QA Assignment 11

#$&*

course PHY 121

6/26 6

011. Note that there are 12 questions in this set.

.Situations involving forces and accelerations.

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Question: `q001. A cart on a level, frictionless surface contains ten masses, each of

mass 2 kg. The cart itself has a mass of 30 kg. A light weight hanger is attached to

the cart by a light but strong rope and suspended over the light frictionless pulley at

the end of the level surface.

Ignoring the weights of the rope, hanger and pulley, what will be the acceleration of

the cart if one of the 2 kg masses is transferred from the cart to the hanger?

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Your solution:

I know that gravity will pull on the weight at the rate of 9.8 m/s^2. Since the weight

is 2 kg, you would multiply that by 2 for a Force of 19.6 Newtons.

The next part is unclear. I took that force of 19.6 Newtons and divided it by 48 kg,

which is the weight of the cart (30 kg) plus the remaining weights (9 x 2 kg=18 kg).

This gives you .408 m/s^2 for acceleration.

@&

30 kg is the mass of the cart, not its weight.

On these problems that is an imprortant distinction. Weight is a force, capable of accelerating things. Mass is something that has weight, but mass also resists acceleration.

*@

@&

All the mass is being accelerated by the net force, wherever it is in the system.

*@

confidence rating #$&*:

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Given Solution:

At the surface of the Earth gravity, which is observed here near the surface to

accelerate freely falling objects at 9.8 m/s^2, must exert a force of 2 kg * 9.8 m/s^2 =

19.6 Newtons on the hanging 2 kg mass.

This force will tend to accelerate the system consisting of the cart, rope, hanger and

suspended mass, in the 'forward' direction--the direction in which the various

components of the system must accelerate if the hanging mass is to accelerate in the

downward direction. This is the only force accelerating the system in this direction.

All other forces, including the force of gravity pulling the cart and the remaining

masses downward and the normal force exerted by the level surface to prevent gravity

from accelerating the cart downward, are in a direction perpendicular to the motion of

these components of the system; furthermore these forces are balanced so that they add

up to 0.

The mass of the system is that of the 30 kg cart plus that of the ten 2 kg masses, a

total of 50 kg.

The net force of 19.6 Newtons exerted on a 50 kg mass therefore results in acceleration

a = Fnet / m = 19.6 Newtons / (50 kg)

= 19.6 kg m/s^2 / (50 kg)

= .392 m/s^2.

STUDENT QUESTION:

Our answers are close, but wouldn’t the 50kg actually by 48kg since the 2kg weight was

taken out of the kart?

INSTRUCTOR RESPONSE:

The 2 kg weight is still part of the mass that's being accelerated. It's been moved from

the cart to the hanger, but it's still there. All 50 kg are being accelerated by that

net force.

STUDENT QUESTION

Am I allowed to divide Newtons by Kg? Or do I have to change the Newtons to kg*m/s^2?

INSTRUCTOR RESPONSE

You need to reduce everything to fundamental units. How would you know what N / kg is

unless they are both expressed in compatible units?

You could of course memorize the fact that N / kg gives you m/s^2, along with about 200

other shortcuts, but that would be a waste of time and wouldn't contribute much to your

insight or your ability to work out units in unfamiliar situations.

You can count the number of fundamental units in all of physics on one hand. If you

know the definitions of the quantities, this makes it very easy to deal with questions

of units. Unit calculations come down to the simple algebra of multiplying and dividing

fractions, whose numerators and denominators are just products and powers of a few

simple units.

STUDENT QUESTION

Does it matter if you define aGravity as 9.8 m/s^2 or -9.8 m/s^2? Would -.39m/s^2 still

be

correct since we just know it's perpendicular, and not which perpendicular direction is

positive?

INSTRUCTOR RESPONSE:

In your solution you said

'Fnet = 2kg * -9.8m/s^2 = -19.6 N in the downward direction'

You weren't completely specific about what the - sign meant and what direction is

positive. If the upward direction is considered positive, then we would simply say that

the gravitational force is -19.6 N. We wouldn't add 'in the downward direction' because

having declared upward as positive, the negative sign already tells us that the force is

downward. If we were to say that the force is 19.6 N in the downward direction, this

would be a true statement, without the use of the -sign.

