Post by: atom on February 22, 2012, 03:53:56 pm
A student whirls a rubber weight on the end of a string in a horizontal clockwise circle at a constant speed of 6.0 m/s. From an instant when the stopper is moving in a northerly direction, find its change in velocity after moving around one-ninth of a turn.
Thanks in advance :)
Post by: Aurelian on February 22, 2012, 04:52:17 pm
The first instance the speed is 6.0 m/s, and the direction is northerly.
The stopper then moves around one-ninth of a turn - i.e., 20 degrees. The speed is still 6.0 m/s. We therefore have two vector quantities. Bring them together, and use trig to find the change in velocity.
I'll admit this is a little unintuitive, which is why I wont continue the question to completion in case I misplace the angle when setting up my vector diagram. Usually any question like this will be such that the two instances are at right angles to each other, and so it's simple pythagoras and relatively easy to determine; I doubt you'd see this kind of question on any VCAA endorsed assessment.
Sorry I can't conclude the question (someone with spesh knowledge might be more helpful) but I hope this points you in the right direction.
Post by: rife168 on February 22, 2012, 05:24:54 pm
Sorry if it's a bit messy.. Also, I have worked from right to left for some reason, so the diagram on the right is what I started with, then I resolved the vectors into a triangle and used the fact that
Do you know what the solution is? I could be completely wrong, but that is how I would generally go about it.
edit: Bearing should be
oops -.-
Post by: atom on February 22, 2012, 07:36:31 pm
Post by: rife168 on February 22, 2012, 08:30:46 pm
I've got the answer here, but it's a little different from yours. Look at the attached picture.
Yeah how clumsy of me! Just a mistake with my bearings -.- I don't use them very often.
That is also a much faster and efficient way of solving the problem, it is a more geometric solution than mine.
Post by: yawho on February 22, 2012, 09:48:36 pm
Post by: paulsterio on February 22, 2012, 10:05:47 pm
An inclined plane question. A 1 kg object at rest on an inclined plane. Find the magnitude and direction of the reaction force of the inclined plane on the object.
Direction is perpendicular to the plane.
I can't actually work it out, because you didn't include the angle.
Post by: Bhootnike on February 22, 2012, 10:27:18 pm
Post by: yawho on February 22, 2012, 10:49:43 pm
i hope im right haHAWhat is the answer?
Post by: Phy124 on February 22, 2012, 10:56:48 pm
To calculate the answer you need to give the angle at which the plane is inclined to the horizontal. That isi hope im right haHAWhat is the answer?
Post by: yawho on February 23, 2012, 03:45:40 pm
No angle is given in the assignment.To calculate the answer you need to give the angle at which the plane is inclined to the horizontal. That isi hope im right haHAWhat is the answer?in the equation.
Post by: paulsterio on February 23, 2012, 05:13:06 pm
An inclined plane question. A 1 kg object at rest on an inclined plane. Find the magnitude and direction of the reaction force of the inclined plane on the object.
Do you know the co-efficient of friction then? There has to be more information that what you've given us.
Post by: Bhootnike on February 23, 2012, 06:50:49 pm
i hope im right haHAWhat is the answer?
Well, you said reaction force, so From the diagram, i guess reaction force refers to the normal reaction force, which is 10sinTHETA, and you say that there is no angle so maybe it's just that?
Post by: yawho on February 23, 2012, 08:34:31 pm
Post by: rife168 on February 23, 2012, 08:41:04 pm
Another question: A 0.5 kg book is resting on a horizontal table has a weight force acting on it. Name the reaction force to this weight force in Newton's third law.Are these questions looking for a qualitative or quantitative answer? What is the nature of some other questions that are on the same page/assignment?
Post by: yawho on February 23, 2012, 08:49:53 pm
Post by: rife168 on February 23, 2012, 08:54:18 pm
Second part of the same question: Now the 0.5 kg book is falling off the table. Assuming there is no air resistance, name the reaction force to the weight force on the book in Newton's third law.
edit: oops, I was incorrect, see queendaisy's response -.-
Post by: queendaisy on February 23, 2012, 09:10:02 pm
Another question: A 0.5 kg book resting on a horizontal table has a weight force acting on it. Name the reaction force to this weight force in Newton's third law.
Action reaction forces always act on different bodies. Therefore, the gravitational force of the earth on the book (I.e this weight force you mentioned) has a reaction pair that is the gravitational force of the book on the earth: the earth is pulled towards the book as well, but it's not noticeable cos the mass of the book is so small in comparison to the mass of the earth!
Normal reaction and weight are not action reaction pairs cos they both act on the book.
