VCE Specialist 3/4 Question Thread!

[Homer]

Question 2
A box of mass 5kg rests on a rough horizontal floor. The coefficient of friction between the
box and the floor is 0 ⋅1. A boy applies a horizontal dragging force of D newtons to the box in
an attempt to move it.
a. Find the values of D if the box is not at the point of moving across the floor.

The solutions say that D is smaller than 4.9N (the value of D for when it is equal to maximal friction). Just a couple of questions about this:

- Is it better to write D is between 0 and 4.9N, not inclusive of end points?
- Can someone explain the point of moving? I assumed the answer would be R+\{4.9}, as I thought it'd only be on the point of moving when D is equal to maximal friction. Is the solution wrong or am I just misunderstanding what 'point of moving' means? I'm aware that friction is variable and Fr =< mju(N), but I wouldn't have thought an object was considered to be on the point of moving unless D was equal to maximal friction.

I am pretty sure the answers right. since the question asks for values of D if the box is NOT at the point of moving across the floor. I guess you can write it as [0,4.9)

Also, really petty point. I was doing an exam today, I wrote 1.5 instead of 3/2 and I realise VCAA takes marks off for non-exact answers. Would they take a mark off if I write 1.5 instead of 3/2 since 1.5 is technically non-exact?

1.5 is exact

[psyxwar]

I am pretty sure the answers right. since the question asks for values of D if the box is NOT at the point of moving across the floor. I guess you can write it as [0,4.9)

Yeah, but isn't the point of moving when the forwards force exactly equals the maximum friction? A stationary object with no force applied is not on the point of moving is it, and hence the answer should be all positive real values except for 4.9?

[keltingmeith]

Yeah, it's an Insight exam 1 2010

I am sort of getting the gist of this. Can you explain a little bit more?
When will friction act in the same direction as the pushing force/thrust (whatever you call it)?

I always thought friction acts in the opposite direction of motion.

Also, really petty point. I was doing an exam today, I wrote 1.5 instead of 3/2 and I realise VCAA takes marks off for non-exact answers. Would they take a mark off if I write 1.5 instead of 3/2 since 1.5 is technically non-exact?

Thanks (sorry for two questions at once)/

Firstly, on the decimal thing:
Scientifically, I would say that 3/2 is much more informative than 1.5. However, in terms of what VCAA want, you should be okay to put in decimals if they're exact representations of fractions.

On the friction thing:
You're exactly right - friction will ALWAYS oppose motion. So, let's consider our box.
Horizontally, there are three forces acting on it - the first is the pulling force, P. The second is the tension in the string, T. And the final force is friction, Fr. Now, Fr will depend on the motion of the box, however P and T will act independent of what the box is doing (in fact, they'll control what the box is doing).

So, since T and P oppose each other, we have three options (note: even though these are vector quantities, T and P in this situation refers to the respective force's magnitudes):

1) T>P. In this case, there will be net force to the right, which means that the box will be moving to the right. So, friction will oppose this and point to the left.

2) T=P. In this case, the box will sit in equilibrium, and the effects of friction are negligible.

3) T<P. In this case, there will be net force to the left, which means that the box will be moving to the left. So, friction will oppose this and point to the right.

Essentially, you can't assume friction will point some way until you've looked at which way the net motion is. The solutions have assumed that the box is moving to the right, but the question did not indicate which way movement was, nor that T>P, and so you can't make that assumption. In fact, if it was an exam, they'd have to pay you the marks no matter which way you pointed your friction, since they didn't give you enough information.

[M_BONG]

Firstly, on the decimal thing:
Scientifically, I would say that 3/2 is much more informative than 1.5. However, in terms of what VCAA want, you should be okay to put in decimals if they're exact representations of fractions.

On the friction thing:
You're exactly right - friction will ALWAYS oppose motion. So, let's consider our box.
Horizontally, there are three forces acting on it - the first is the pulling force, P. The second is the tension in the string, T. And the final force is friction, Fr. Now, Fr will depend on the motion of the box, however P and T will act independent of what the box is doing (in fact, they'll control what the box is doing).

So, since T and P oppose each other, we have three options (note: even though these are vector quantities, T and P in this situation refers to the respective force's magnitudes):

1) T>P. In this case, there will be net force to the right, which means that the box will be moving to the right. So, friction will oppose this and point to the left.

2) T=P. In this case, the box will sit in equilibrium, and the effects of friction are negligible.

3) T<P. In this case, there will be net force to the left, which means that the box will be moving to the left. So, friction will oppose this and point to the right.

Essentially, you can't assume friction will point some way until you've looked at which way the net motion is. The solutions have assumed that the box is moving to the right, but the question did not indicate which way movement was, nor that T>P, and so you can't make that assumption. In fact, if it was an exam, they'd have to pay you the marks no matter which way you pointed your friction, since they didn't give you enough information.

Cheers! Best explanation I have received on this thread so far and your explanation made it sound simple Thanks for that!

[keltingmeith]

Question 2
A box of mass 5kg rests on a rough horizontal floor. The coefficient of friction between the
box and the floor is 0 ⋅1. A boy applies a horizontal dragging force of D newtons to the box in
an attempt to move it.
a. Find the values of D if the box is not at the point of moving across the floor.