However this system doesn't move in just the up-down direction, and we have to be

careful how we define our positive direction:

The signs of the displacement, velocity and acceleration depend on the direction we

choose as positive.

We consider the system to consist of the cart and the hanging weights. Parts of this

system are moving in the vertical

direction and parts are moving in the horizontal direction, so neither vertical nor

horizontal can be regarded as the

positive direction.

Let's assume that we are oriented so that from our position the cart moves to the right

as the hanging weights descend. If

the cart moves to the right, the hanging weights move downward; if the cart moves to the

left the hanging weight move upward.

We can describe these motions as 'right-down' and 'left-up'.

We have to declare our choice of positive direction. We can choose either 'right-down'

or 'left-up' to be positive,

whichever we find more convenient; having made this choice the opposite direction will

be regarded as negative.

Now gravity will pull the system in the 'right-down' direction. You might prefer to

regard the gravitational force as

negative, in which case you will choose 'left-up' as the positive direction. You have

implicitly done so by making your

acceleration negative.

So choosing 'left-up' as the positive direction the force exerted by gravity on the

hanging mass, which pulls the system in

the 'right-down' direction, will be negative. The resulting acceleration will in this

case be a = F_net / m = -19.6 N / (50

kg) = -.392 m/s^2. This means that acceleration will be opposite the direction of

motion--i.e., in the 'right-down'

direction.

It would have been equally valid to choose 'right-down' as the positive direction. The

gravitational force on the hanging

mass would be +19.6 N and the acceleration +.392 N.

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Self-critique (if necessary):

I couldn't remember if the weight was left or removed in the calculations. It almost

feels like it is being counted twice, but I understand that it is still part of the

""system.""

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Self-critique rating: 2

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Question: `q002. Two 1-kg masses are suspended over a pulley, one on either side. A

100-gram mass is added to the 1-kg mass on one side of the pulley. How much force does

gravity exert on each side of the pulley, and what is the net force acting on the entire

2.1 kg system?

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Your solution:

On one side, there is 1.0 kg * 9.8 m/s^2 = 9.8 Newtons

On the other side, there is 1.1 kg * 9.8 m/s^s = 10.78 Newtons

I am not sure if the net force acting on the whole system is .98 Newtons (the

difference) or 20.58 Newtons (the sum).

confidence rating #$&*:

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Given Solution:

The 1-kg mass experiences a force of F = 1 kg * 9.8 m/s^2 = 9.8 Newtons. The other side

has a total mass of 1 kg + 100 grams = 1 kg + .1 kg = 1.1 kg, so it experiences a force

of F = 1.1 kg * 9.8 Newtons = 10.78 Newtons.

Both of these forces are downward, so it might seem that the net force on the system

would be 9.8 Newtons + 10.78 Newtons = 20.58 Newtons. However this doesn't seem quite

right, because when one mass is pulled down the other is pulled up so in some sense the

forces are opposing. It also doesn't make sense because if we had a 2.1 kg system with

a net force of 20.58 Newtons its acceleration would be 9.8 m/s^2, since we know very

well that two nearly equal masses suspended over a pulley won't both accelerate downward

at the acceleration of gravity.

So in this case we take note of the fact that the two forces are indeed opposing, with

one tending to pull the system in one direction and the other in the opposite direction.

We also see that we have to abandon the notion that the appropriate directions for

motion of the system are 'up' and 'down'. We instead take the positive direction to be

the direction in which the system moves when the 1.1 kg mass descends.

We now see that the net force in the positive direction is 10.78 Newtons and that a

force of 9.8 Newtons acts in the negative direction, so that the net force on the system

is 10.78 Newtons - 9.8 Newtons =.98 Newtons.

The net force of .98 Newtons on a system whose total mass is 2.1 kg results in an

acceleration of .98 Newtons / (2.1 kg) = .45 m/s^2, approx.. Thus the system

accelerates in the direction of the 1.1 kg mass at .45 m/s^2.

Additional note on + and - directions:

One force tends to accelerate the system in one direction, the other tends to accelerate

it in the opposite direction.

So you need to choose a positive direction and put a + or - sign on each force,

consistent with your chosen positive direction.

The positive direction can't be 'up' or 'down', since part of the system moves up while

another part moves down.

The easiest way to specify a positive direction is to specify the direction of one of

the masses.

