PHY 201 QA20

#$&*

course PHY 201

20.  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?

 

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

 

 Weight of block = 5 kg * 9.8 m/s^2 = 49 N, the Normal force =-49 N. The net_force of this system is = 49N - 49N = 0

The weight of the hanging block is 2 kg * 9.8 m/s^2 = 19.6 N. There are no other forces acting on this block other than gravity.

Force = m * a, a = Force / mass = 19.6 N / (5 kg + 2 kg) = 2.8 m/s^2. The total mass of the system is the 5 kg block on the tabletop and the 2 kg block hanging on the string. These two masses are tied to together to create a single system, therefore, the total mass must include both of these.

 

 

confidence rating #$&*:#8232; 

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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):

 

 

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Self-critique rating: 

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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.

 

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

net_force = Fweight1 - Fn + Fweight2 - Ffriction,

Fweight1 = 49 N, Fn = 49 N, Fweight2 = 19.6 N, Ffrict = 49 N * 0.10 = 4.9 N

net_force = 49 N - 49 N + 19.6 N - 4.9 N = 14.7 N

F = m * a, a = F / m = 14.7 N / (7 kg) = 2.1 m/s^2

 

confidence rating #$&*:8232; 

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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):

 

 

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Self-critique rating: 

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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.

 

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

x component = 49 N cos (270 + 12) = 10.2 N

y component = 49 N sin (270 + 12) = -47.9 N

The elastic supporting friction of the incline is equal and opposite of the y component (perpendicular force), 47.9 N

net_force = Fweight1 + F// - Fperpendicular + Fweight2 - Ffriction

net_force = 49 N + 10.2 N - 47.9 N + 19.6 N - 4.9 N = 26 N

a = F / m = 26 N / 7 kg = 3.7 m/s^2

 

 

confidence rating #$&*:8232; 

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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 &&&&).

1. 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.

2. The negative y axis of this system is at 270 degrees as measured counterclockwise from the positive x axis.

3. 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.

4. 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.

5. The gravitational force is no longer at 270 degrees, since the rotation of the system has rotated the y axis away from vertical.

6. 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: 

 

 m1 = 80 kg, theta = 15

weight = 80 kg * 9.8 m/s^2 = 784 N

x-component = 784 N cos 15 = 757 N (parallel weight)

y-component = 784 N sin 255 = 203 N (perpendicular weight)

Frictional force will tend to oppose the weight sliding down the incline. Up to a certain angle of incline, the frictional force can be anything up to and including some multiple of the normal force.

For example the frictional force for certain incline and a certain object could be anything up to a limit of 15% of the normal force; using f for the magnitude of the frictional force and N for the magnitude of the normal force we would then say that f <= .15 N (the <= stands for ' less than or equal to').

In general we use the Greek letter `mu to stand for the limiting proportion of the normal force which is available as friction.

We thus write f <= `mu N (no how the Greek letter `mu is written in the figure below).

`mu is called the coefficient of friction for the object and the incline.

 

The normal force is the perpendicular force, therefore, F <= .15(203 N) = 30.5 N

Net_force = W + W// + Wperpendicular + Fn - Ffrict = 784 N + 757 N + 203 N - 203 N - 30.5 N = 1511 N

a = F / m= 1511 N / 80 kg = 19 m/s^2

 

Calculating the frictional force on the incline is a little vague to me, didn't really understand fully what the class notes was trying to tell me.

 

 

confidence rating #$&*:

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Self-critique (if necessary):

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Self-critique rating:

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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.

 

 

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

x component = 49 N cos (270 + 12) = 10.2 N

y component = 49 N sin (270 + 12) = -47.9 N

The elastic supporting friction of the incline is equal and opposite of the y component (perpendicular force), 47.9 N

net_force = Fweight1 + F// - Fperpendicular + Fweight2 - Ffriction

net_force = 49 N + 10.2 N - 47.9 N + 19.6 N - 4.9 N = 26 N

a = F / m = 26 N / 7 kg = 3.7 m/s^2

 

 

confidence rating #$&*:8232; 

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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 &&&&).

1. 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.

2. The negative y axis of this system is at 270 degrees as measured counterclockwise from the positive x axis.

3. 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.

4. 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.

5. The gravitational force is no longer at 270 degrees, since the rotation of the system has rotated the y axis away from vertical.

6. 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 

 

 m1 = 80 kg, theta = 15

weight = 80 kg * 9.8 m/s^2 = 784 N

x-component = 784 N cos 15 = 757 N (parallel weight)

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Your components don't fit the situation. The weight vector is much closer to the y axis than to the x axis, so the magnitude of its y component will clearly be greater than that of its x component.

The incline is only 15 degrees above horizontal, the weight 784 N. The component of the weight vector along the incline will be much less than the weight, so for this reason as well a 757 N weight component parallel to the incline does not correspond to the situation.

The weight vector makes angle 255 degrees as measured counterclockwise from the positive x axis. So the x component of this vector is found by multiplying the magnitude of the vector, which is in this case the weight, by the cosine of the 255 degree angle.

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y-component = 784 N sin 255 = 203 N (perpendicular weight)

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784 N * sin(255 deg) is the right calculation, but this result is not equal to 203 N.

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Frictional force will tend to oppose the weight sliding down the incline. Up to a certain angle of incline, the frictional force can be anything up to and including some multiple of the normal force.

For example the frictional force for certain incline and a certain object could be anything up to a limit of 15% of the normal force; using f for the magnitude of the frictional force and N for the magnitude of the normal force we would then say that f <= .15 N (the <= stands for ' less than or equal to').

In general we use the Greek letter `mu to stand for the limiting proportion of the normal force which is available as friction.

We thus write f <= `mu N (no how the Greek letter `mu is written in the figure below).

`mu is called the coefficient of friction for the object and the incline.

 

The normal force is the perpendicular force, therefore, F <= .15(203 N) = 30.5 N

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GIven the correct y component of the weight vector, this calculation would yield the correct result.

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Net_force = W + W// + Wperpendicular + Fn - Ffrict = 784 N + 757 N + 203 N - 203 N - 30.5 N = 1511 N

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You can use regular addition to add vectors that act along the same line, but you can't do so with vectors that do not.

If you know the x and y components of the net force, you can use the Pythagorean Theorem and the arctangent to find the magnitude and angle of the resultant vector.

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a = F / m= 1511 N / 80 kg = 19 m/s^2

 

Calculating the frictional force on the incline is a little vague to me, didn't really understand fully what the class notes was trying to tell me.

 

 

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Overall your thinking isn't bad, but you are making errors in detail.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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