QA 20

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course PHY 121

7/23 2

020. Forces (inclines, friction)

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Question: `q001. Note that this assignment contains 4 questions.

. A 5 kg block rests on a tabletop. A string runs horizontally from the block over a

pulley of negligible mass and with negligible friction at the edge of the table. There

is a 2 kg block hanging from the string. If there is no friction between the block in

the tabletop, what should be the acceleration of the system after its release?

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Your solution:

2 kg * 9.8 m/s = 19.6 N

Total mass = 5 kg + 2 kg = 7 kg

Acceleration of the system: 19.6 N / 7 kg = 2.8 m/s^2 in the direction of the 2 kg mass

confidence rating #$&*:

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Given Solution:

Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but

presumably the tabletop is strong enough to support the block and so exerts exactly

enough force, 49 Newtons upward, to support the block. The total of this supporting

force and the gravitational force is zero.

The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by

any force acting on the two mass system, so we have a system of total mass 7 kg subject

to a net force of 19.6 Newtons.

The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8

meters/second ^ 2.

STUDENT QUESTION

since it is a frictionless system i thought the answer would the acceleration of gravity

since that would be the force pulling the objects down.

INSTRUCTOR RESPONSE

A mass resting by itself on a level tabletop does now accelerate downward in response to

the gravitational force exerted on it. This is because the tabletop pushes up on it.

This occurs whether friction is high, low or absent. Of course if friction is very low,

you have to be sure the tabletop is very nearly level, and if friction is absent you'd

best be sure it is completely level.

These forces are still present if you add the hanging mass to the system.

The mass on the tabletop does experience the downward pull of gravity, but that force is

balanced by the supporting force of the tabletop. The mass therefore does not move in

the direction of the gravitational pull. It is nevertheless part of the system being

accelerated by the gravitational force on the hanging mass.

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `q002. Answer the same question as that of the previous problem, except this

time take into account friction between the block in the tabletop, which exerts a force

opposed to motion which is .10 times as great as the force between the tabletop and the

block. Assume that the system slides in the direction in which it is accelerated by

gravity.

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Your solution:

You still have 2 kg * 9.8 m/s = 19.6 Newtons of force

The total mass is still 5 kg + 2 kg = 7 kg

The force for the mass on the table is 5 kg * 9.8 m/s^2 = 49 Newtons

Since the friction is working on this mass, 10% of 49 Newtons is 4.9 Newtons

Net force will be 19.6 N - 4.9 N = 14.7 N

14.7 N / 7 kg = 2.1 m/s^2

confidence rating #$&*:

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Given Solution:

Again the weight of the object is exactly balance by the upward force of the table on

the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of

.10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of

the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted

by gravity on the 2 kg object.

The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The

system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^

2.

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `q003. Answer the same question as that of the preceding problem, but this

time assume that the 5 kg object is not on a level tabletop but on an incline at an

angle of 12 degrees, and with the incline descending in the direction of the pulley.

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Your solution:

5 kg * 9.8 m/s = 49 N

2 kg * 9.8 m/s = 19.6 N

Total mass = 5 kg + 2 kg = 7 kg

49 N * .12 = 5.88 N

Friction = 4.9 N

Net force = 19.6 N - 4.9 N + 5.88 N = 20.58 N

20.58 N / 7 kg = 2.94 m/s^2

confidence rating #$&*:

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Given Solution:

In this case you should have drawn the incline with the x axis pointing down the incline

and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees

below horizontal. As a result the weight vector, rather than being directed along the

negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12

degrees with the negative y axis. So the weight vector makes an angle of 270 degrees +

12 degrees = 282 degrees with the positive x axis.

The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons,

therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately.

Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately.

The incline exerts sufficient force that the net y component of the force on the block

is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force

which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction

of motion. Assuming that the direction of motion is down the incline, frictional force

is therefore -4.8 Newtons in the x direction.

The weight component in the x direction and the frictional force in this direction

therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to

accelerate the system in the same direction as does the weight of the 2 kg mass. This

results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system.

The system therefore accelerates at rate

a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately.

STUDENT COMMENT:

i get confused every time about which angle to use and how to determine it, where did

270 deg come from, i used the right equations and was on to the right answer but used

the wrong numbers in the equations, but i had the right idea

self critique rating:

1

INSTRUCTOR RESPONSE:

The negative y axis lies at 270 deg from the positive x axis, as measured

counterclockwise.

When the coordinate system is rotated 12 degrees in the manner described, the weight

vector stays where it is, but the y negative axis swings 'back' 12 degrees and the angle

of the weight vector becomes 282 degrees.