STUDENT QUESTION

I didn’t understand changing it to making one positive and one negative.

INSTRUCTOR RESPONSE

I understand your question but for your sake, more specific would be better. For

example, instead of 'it' say what it is that's being changed. And specify 'one' what.

In any case:

What do you and do you not understand about each of the following statements:

'The positive direction can't be 'up' or 'down', since part of the system moves up while

another part moves down.'?

Both of the weights pull down, but by the preceding 'down' isn't a direction of motion

for this system.

Two weights tend to move the system in opposite directions; e.g., if there's more weight

on the right side the system will move in a direction opposite the direction is will

move if there's more weight on the left.

If the two weights are equal the system will not accelerate.

The magnitude of the net force is the difference in the weights on the two sides that

determines the net force?

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Self-critique (if necessary):

It makes sense. I hedged my bets with both. My gut told me that it would be the

difference. But, trying to apply all of these ideas made me second-guess myself.

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Self-critique rating: 2

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Question: `q003. If in the previous question there is friction in the pulley, as there

must be in any real-world pulley, the system in the previous problem will not accelerate

at the rate calculated there. Suppose that the pulley exerts a retarding frictional

force on the system which is equal in magnitude to 1% of the weight of the system. In

this case what will be the acceleration of the system, assuming that it is moving in the

positive direction (as defined in the previous exercise)?

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Your solution:

The positive net force is .98 Newtons. 1% of the weight of the system is.021. If I

reduce the net force by .021 Newton, I get .959 Newtons. If I divide .959 Newtons by 2.1

kg (the mass of the system), I get .46 m/s^2.

confidence rating #$&*:

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Given Solution:

We first determine the force exerted a friction. The weight of the system is the force

exerted by gravity on the mass of the system. The system has mass 2.1 kg, so the weight

of the system must be

2.1 kg * 9.8 m/s^2 = 20.58 Newtons.

1% of this weight is .21 Newtons, rounded off to two significant figures. This force

will be exerted in the direction opposite to that of the motion of the system; since the

system is assume to be moving in the positive direction the force exerted by friction

will be

frictional force = -.21 Newtons.

The net force exerted by the system will in this case be 10.78 Newtons - 9.8 Newtons -

.21 Newtons = .77 Newtons, in contrast with the .98 Newton net force of the original

exercise.

The acceleration of the system will be

.77 Newtons / (2.1 kg) = .37 m/s^2, approx..

STUDENT SOLUTION

(this solution and the instructor's commentary address a common error in expression and

in thinking

this is worth a look

the main topic is why it's not appropriate to write an expression like

.98N-.21N=.77N/2.1kg=.37m/s^2

Using my numbers from the previous problem, we had .98N pulling down, so I will subtract

an additional .21

from that to get the Newtons pulling down w/ friction. .98N-.21N=.77N/2.1kg=.37m/s^2.

INSTRUCTOR COMMENTARY

It's clear what you mean by .98N-.21N=.77N/2.1kg=.37m/s^2, and everything you said up to

this point is very good and correct.

However as a mathematical statement .98N-.21N=.77N/2.1kg=.37m/s^2 is incorrect.

If .98N-.21N=.77N/2.1kg=.37m/s^2, then since quantities that are both equal to a third

quantity are equal to one another, .98N-.21N = .37m/s^2.

However N and m/s^2 are complete different units, so the left- and right-hand sides of

this equality are unlike terms. Unlike terms can't possibly be equal.

And of course it's very obvious that

.98N-.21N =. .37m/s^2

is simply an untrue statement.

Untrue statements tend to lead to confusion, e.g., when you review your work and don't

necessarily remember what you were thinking when you wrote the thing down, or when your

statement is viewed by someone else who doesn't already know what to expect.

If you said

F_net = .98N-.21N=.77N

so

a = F_net / m = .77 N / (2.1kg) =.37m/s^2

then you would not have any false statements in your solution, your solution would be

clear to anyone who understands Newton's Second Law, and would be much more likely

useful to you when reviewing your work.

(Note also that N / kg = (kg m/s^2) / kg = m/s^2

It's important to maintain the habit of reducing units to fundamental units and doing

the algebra of the units.)

STUDENT QUESTION

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Self-critique (if necessary):

I started with the original net force instead of the original total mass. This caused a

big mistake. I understand the steps for solving this problem, at the moment.