In response to your second q (with book falling) the reaction force is the same as the first question. There should always be a reaction force.. Otherwise newtons third law would not work!
Post by: yawho on February 23, 2012, 09:13:16 pm
Post by: queendaisy on February 23, 2012, 09:19:51 pm
What is your opinion on the inclined plane question a few posts ago?
I agree with what has been said, you can't get a numerical answer for the magnitude of the normal force without the angle (N=mgcosx) or the coefficient of static friction (you can determine the angle from there). The direction is perpendicular to the inclined plane. Sorry I couldn't help you further, but I am pretty sure there is no way of working it out without one of those two!
Post by: yawho on February 23, 2012, 09:23:04 pm
Post by: yawho on February 23, 2012, 09:41:15 pm
Post by: paulsterio on February 23, 2012, 09:45:31 pm
I'll ask my tutor she finished 12 last year and let you know in 10 min.
Mind posting a scan of the question or a reference as to where the question could be found? Maybe you didn't pick up something crucial.
Post by: yawho on February 24, 2012, 05:06:54 pm
What is your opinion on the inclined plane question a few posts ago?
I agree with what has been said, you can't get a numerical answer for the magnitude of the normal force without the angle (N=mgcosx) or the coefficient of static friction (you can determine the angle from there). The direction is perpendicular to the inclined plane. Sorry I couldn't help you further, but I am pretty sure there is no way of working it out without one of those two!
My teacher said the reaction force of the plane on the object is 10 N vertically upward when the angle is less than 90 deg and the object is at rest.
Post by: queendaisy on February 24, 2012, 05:43:45 pm
What is your opinion on the inclined plane question a few posts ago?
I agree with what has been said, you can't get a numerical answer for the magnitude of the normal force without the angle (N=mgcosx) or the coefficient of static friction (you can determine the angle from there). The direction is perpendicular to the inclined plane. Sorry I couldn't help you further, but I am pretty sure there is no way of working it out without one of those two!
My teacher said the reaction force of the plane on the object is 10 N vertically upward when the angle is less than 90 deg and the object is at rest.
.. I guess that makes sense, but isn't that really only the vertical component of the normal force?
Post by: Phy124 on February 24, 2012, 06:18:45 pm
.. I guess that makes sense, but isn't that really only the vertical component of the normal force?I would have thought so. The actual normal reaction force is perpendicular to the plane as far as I know.
Post by: queendaisy on February 24, 2012, 08:03:22 pm
What is your opinion on the inclined plane question a few posts ago?
I agree with what has been said, you can't get a numerical answer for the magnitude of the normal force without the angle (N=mgcosx) or the coefficient of static friction (you can determine the angle from there). The direction is perpendicular to the inclined plane. Sorry I couldn't help you further, but I am pretty sure there is no way of working it out without one of those two!
My teacher said the reaction force of the plane on the object is 10 N vertically upward when the angle is less than 90 deg and the object is at rest.
.. I guess that makes sense, but isn't that really only the vertical component of the normal force?
Actually no, I don't think that works, because that doesn't take into account the friction, which there obviously must be if the object is at rest. If you are going to resolve the forces into vertical and horizontal (which would be must more time consuming than resolving along and perpendicular to plane btw) then there is a vertical component to the friction, so vertical comp. of normal cannot equal weight.
Post by: paulsterio on February 26, 2012, 10:10:04 pm
Look at your cos graph, it is greatest when theta = 0 and will gradually decrease to 0 when theta = 90 degrees.
Post by: yawho on February 27, 2012, 08:42:00 am
How is the reaction force 10N? Its mass is 1kg, giving it a weight of 10N (g = 10N/kg), if it is on an incline, the reaction force will be LESS than 10N. As the incline gets greater (i.e. higher angle) the reaction force will decrease. Think about it logically. As the slope gets higher, less force is pushing upward perpendicular to the surface. You can also look at it mathematically.My teacher is smart. She is right 99.9...% time. May be you got the wrong idea.
Look at your cos graph, it is greatest when theta = 0 and will gradually decrease to 0 when theta = 90 degrees.
Post by: paulsterio on February 27, 2012, 10:17:57 am
I can assure you, there's a misunderstanding somewhere along the lines. I guarantee you that the Normal Reaction Force is less than the Weight force.
Post by: yawho on February 27, 2012, 02:02:21 pm
Maybe she's got the wrong idea.Some one must be wrong. I tend to side with my teacher. She answered the question but you didn't.
I can assure you, there's a misunderstanding somewhere along the lines. I guarantee you that the Normal Reaction Force is less than the Weight force.
Post by: rife168 on February 27, 2012, 05:07:02 pm
Maybe she's got the wrong idea.Some one must be wrong. I tend to side with my teacher. She answered the question but you didn't.