MORE EXTENSIVE EXPLANATION

The weight is in the downward vertical direction, which matches the direction of the

original vertical-horizontal coordinate system. So in the original system the weight

vector is at 270 degrees. However the positive x axis of the original coordinate system

doesn't match the direction of motion along the incline.

It's generally simpler to have the x axis parallel to the direction of motion. To

accomplish this we rotate the coordinate system 12 degrees in the clockwise direction.

As the coordinate system is rotated, the positive x axis rotates to an orientation 12

degrees below horizontal, and the negative y axis 'swings out' 12 degrees from its

original vertical orientation. This leaves the vertical weight vector in the fourth

quadrant, 12 degrees from the newly oriented negative y axis. As measure

counterclockwise from the positive x axis, the weight vector is now at angle 282

degrees.

The first figure below depicts a weight vector with its initial point at the origin of

an x-y coordinate system in standard vertical-horizontal orientation. It should be

clear that the vector is at angle 270 degrees, as measured counterclockwise from the

positive x axis.

The second figure shows the same weight vector, which is still vertical, but with the

coordinate system rotated so that the positive x axis is directed down a 12-degree

incline. It should be clear that negative y axis will have rotated 12 degrees from its

original position so that the weight vector is now at 282 degrees.

STUDENT QUESTION

I have read both of the explanations I am still not sure where the 270 degrees is coming

from, how do we calculate this

INSTRUCTOR RESPONSE

The above discussion is fairly long, and involves an idea that is difficult for many

students.

The following should provide a convenient summary of the key ideas.

If this point-by-point summary doesn't clarify everything for you, please submit a copy

of these questions along with inserted comments, questions, etc. (insertions should be

marked with &&&&).

If the x-y coordinate system is in its 'standard orientation', with the x axis

horizontal, pointing to the right, and the y axis vertical and upward, then gravity

pulls straight down, along the negative y axis.

The negative y axis of this system is at 270 degrees as measured counterclockwise from

the positive x axis.

If the coordinate system is rotated 12 degrees in the clockwise direction, then the x

axis now points slightly below horizontal, and the y axis is 12 degrees from vertical.

The gravitational force is still vertical, so the gravitational force is no longer in

the direction of the negative y axis. It is now directed at an angle of 12 degrees from

the negative y axis.

The gravitational force is no longer at 270 degrees, since the rotation of the system

has rotated the y axis away from vertical.

The gravitational force is now in the fourth quadrant of the rotated system, directed at

282 degrees.

STUDENT QUESTION

i am really lost on these problems involving forces and angles. i don't understand the

concepts of cosine and sin and what

the mean with relationship to force.

i look at the answer below and i sort of understand how to derive the angle but what i

am unclear about is how you draw. it

makes sense the way you have it drawn but what if you drew the incline mirrored to what

is now? wouldn't that put the force

vector in the third quandrant instead of the forth? so how do you know how to correctly

draw it.

i don't understand what the 49cosine(282) and 49 sin ( 282) actually represents. i know

that we use it becase the 49N is the

force resting on the incline but I dont understand what the other numbers do. I don't

want to just know that I am suppose to

use them .

I also don't understand how you just know that """"The incline exerts sufficient force

that the net y component of the force

on the block is zero"""" Can you please explain?

INSTRUCTOR RESPONSE

You could draw the incline in the opposite direction, with the incline going up and to

the right. The x axis would be directed up the incline. If you did this the force

vector would be in the third quadrant. The angle would be 258 deg, and the x component

of the force would therefore be negative, meaning that the x component of the

gravitational force is down the incline (since the x axis points up the incline). Of

course objects do tend to accelerate down inclines, so this is exactly what would be

expected.

A student having completed the prerequisite courses for Phy 201, 231 or 241 should have

had enough trigonometry to know what the sine and cosine mean. However it has been

observed over many years of teaching that this often not the case. As a result, the

vector analysis in this course is built around one simple model, the circular model.

This model has four rules, as presented in Introductory Problem Set 5. You should

review those rules, which were part of a previous assignment. Briefly stated the rules

say the following:

A vector of magnitude A directed at angle theta as measured counterclockwise from the

positive x axis of a right-hand coordinate system has x component A cos(theta) and y

component A sin(theta).

A vector with x component R_x and y component R_y has magnitude R = sqrt(R_x^2 + R_y^2)

and is directed at angle arcTan(R_y / R_x), plus 180 degrees if R_x is negative.

It isn't difficult to make sense of the rules, but for now we will simply apply them and

try to make sense of the results.

Applying the rules to the weight vector we obtain the x and y components in the given

solution.