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Self-critique rating: 2

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Question: `q004. Why was it necessary in the previous version of the exercise to

specify that the system was moving in the direction of the 1.1 kg mass. Doesn't the

system have to move in that direction?

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Your solution:

I guess that you have to specify that the system was moving in the direction of the 1.1

kg mass because there could be other forces working on the system that causes that not

to happen. It's not a given.

confidence rating #$&*:

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Given Solution:

If the system is released from rest, since acceleration is in the direction of the 1.1

kg mass its velocity will certainly always be positive.

However, the system doesn't have to be released from rest. We could give a push in the

negative direction before releasing it, in which case it would continue moving in the

negative direction until the positive acceleration brought it to rest for an instant,

after which it would begin moving faster and faster in the positive direction.

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Self-critique (if necessary): OK

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Self-critique rating: 2

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Question: `q005. If the system of the preceding series of exercises is initially

moving in the negative direction, then including friction in the calculation what is its

acceleration?

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Your solution:

If the friction is going in the positive direction, then it's possible that it's

""helping"" in the negative direction. If that is the case, I have calculated the

following:

20.58 Newtons + .21 Newtons = 20.79 Newtons

20.79 Newton / 2.1 kg would mean that the acceleration is 9.9 m/s^2.

That being said, I am not sure I've interpreted the question correctly.

confidence rating #$&*:

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Given Solution:

Its acceleration will be due to the net force. This net force will include the 10.78

Newton force in the positive direction and the 9.8 Newton force in the negative

direction. It will also include a frictional force of .21 Newtons in the direction

opposed to motion.

Since motion is in the negative direction, the frictional force will therefore be in the

positive direction. The net force will thus be

Fnet = 10.78 Newtons - 9.8 Newtons + .21 Newtons = +1.19 Newtons,

in contrast to the +.98 Newtons obtained when friction was neglected and the +.77

Newtons obtained when the system was moving in the positive direction. ,

The acceleration of the system is therefore

a = 1.19 N / (2.1 kg) = .57 m/s^2.

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Self-critique (if necessary):

I understand. It's just going to take some time to sink in.

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Self-critique rating: 2

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Question: `q006. If friction is neglected, what will be the result of adding 100 grams

to a similar system which originally consists of two 10-kg masses, rather than the two

1-kg masses in the previous examples?

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Your solution:

If you have two 10-kg masses, and then add 100 more grams to one side, it will have the

same effect as adding 100 grams to two 1-kg masses. The 10 kg on each side balance

either other and, in effect, cancel either other out. If you multiplied it all out, you

would get 10 kg * 9.8 m/s^2 = 98 N. 10.1 kg * 9.8 m/s^2 = 98.98 N. The difference is

.98 Newtons, just like the earlier problem.

confidence rating #$&*:

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Given Solution:

In this case the masses will be 10.1 kg and 10 kg. The force on the 10.1 kg mass will

be 10.1 kg * 9.8 m/s^2 = 98.98 Newtons and the force on the 10 kg mass will be 10 kg *

9.8 m/s^2 = 98 Newtons.

The net force will therefore be .98 Newtons, as it was in the previous example where

friction was neglected.

We note that this.98 Newtons is the result of the additional 100 gram mass, which is the

same in both examples.

The total mass of the system is 10 kg + 10.1 kg = 20.1 kg, so that the acceleration of

the system is

a = .98 Newtons / 20.1 kg = .048 m/s^2, approx..

Comparing this with the preceding situation, where the net force was the same (.98 N)

but the total mass was 2.1 kg, we see that the same net force acting on the

significantly greater mass results in significantly less acceleration.

Note on the direction of the frictional force: It's not quite accurate to say that the

frictional force is always in the direction opposite motion. I'm not really telling you

the whole story here--trying to keep things simple. Friction can indeed speed things up,

depending on your frame of reference.

The more accurate statement is that forces exerted by kinetic friction act in the

direction opposite the relative motion of the two surfaces. (Forces exerted by static

friction act in the direction opposite the sum of all other forces).