I can assure you, there's a misunderstanding somewhere along the lines. I guarantee you that the Normal Reaction Force is less than the Weight force.
I agree with Paul, the magnitude of the normal force on an inclined plane varies with the angle of the slope. If it was just 10N then the object wouldn't ever slide down the slope.
I think that your teacher may be referring to the reaction force due to gravity. The Earth pulls on the object with a force of 10N, and in return the object pulls back on the earth with a force of 10 N in the opposite direction.
Either that or the plane and the object have a very high coefficient of friction, so the object does not slide down the plane. If it is at rest, then the net force must be zero, which means that there must be an opposing force of 10N, though this wouldn't be the normal reaction force, it would be the normal reaction + resistance force.
I think your teacher may have misunderstood your question or you have misunderstood her answer, or maybe you missed a bit of information in the original question.
Post by: paulsterio on February 27, 2012, 06:03:18 pm
Some one must be wrong. I tend to side with my teacher. She answered the question but you didn't.No, I'm not saying that someone is wrong, I'm saying that there's a misunderstanding, are you sure you wrote the question correctly and are you sure she saw the question correctly...etc. there's a lot of things which can go wrong in between, I'm not saying anyone's right or wrong, I'm saying we're obviously not all on the same page.
Yes, I agree with fletch on this, however, I'd also like to add this to clarify a little.
You mentioned originally that the object is stationary on the plane. This means that the net force is equal to 0. Hence, if you resolve the Weight force into two components, one parallel to the slope and one perpendicular, the friction force will cancel out the one parallel to the slope, hence it won't move down the incline. However, the one perpendicular to the slope is the Normal Reaction. This is what the Normal force is defined as, the force perpendicular to the slope. It will vary depending on the size of the incline. The larger the incline is, the greater the force down the slope, the less the normal reaction. Since they have to add to give the weight force, they will always be inversely proportional. A decrease in the force down the slope will cause an increase in the normal reaction and vice versa.
The maximum Normal Reaction force possible is 10N, which occurs when there is no incline, it is flat. The minimum Normal Reaction is 0N, which occurs when the object is in freefall, i.e. when the slope is 90 degrees and hence the object just falls vertically downwards.
There is no reason to assume that the Normal Reaction force is 10N. That's all I'm saying.
Post by: yawho on February 27, 2012, 06:37:41 pm
My teacher's explanation: The weight of the object is 10 N downward, so it exerts a force of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.Maybe she's got the wrong idea.Some one must be wrong. I tend to side with my teacher. She answered the question but you didn't.
I can assure you, there's a misunderstanding somewhere along the lines. I guarantee you that the Normal Reaction Force is less than the Weight force.
I agree with Paul, the magnitude of the normal force on an inclined plane varies with the angle of the slope. If it was just 10N then the object wouldn't ever slide down the slope.
I think that your teacher may be referring to the reaction force due to gravity. The Earth pulls on the object with a force of 10N, and in return the object pulls back on the earth with a force of 10 N in the opposite direction.
Either that or the plane and the object have a very high coefficient of friction, so the object does not slide down the plane. If it is at rest, then the net force must be zero, which means that there must be an opposing force of 10N, though this wouldn't be the normal reaction force, it would be the normal reaction + resistance force.
I think your teacher may have misunderstood your question or you have misunderstood her answer, or maybe you missed a bit of information in the original question.
Post by: Bhootnike on February 27, 2012, 07:13:02 pm
Quote
The weight of the object is 10 N downward, so it exerts aforce of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.
the bit where you say " so it exerts a force of 10N on the plane downward" , sounds odd
because if you picture the incline,
/|
* / |
/ |
/___|
^ sort of haha, but yeah, i imagined that the force that the object exerts on the INCLINED plane be 10cosTHETA.
i dont see what;d you gain from finding out the reaction force that your teacher has mentioned, i associated reaction with the normal reaction mgsintheta.
weirdd question
Post by: yawho on February 27, 2012, 08:54:11 pm
My guess is the question tests your understanding of the action force of the object on the plane and the reaction force of the plane on the object in newton's 3 law.QuoteThe weight of the object is 10 N downward, so it exerts aforce of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.
the bit where you say " so it exerts a force of 10N on the plane downward" , sounds odd
because if you picture the incline,
/|
* / |
/ |
/___|
^ sort of haha, but yeah, i imagined that the force that the object exerts on the INCLINED plane be 10sinTHETA.
i dont see what;d you gain from finding out the reaction force that your teacher has mentioned, i associated reaction with the normal reaction mgsintheta.
weirdd question
Post by: paulsterio on February 27, 2012, 10:25:29 pm
My teacher's explanation: The weight of the object is 10 N downward, so it exerts a force of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.
