravity and air

resistance.

confidence rating #$&*:

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Given Solution:

`a** The net force will consist of the downward

gravitational force and the force of air

resistance opposing the motion.

If the ball is rising the y component of the

air resistance will be in the downward

direction, reinforcing the gravitational force

and giving a net downward y component slightly

exceeding that of gravity.

If the ball is falling the y component will be

in the upward direction, opposing the

gravitational force and giving a net downward y

component slightly less than that of gravity.

In either case the x component will be in the

direction opposite to the 'forward' motion of

the ball, so the net force will be directed

mostly downward but also a bit 'backward'.

There are also air pressure forces related to

the spinning of the ball; the net force exerted

by air pressure causes the path of the ball to

curve a bit, but these forces won't be

considered here. **

STUDENT QUESTION

What about as the ball is moving forward, is

ther no air resistance being pushed against the

ball horizontally as it flies to the outfield?

INSTRUCTOR RESPONSE

The ball typically experiences air resistance

with components in both the x and the y

direction.

If it's rising the y component of the air

resistance is downward, if it's falling the y

component is upward. If it's at the very top of

its arc, then for an instant it is neither

rising nor falling and there is no air

resistance in the y direction.

The x component of the air resistance is in the

direction opposite the 'forward' motion of the

ball. I believe this is the force you asked

about in your question.

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Question: `qUniv. 5.88 (5.84 10th edition).

Elevator accel upward 1.90 m/s^2; 28 kg box;

coeff kin frict 0.32. How much force to push at

const speed?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

F=m*a=28kg*1.9m/s^2=53.2N

F=28kg*9.8m/s^2=274.4N

Fnet=274.4+53.2N=330N approx

330N*.32=105N

confidence rating #$&*:

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Given Solution:

`aSTUDENT SOLUTION AND INSTRUCTOR COMMENT: The

magnitude of kinetic friction force is fk =

mu-sub k * N. First we add the 1.9 to 9.8 and

get 11.7 as the acceleration and times that by

the 28 kg and get 327.6 as the force so

plugging in we get fk = 0.32 * 327.6 = 104.8 N.

** Good.

The net force Fnet on the box is Fnet = m a =

1.90 m/s^2 * 28 kg.

The net force is equal to the sum of the forces

acting on the box, which include the weight mg

acting downward and the force of the floor on

the box acting upward. So we have

Fnet = Ffloor - m g = m a.

Thus Ffloor = m g + m a = 28 kg * 9.8 m/s^2 +

28 kg * 1.90 m/s^2 = 28 kg * 11.7 m/s^2 = 330

N, approx.

Being pushed at constant speed the frictional

force is f = `mu * N, where N is the normal

force between the box and the floor. So we have

f = .32 * 330 Newtons = 100 N, approx. **

STUDENT QUESTION:

I don't understand why the net force is the

weight of the box + the upward force of the

elevator. Since the weight is

directed downwards, and the elevator is going

upwards, shouldn't it be the force of the

elevator - weight? ????

INSTRUCTOR RESPONSE

There are two forces acting on the box in the

vertical direction, the weight (acting

downward) and the normal force exerted by floor

of the the elevator on the box (which acts

upward).

Using Ffloor for the normal force, the net

vertical force is therefore

net vertical force = Ffloor - weight.

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Self-critique (if necessary):

I apparently got the correct answer but I

thought I was quitting half way through the

problem because I did not know what to do next.

What is “fk = mu-sub k * N”???

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Self-critique Rating:1

@&

In the given solution it is stated that

f = `mu * N,

the general expression for the frictional force. `mu is the coefficient of friction and N is the normal force between the surfaces.

A further distinction is often necessary between kinetic and static friction.

`mu_k stands for coefficient of kinetic friction, `mu_s for coefficient of static friction.

f_k would the the force of kinetic friction, which would be mu_k * N.

*@

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&#Good work. See my notes and let me know if you have questions. &#