For example a concrete block, free to slide around in the bed of a pickup truck which is

accelerating forward, is accelerated by the frictional force between it and the

truckbed. So the frictional force is in its direction of motion. If the block doesn't

slide, it is static friction that accelerates it and there is no relative motion between

the surfaces of the block and the truckbed. If the block does slide, the frictional

force is still pushing it forward relative to the road, and relative to the road it

accelerates in its direction of motion, but the frictional force isn't sufficient to

accelerate it at the same rate as the truck; it therefore slides backward relative to

the truckbed. Relative to the truckbed the block slides backward while the frictional

force pushes it forward--the frictional force is in the direction opposite the relative

motion.

If the block is sliding, it is moving toward the back of the truck while friction is

pushing it toward the front. So in this case the frictional force acts in the direction

opposite the relative motion of the two surfaces.

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Self-critique (if necessary):

OK

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Self-critique rating: 3

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Question: `q007. If friction is not neglected, what will be the result for the system

with the two 10-kg masses with .1 kg added to one side? Note that by following what has

gone before you could, with no error and through no fault of your own, possibly get an

absurd result here, which will be repeated in the explanation then resolved at the end

of the explanation.

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Your solution:

The frictional force of this weight would be 20.1 kg * 9.8 m/s^2 = 196.98 Newtons.

1% of this is 1.97 N.

When you take the force from the 10.1 kg side and subtract the force from the 10 kg side

minus the frictional force, you get:

98.98 Newtons - 98 Newtons - 1.97 Newtons = -.99 Newtons

confidence rating #$&*:

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Given Solution:

If friction is still equal to 1% of the total weight of the system, which in this case

is 20.1 kg * 9.8 m/s^2 = 197 Newtons, then the frictional force will be .01 * 197

Newtons = 1.97 Newtons. This frictional force will oppose the motion of the system.

For the moment assume the motion of the system to be in the positive direction. This

will result in a frictional force of -1.97 Newtons. The net force on the system is

therefore 98.98 Newtons - 98 Newtons - 1.97 Newtons = -.99 Newtons.

This net force is in the negative direction, opposite to the direction of the net

gravitational force. If the system is moving this is perfectly all right--the

frictional force being greater in magnitude than the net gravitational force, the system

can slow down.

Suppose the system is released from rest. Then we might expect that as a result of the

greater weight on the positive side it will begin accelerating in the positive

direction. However, if it moves at all the frictional force would result in a -.99

Newton net force, which would accelerate it in the negative direction and very quickly

cause motion in that direction. Of course friction can't do this--its force is always

exerted in a direction opposite to that of motion--so friction merely exerts just enough

force to keep the object from moving at all.

Friction acts as though it is quite willing to exert any force up to 1.97 Newtons to

oppose motion, and up to this limit the frictional force can be used to keep motion from

beginning.

In fact, the force that friction can exert to keep motion from beginning is usually

greater than the force it exerts to oppose motion once it is started.

STUDENT QUESTION

I understand how we got the 197 newtons, but I do not understand why we need to find the

1 percent, Is

friction always 1 percent of of the total weight of the object, or were we just to

assume from the previous problem?????? I understand why it is negative and how we get

1.97 newtons

INSTRUCTOR RESPONSE

Friction is a percent of the force 'pressing' two objects together; the percent depends

on the nature of the two surfaces (e.g., ice sliding over smooth plastic will be a small

percent, while rubber sliding over asphalt is much higher).

That percent is usually called the 'coefficient of friction', and is generally expressed

in decimal form (e.g., a 1% coefficient of friction would be .01).

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Self-critique (if necessary): OK

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Self-critique rating: 2

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Question: `q008. A cart on an incline is subject to the force of gravity. Depending

on the incline, some of the force of gravity is balanced by the incline. On a

horizontal surface, the force of gravity is completely balanced by the upward force

exerted by the incline. If the incline has a nonzero slope, the gravitational force

(i.e., the weight of the object) can be thought of as having two components, one

parallel to the incline and one perpendicular to the incline. The incline exerts a

force perpendicular to itself, and thereby balances the weight component perpendicular

to the incline. The weight component parallel to the incline is not balanced, and tends

to accelerate the object down the incline. Frictional forces tend to resist this

parallel component of the weight and reduce or eliminate the acceleration.