Firstly, Bhootnike, N = mgcos(theta) not mgsin(theta) :P
Secondly, I can see where you're coming from yawho but I agree with Bhootnike, it's a weird and very ambiguous question in that why would it want you to find the force that is acting vertically upwards?
Post by: Bhootnike on February 27, 2012, 10:29:24 pm
My teacher's explanation: The weight of the object is 10 N downward, so it exerts a force of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.
Firstly, Bhootnike, N = mgcos(theta) not mgsin(theta) :P
Secondly, I can see where you're coming from yawho but I agree with Bhootnike, it's a weird and very ambiguous question in that why would it want you to find the force that is acting vertically upwards?
woops! haha
my bad, but yeahh, you guys know what i meant :p
Post by: yawho on February 28, 2012, 10:31:44 pm
Why did you say it's a weird and very ambiguous question?My teacher's explanation: The weight of the object is 10 N downward, so it exerts a force of 10 N on the plane downward, newton's third law, the plane exerts a force of 10 N on the object upward. The question asked for reaction force, not the normal component of the reaction force.
Firstly, Bhootnike, N = mgcos(theta) not mgsin(theta) :P
Secondly, I can see where you're coming from yawho but I agree with Bhootnike, it's a weird and very ambiguous question in that why would it want you to find the force that is acting vertically upwards?
Post by: Bhootnike on February 28, 2012, 10:42:06 pm
ambiguous since reaction force is usually associated with the normal reaction force.
thats why we were all asking if there was an angle :)
weird because its not something you'd usually wanna work out. like in practical sense, what would you gain. by having an angle present and thus figuring out normal, you can figure out the component of the weight force of the object acting on the plan. and those types r the questions ive seen ! but yeah as you said, if its testing your understanding , i guess its legit!
p.d.s.n.o.a.f.y.v.
Post by: paulsterio on February 28, 2012, 10:46:39 pm
I agree with Bhootnike, it's a little random, like you're not even figuring out anything. The fact that it's called the "Reaction Force" is also ambiguous, it should have been the force upwards or something like that which is more correct.
Post by: yawho on February 28, 2012, 10:52:57 pm
my view -In general, reaction force is not usually associated with normal reaction force in N3L
ambiguous since reaction force is usually associated with the normal reaction force.
thats why we were all asking if there was an angle :)
weird because its not something you'd usually wanna work out. like in practical sense, what would you gain. by having an angle present and thus figuring out normal, you can figure out the component of the weight force of the object acting on the plan. and those types r the questions ive seen ! but yeah as you said, if its testing your understanding , i guess its legit!
p.d.s.n.o.a.f.y.v.
The 10N upward reaction force on the object explains in a simple way why the object is at rest on the plane.
Post by: queendaisy on February 28, 2012, 10:57:44 pm
Friction explains why the object is at rest on an inclined plane.
But yes, this question is silly and not worded very well, and took way too long to answer haha :P
Post by: yawho on February 28, 2012, 10:59:29 pm
Can we stop arguing about this question?The force upwards? Where is this force upwards coming from?
I agree with Bhootnike, it's a little random, like you're not even figuring out anything. The fact that it's called the "Reaction Force" is also ambiguous, it should have been the force upwards or something like that which is more correct.
Post by: yawho on February 28, 2012, 11:01:58 pm
Actually no the 10N upward reaction force does not explain why the object is at rest because this upward force acts on the earth, not on the object... Action-reaction pairs always always always act on different bodies. Therefore you can't add the the pairs and say that because they add to zero, the object is at rest.. because they act on different bodies!It does explain why the object is at rest. The sum of this upward reaction force and the downward weight force is zero.
Friction explains why the object is at rest on an inclined plane.
But yes, this question is silly and not worded very well, and took way too long to answer haha :P
Post by: paulsterio on February 28, 2012, 11:07:14 pm
It does explain why the object is at rest. The sum of this upward reaction force and the downward weight force is zero.
the reaction force is not upwards?
and it is at rest due to friction which acts parallel to the slope up the slope. this is equal to the component of the weight force which is acting down the slope?
The force upwards? Where is this force upwards coming from?
the ground pushes up on the block
Post by: yawho on February 28, 2012, 11:11:50 pm
What are the forces from the plane acting on the object?It does explain why the object is at rest. The sum of this upward reaction force and the downward weight force is zero.
the reaction force is not upwards?
and it is at rest due to friction which acts parallel to the slope up the slope. this is equal to the component of the weight force which is acting down the slope?The force upwards? Where is this force upwards coming from?
the ground pushes up on the block
the ground pushes up on the block? Did you mean the plane? If that was what you meant, it is not called reaction force of the plane on the object?