Had the figure been drawn with the incline rising as we move to the right the angle of

the weight vector would be 270 deg - 12 deg = 258 deg, as discussed earlier in this

response. The components would then be

F_x = F cos(theta) = 49 N * cos(258 deg) = -10 N (note the contrast between the given

solution, in which the x axis was directed down the incline and the x component of the

force was +10 N; in both cases, however, the x component of the gravitational force is

directed down the incline)

F_y = F sin(theta) = 49 N * sin(258 deg) = -48 N (just as before the y component is

negative; in both cases the y component acts perpendicular to the incline, tending to

compress or bend the incline inward).

These results make sense. The gravitational force is quite a bit closer to

perpendicular than parallel to the incline. The component of the gravitational force

parallel to the incline is therefore quite a bit greater in magnitude than its component

parallel to the incline.

If the angle of the incline was increased a little bit the gravitational force would

still be closer to perpendicular than to parallel, but not as much as before. The

parallel component would increase and the perpendicular component would decrease.

If the incline was at 45 degrees, the gravitational force would be just as close to

parallel as to perpendicular. The magnitudes of the perpendicular and parallel

components would be equal. The angle of the gravitational force would be either 315

degrees or 225 degrees, depending on which way you sketch the incline. You can check to

see that this is consistent with the behavior of the sines and cosines: The magnitudes

of the sine and cosine of 315 degrees are equal to one another, and are also equal to

the magnitudes of the sine and cosine of 225 degrees.

As the angle of the incline approaches 90 degrees, the perpendicular component shrinks

toward zero while the parallel component approaches the magnitude of the gravitational

force. At 90 degrees the component parallel to the incline is equal to the

gravitational force, and as a result the object accelerates parallel to the incline with

an acceleration equal to that of gravity. The x axis will now be pointing straight up,

so that the gravitational force is at angle 180 degrees with respect to the x axis. You

can check consistency again: cos(180 deg) = -1 and sin(180 deg) = 0, so that F_x would

now be -49 N and F_y would be 0.

Finally in response to your question about the sentence 'The incline exerts sufficient

force that the net y component of the force on the block is zero.':

We assume that the weight doesn't collapse the incline or cause it to bend or compress

enough to significantly change its angle of elevation.

Since the weight must therefore move along the incline, it doesn't move perpendicular to

the incline.

It therefore does not accelerated in the direction perpendicular to the incline.

It follows that the net force on the weight, in the y direction, is zero.

Gravity exerts a force perpendicular to the incline (this is the 'perpendicular

component' discussed here), so gravity does exert a nonzero force in the y direction.

But the net force in the y direction is zero.

So something else is exerting a force in the y direction.

That something, in this context, is the incline.

Thus the incline exerts sufficient force that the net y component of the force on the

block is zero.

STUDENT QUESTION

I looked at given solution etc. still not sure why or how to alter the angle instead of

using just a normal 12 degree angle why shift it down and add it to 270 deg?

INSTRUCTOR RESPONSE

The gravitational force is directed downward, which with the x and y axes respectively

directed in the horizontal and vertical directions puts in along the negative y axis, at

270 degrees as measured counterclockwise from the positive x axis.

The analysis of the vectors is simplest if one of your axes is in the direction of

motion. The motion will be along the incline, so we rotate the x axis until it is

parallel parallel to the incline.

Thus you will rotate the x axis 12 degrees while keeping the gravitational force where

it is. The result, assuming the incline to be descending toward the right, will be that

the coordinate system is rotated clockwise until the x axis is oriented at 12 degrees

below the horizontal. In the process the negative y axis rotates 12 degrees clockwise

from the vertical, leaving the weight vector in the fourth quadrant at 270 deg + 12 deg

= 282 deg from the positive x axis.

If you understand the assignment and were able to solve the previously given problems

from your worksheets, you should be able to complete most of the following problems

quickly and easily. If you experience difficulty with some of these problems, you will

be given notes and we will work to resolve difficulties.

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Question: `q004. An 80 kg person sits on a 15-degree incline. On a coordinate system

whose x axis is directed parallel to the incline, rising from left to right, the vector

representing the person's weight makes an angle of 255 degrees as measured

counterclockwise from the positive x axis.

What are the x and y components of the person's weight?

To keep the person from sliding down the incline, friction has to exert a force up the

incline. How much frictional force is required?

What is the frictional force as a percent of the normal force?

If friction can exert a force only 10% as great as the normal force, the person will

slide down the incline. What will he the resulting acceleration?

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Your solution:

80 kg * 9.8 m/s^2 = 784 Newtons

255 deg + 15 deg = 270 deg

x = 784 N (cos) (270 deg) = 0

y = 784 N (sin) (270 deg) = -784

Friction - .1 * 784 N = 78.4 N

a = F/m

F = 784 N - 78.4 N = 705.6 N

a = 705.6 N / 80 kg

a = 8.82 m/s^2

I'm not sure what to do with the frictional force required to keep the person from

moving.

confidence rating #$&*:

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