A complete analysis of these forces is best done using the techniques of vectors, which

will be encountered later in the course. For now you can safely assume that for small

slopes (less than .1) the component of the gravitational force parallel to the incline

is very close to the product of the slope and the weight of the object. [If you

remember your trigonometry you might note that the exact value of the parallel weight

component is the product of the weight and the sine of the angle of the incline, that

for small angles the sine of the angle is equal to the tangent of the angle, and that

the tangent of the angle of the incline is the slope. The product of slope and weight

is therefore a good approximation for small angles or small slopes.]

What therefore would be the component of the gravitational force acting parallel to an

incline with slope .07 on a cart of mass 3 kg?

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Your solution:

If I understand this correctly, it would be .07 * 3 kg = .21 (don't know what the units

will be)

confidence rating #$&*:

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Given Solution:

The gravitational force on a 3 kg object is its weight and is equal to 3 kg * 9.8 m/s^2

= 29.4 Newtons. The weight component parallel to the incline is found approximately as

(parallel weight component) = slope * weight = .07 * 29.4 Newtons = 2.1 Newtons

(approx.).. STUDENT COMMENT: It's hard to think of using the acceleration of 9.8 m/s^2

in a situation where the object is not free falling. Is weight known as a force measured

in Newtons? Once again I'm not used to using mass and weight differently. If I set a 3

kg object on a scale, it looks to me like the weight is 3 kg.INSTRUCTOR RESPONSE: kg is

commonly used as if it is a unit of force, but it's not. Mass indicates resistance to

acceleration, as in F = m a. {}{}eight is the force exerted by gravity. {}{}The weight

of a given object changes as you move away from Earth and as you move into the proximity

of other planets, stars, galaxies, etc.. As long as the object remains intact its mass

remains the same, meaning it will require the same net force to give it a specified

acceleration wherever it is.{}{}An object in free fall is subjected to the force of

gravity and accelerates at 9.8 m/s^2. This tells us how much force gravity exerts on a

given mass: F = mass * accel = mass * 9.8 m/s^2. This is the weight of the object. **

STUDENT NOTE:

i didn't think to use the acceleration of gravity for this one, it said the object was

paralell

INSTRUCTOR RESPONSE:

The situation talks about the weight having two components, one being parallel to the

incline, and the instruction tells you how to find that parallel component when the

slope of the incline is small.

We know from experience that an object will pick up speed along the incline, as opposed

to the direction perpendicular to the incline (to move in the perpendicular direction

the object would have to leave the incline, either burrowing down into the incline or

levitating up off the incline).

The direction along the incline is parallel to the incline. So its acceleration is

parallel to the incline, and the net force must be parallel to the incline.

In the absence of other forces, only gravity has a component parallel to the incline.

Therefore in this ideal case the gravitational component parallel to the incline is the

net force. In reality there are other forces present (e.g., friction) but the parallel

gravitational component is nevertheless present, and contributes to the net force in the

direction of motion.

STUDENT QUESTION

If we are finding the force at 29.4 newtons is that the perpendicular, and the slope *

weight is parallel, why is

this considered the weight instead of force in the calculation, is this because the

weight is the force that is moving the

object

INSTRUCTOR RESPONSE

The weight of an object is the force exerted on it by gravity.

Since objects near the surface of the Earth accelerates downward, if free to do so, at

9.8 m/s^2, the weight of an object is 9.8 m/s^2 multiplied by its mass.

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Self-critique (if necessary):

I interpreted it as slope times weight. Now I know that it's weight times gravitational

pull. That product times the slope. So, 2.1 Newtons is the force of gravity on this

cart?

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Self-critique rating: 2

@&

Gravity exerts about 30 Newtons of force on the 3 kg cart, obtained by multiplying the cart's mass by the acceleration of gravity.

The gravitational force is mostly perpendicular to the incline, but a small component of the force is parallel to the incline.

In this case .07 of the gravitational force is parallel to the incline and tends to accelerate the mass down the incline.

We call that the 'parallel component' of the gravitational force, and it is equal to approximately .07 * 30 Newtons = 2.1 Newtons.

*@

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Question: `q009. What will be the acceleration of the cart in the previous example,

assuming that it is free to accelerate down the incline and that frictional forces are

negligible?

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Your solution:

a * m = Fnet

a = Fnet / m

a = 2.1 Newtons (?) / 3 kg

a = .7 m/s^2

confidence rating #$&*:

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Given Solution:

The weight component perpendicular to the incline is balanced by the perpendicular force

exerted by the incline. The only remaining force is the parallel component of the

weight, which is therefore the net force. The acceleration will therefore be a = F / m

= 2.1 Newtons / (3 kg) = .7 m/s^2.