Post by: paulsterio on February 28, 2012, 11:30:05 pm
The plane is pushing up on the object because the object is pushing down on the plane. The force which the plane acts on the object is called the reaction force?
What are you talking about?
Post by: yawho on February 29, 2012, 07:29:45 am
This discussion is going nowhere.Are you now saying the plane pushes the object with a upward force of 10 N? If that is the case, then it is called the reaction force of the plane on the object in N3L because the object pushes the plane with a downward force of 10 N. This was what my teacher said. But you said the reaction force is the normal force and not the 10 N upward force on the object.
The plane is pushing up on the object because the object is pushing down on the plane. The force which the plane acts on the object is called the reaction force?
What are you talking about?
Post by: thushan on February 29, 2012, 10:04:35 am
Anyways, to answer that inclined plane question:
The 1 kg object on the inclined plane is AT REST. This means that the resultant force on the object is 0 N, via F = ma (Newton's 2nd Law).
There is the gravitational force from the Earth ON the object of 10 N going vertically down.
There is the frictional force on the object going up the plane.
There is the NORMAL reaction force on the object going perpendicular to the direction of the plane. Btw - normal = perpendicular.
What your teacher might have meant is that the frictional force is a reaction force of the plane on the block in a way because friction arises from the rubbing of two rough surfaces where part of the object gets 'caught' within a microscopic cavity of the plane and pushes at it, and the plane pushes back up the object (reaction force) - and that constitutes friction. So the total reaction force is the sum of the normal reaction force + the frictional force (which one can argue is a reaction force too) = 10 N upwards - because the sum of those two forces must be equal and opposite to the gravitational force downwards (10 N) as the resultant force on object is 0 N.
Post by: yawho on February 29, 2012, 04:45:41 pm
Yawho - I'm sure you don't mean it, but you're coming across as a tad aggressive - so just watch your tone :PI don't think I was aggressive in pointing out the inconsistency. Was my teacher right in saying the reaction force of the plane on the object is 10 N upward?
Anyways, to answer that inclined plane question:
The 1 kg object on the inclined plane is AT REST. This means that the resultant force on the object is 0 N, via F = ma (Newton's 2nd Law).
There is the gravitational force from the Earth ON the object of 10 N going vertically down.
There is the frictional force on the object going up the plane.
There is the NORMAL reaction force on the object going perpendicular to the direction of the plane. Btw - normal = perpendicular.
What your teacher might have meant is that the frictional force is a reaction force of the plane on the block in a way because friction arises from the rubbing of two rough surfaces where part of the object gets 'caught' within a microscopic cavity of the plane and pushes at it, and the plane pushes back up the object (reaction force) - and that constitutes friction. So the total reaction force is the sum of the normal reaction force + the frictional force (which one can argue is a reaction force too) = 10 N upwards - because the sum of those two forces must be equal and opposite to the gravitational force downwards (10 N) as the resultant force on object is 0 N.
I read the original question again. It asked for the magnitude and direction of the reaction force, not for the analysis of the reaction force.
Post by: thushan on February 29, 2012, 05:15:09 pm
Direction = upwards.
That's what I said before. And your teacher was right.
I was just explaining how the reaction force was 10 N upwards and wanted to clarify that it was the TOTAL reaction force, not the NORMAL reaction force (which you come across more often).
Post by: yellowsone31 on March 07, 2012, 05:33:20 pm
Post by: paulsterio on March 07, 2012, 05:49:01 pm
it means that you will get the highest exam mark in your cohort as your sac mark
Post by: Phy124 on March 07, 2012, 08:45:38 pm
What does it actually mean to be rank 1 in a SAC? Cause I was rank 1 for the first physics SAC and I have no idea what it actually means lol :)If you maintain rank 1 you will get the highest scaled SAC scores (Highest General Assessment 1 score), within your school at the end of the year.
Never mind, beaten.
Post by: yellowsone31 on March 07, 2012, 09:36:32 pm
means next to nothing in one sac, you want to be ranked 1 overall
it means that you will get the highest exam mark in your cohort as your sac mark
thanks !
Post by: yellowsone31 on March 07, 2012, 09:36:49 pm
What does it actually mean to be rank 1 in a SAC? Cause I was rank 1 for the first physics SAC and I have no idea what it actually means lol :)If you maintain rank 1 you will get the highest scaled SAC scores (Highest General Assessment 1 score), within your school at the end of the year.
Never mind, beaten.
thanks for replying :)