STUDENT QUESTION

I’m not sure where the perpendicular weight component and perpendicular force came from.

Can you explain this, and how that leaves only the parallel component of the weight?

INSTRUCTOR RESPONSE

If you stand on a level floor, the floor exerts an upward force equal and opposite your

weight, which supports you.

If the floor was tilted the force it exerts wouldn't be directly upward, but rather

tilted with respect to vertical. The floor can only exert a force perpendicular to

itself. This force exerted by the floor is called the normal force.

Of course if the floor isn't too steep and/or too slippery, you won't slide downward,

and it will seem that the floor is exerting an upward force to counter your weight.

However the floor only exerts the force perpendicular to itself, called the normal

force. That force doesn't keep you from sliding downward.

The force that prevents you from sliding downward is the frictional force between you

and the floor. The normal force between you and the floor tends to push you and the

floor together, and this is where the frictional force comes from.

You might object that it's the floor that exerts the frictional force, so that it really

is the floor that's holding you up.

Even if I were to accept your objection, I would insist on separating the normal force

exerted by the floor from the frictional force. The normal force and the frictional

force have different sources and are different kinds of forces.

Now in relation to this problem, the normal force counters the component of the

gravitational force which is perpendicular to the incline. So there's zero net force

perpendicular to the incline, as you can tell since you don't accelerate down into the

incline or up off the incline.

The gravitational force isn't perpendicular to the incline, though. It has a component

parallel to the incline, and the normal force doesn't act in that direction. Friction

can act parallel to the incline, and will do so if it is present, but if there's no

friction the component of the gravitational force parallel to the incline accelerates

you down the incline.

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Self-critique (if necessary): OK

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Self-critique rating: 2

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Question: `q010. What would be the acceleration of the cart in the previous example if

friction exerted a force equal to 2% of the weight of the cart, assuming that the cart

is moving down the incline? [Note that friction is in fact a percent of the

perpendicular force exerted by the incline; however for small slopes the perpendicular

force is very close to the total weight of the object].

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Your solution:

3 kg * 9.8 m/s^2 = 29.4 Newtons

2% of 29.4 Newtons = .59 Newtons

29.4 Newtons * .07 = 2.1 Newtons

Fnet = 2.1 Newtons - .49 Newtons

Fnet = 1.51 Newtons

a * m = Fnet

a = Fnet / m

a = 1.51 Newtons/3 kg

a = .5 m/s^2

confidence rating #$&*:

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Given Solution:

The weight of the cart was found to be 29.4 Newtons. The frictional force will

therefore be .02 * 29.4 Newtons = .59 Newtons approx.. This frictional force will

oppose the motion of the cart, which is down the incline.

If the downward direction along the incline is taken as positive, the frictional force

will be negative and the 2.1 Newton parallel component of the weight will be positive.

The net force on the object will therefore be

net force = 2.1 Newtons - .59 Newtons = 1.5 Newtons (approx.).

This will result in an acceleration of

a = Fnet / m = 1.5 Newtons / 3 kg = .5 m/s^2.

STUDENT QUESTION (instructor comments in bold)

ok, so the weight is equal to the force that is making the onject go down the incline

(because force and mass are equal

here????)

the force is equal to the mass times the acceleration of gravity

the mass is 3 kg, the weight is 29.4 Newtons

the mass or weight was 29.4newtons. 2 percent of this is .02 * 29.4 = .59 newtons which

is the frictional force, and that should

be -.59 since friction works in the opposite direction of the system. To find the net

force of the system with add -.59 + 2.1Newtons

= 1.5 newtons

So the mass of the systme = 3kg

a= f/m = 1.5/3kg = .5m/s^2

I am a little confused, I am not sure about the weight (being the gravity 9.8m/s^2 * the

mass 3kg= 29.4newtons) is related to

the force of the system slope .07 * weight 29.4?????

That goes back to the statement in the preceding problem, about how the gravitational

force splits into two components, one being parallel to the incline and equal to slope *

weight (provided slope is small).

We will see soon, in terms of vectors, why this is so, and also how to deal with the

situation where the slope isn't small.

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Self-critique (if necessary): OK

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Self-critique rating: 2

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Question: `q011. Given the conditions of the previous question, what would be the

acceleration of the cart if it was moving up the incline?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I am getting very confused with this.

I am assuming that the force exerted by gravity on the 3 kg cart is still 29.4 Newtons.

I am also assuming that the friction is .59 Newtons. Going up the slope instead of down

the slope should increase the amount of energy needed to get the cart up the hill. If I

make the slope negative, then I get a bunch of negative numbers, including a negative

acceleration. This does not make sense.

confidence rating #$&*:

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Given Solution:

In this case the frictional force would still have magnitude .59 Newtons, but would be

directed opposite to the motion, or down the incline. If the direction down the incline

is still taken as positive, the net force must be

net force = 2.1 Newtons + .59 Newtons = 2.7 Newtons (approx).

The cart would then have acceleration

a = Fnet / m = 2.7 Newtons / 3 kg = .9 m/s^2.

STUDENT QUESTION

?????????????????????????????????if the cart is moving downhill and this is considered

in the positive direction parallel to

the weight of the object which is also considered to be positve, why if it is going in

the opposite direction would that not

be considered negative since the systme is moving against gravity and the force of it

own weight ( which made it go down the

incline in the previous problem) ??????????????????????????

INSTRUCTOR RESPONSE

The direction of gravity does not determine the positive direction; the positive

direction is simply declared in the solution, and you would be free to use either

direction as positive.

Once the positive direction is declared, all forces, displacements, velocities and

accelerations will be positive or negative depending on whether they are in the positive

direction, or opposite to it.

The positive direction as chosen in the given solution is down the incline. The

displacement is up the incline, as is the velocity, so both displacement and velocity

are negative.

The frictional force and the component of the gravitational force along the incline are

both positive, according to the choice of positive direction.

So the object has a negative velocity and a positive acceleration, meaning that it is

moving in the negative direction but slowing.

It takes some thinking to get used to this idea; the idea is far from trivial.

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Self-critique (if necessary):

In addition to having negative numbers, I had multiplied instead of divided, so I was

really wrong. I have written down the correct answer and will continue working on this

business with gravity and friction.

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Self-critique rating: 2

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Question: `q012. Assuming a very long incline, describe the motion of a cart which is

given an initial velocity up the incline from a point a few meters up from the lower end

of the incline. Be sure to include a description of any acceleration experienced by the

cart.

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Your solution:

The cart would have an initial velocity to get it started and would gradually slow as it

moved up the hill, unless there was a force to keep it moving. The force that would be

working on it would be gravity at the rate of 9.8 m/s^2.

confidence rating #$&*:

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Given Solution:

The cart begins with a velocity up the incline, which we still taken to be the negative

direction, and an acceleration of +.9 m/s^2. This positive acceleration tends to slow

the cart while it is moving in the negative direction, and the cart slows by .9 m/s

every second it spends moving up the incline.

Eventually its velocity will be 0 for an instant, immediately after which it begins

moving down the incline as result of the acceleration provided by the weight component

parallel to the incline.

As soon as it starts moving down the incline its acceleration decreases to +.5 m/s^2,

but since the acceleration and velocity are now parallel the cart speeds up, increasing

its velocity by .5 m/s every second, until it reaches the lower end of the incline.

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Self-critique (if necessary):

I got part of it. I did not talk about going up the hill being considered a negative

direction. I am not sure where the .9 m/s^2 (last problem?) or .5 m/s^2 came from. I

stopped my description when the motion stopped, not taking it back down the incline. I

am still working on understanding phrases like ""acceleration provided by the weight

component parallel to the incline"" and ""since the acceleration and velocity are parallel

the cart speeds up, increasing its velocity by .5 m/s.""

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Self-critique rating: 2"

@&

Without friction the acceleration of the cart would be 2.1 N / 3 kg = .7 m/s^2, obtained from the cart's mass and the 2.1 N parallel component of the frictional force. That component is always present and acts down the incline.

Going up, the frictional force is directed down the incline, so both the frictional force and the parallel component of the gravitational force act in the same direction. This is how you end up with acceleration .9 m/s^2 down the incline.

Going down, the frictional force is directed up the incline, and the result is acceleration .5 m/s^2.

*@

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

#*&!

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