The solutions say that D is smaller than 4.9N (the value of D for when it is equal to maximal friction). Just a couple of questions about this:

- Is it better to write D is between 0 and 4.9N, not inclusive of end points?
- Can someone explain the point of moving? I assumed the answer would be R+\{4.9}, as I thought it'd only be on the point of moving when D is equal to maximal friction. Is the solution wrong or am I just misunderstanding what 'point of moving' means? I'm aware that friction is variable and Fr =< mju(N), but I wouldn't have thought an object was considered to be on the point of moving unless D was equal to maximal friction.

The wording for this is so tricky, but I've bolded the relevant bit for you.

The boy is ATTEMPTING to move the box - this implies that the box isn't moving, i.e. the boy can't move the box. Since this is the case, the boy can't be applying enough force to keep it moving. However, you are right - "point of moving" only refers to when D=maximal friction. If D>maximal friction, then the object is no longer on the point of moving - it's just moving.

[psyxwar]

The wording for this is so tricky, but I've bolded the relevant bit for you.

The boy is ATTEMPTING to move the box - this implies that the box isn't moving, i.e. the boy can't move the box. Since this is the case, the boy can't be applying enough force to keep it moving. However, you are right - "point of moving" only refers to when D=maximal friction. If D>maximal friction, then the object is no longer on the point of moving - it's just moving.

Thanks. So are the solutions wrong then, since technically if D<Fr max it is stationary but not on the point of moving?

[keltingmeith]

No, they're right - they want to know when the box is NOT on the point of moving. In reality, this is every number except for 4.9 - however, another "hidden" constraint is that D will not exceed 4.9. So, the appropriate range is 0<D<4.9 (if they forgot their lower bound, they're technically wrong, but it's not that big a deal since one would normally assume D>0 anyway)

[psyxwar]

No, they're right - they want to know when the box is NOT on the point of moving. In reality, this is every number except for 4.9 - however, another "hidden" constraint is that D will not exceed 4.9. So, the appropriate range is 0<D<4.9 (if they forgot their lower bound, they're technically wrong, but it's not that big a deal since one would normally assume D>0 anyway)

sorry I'm not sure if I understand what you mean by the 'hidden constraint' bit.

Why does an attempt necessarily mean it's a failed attempted o.o? and even D is necessarily less than 4.9 because the box isn't moving, isn't the object still not on the point of moving if its less than 4.9? EDIT: dw I just realised the second bit is stupid, think I see what you mean now. But yeah does attempting something necessarily mean you didn't succeed? O___O That doesn't seem right semantically

On that note would it be exclusive or inclusive of 0N, considering if a force was applied doesn't that imply the force is non-zero in magnitude?

[keltingmeith]

sorry I'm not sure if I understand what you mean by the 'hidden constraint' bit.

Why does an attempt necessarily mean it's a failed attempted o.o? and even D is necessarily less than 4.9 because the box isn't moving, isn't the object still not on the point of moving if its less than 4.9?

On that note would it be exclusive or inclusive of 0N, considering if a force was applied doesn't that imply the force is non-zero in magnitude?

I am currently attempting to help you - if I wasn't attempting to help you, either I've given up because I couldn't do it, or I have succeeded in doing so, because if I hadn't succeeded I wouldn't have stopped. This question is in real time, so if he's still attempting, he has yet to succeed.

Often these questions are filled with "hidden constraints" that you need to pick up. For example, since the boy is "attempting to move the box", it must be stationary. On top of that, the force must be non-zero, because the boy is putting an effort into it.

Yes - if D<4.9, the box is NOT on the point of moving - which is what the question is asking for:

a. Find the values of D if the box is not at the point of moving across the floor.

If D=4.9, the box IS on the point of moving, so we don't include that in the range of values. Finally, if D>4.9, then the boy is moving the box, so why would he keep attempting to move it?

And yes - as I said before, since the boy is putting an effort in, that implies the force is non-zero, so we would not include 0.

With all this in mind, we choose 0<D<4.9

[psyxwar]

I am currently attempting to help you - if I wasn't attempting to help you, either I've given up because I couldn't do it, or I have succeeded in doing so, because if I hadn't succeeded I wouldn't have stopped. This question is in real time, so if he's still attempting, he has yet to succeed.

Often these questions are filled with "hidden constraints" that you need to pick up. For example, since the boy is "attempting to move the box", it must be stationary. On top of that, the force must be non-zero, because the boy is putting an effort into it.

Yes - if D<4.9, the box is NOT on the point of moving - which is what the question is asking for:If D=4.9, the box IS on the point of moving, so we don't include that in the range of values. Finally, if D>4.9, then the boy is moving the box, so why would he keep attempting to move it?

And yes - as I said before, since the boy is putting an effort in, that implies the force is non-zero, so we would not include 0.

With all this in mind, we choose 0<D<4.9

thanks for the help! I understand what you mean, but yeah I'm not convinced if I think the wording particularly makes sense. (unless this sort of wording is typical of spesh lol)

"A boy applies a horizontal dragging force of D newtons to the box in an attempt to move it."

idk to me it seems like the 'in an attempt' is there because he wants to move it, and depending on the magnitude on D he may or may not succeed. eg. I did lots of trial questions in an attempt to do well on the SAC -> does this necessarily mean I did badly on the SAC? Idk, maybe english language is working against me ;_;

[keltingmeith]

thanks for the help! I understand what you mean, but yeah I'm not convinced if I think the wording particularly makes sense. (unless this sort of wording is typical of spesh lol)

"A boy applies a horizontal dragging force of D newtons to the box in an attempt to move it."

idk to me it seems like the 'in an attempt' is there because he wants to move it, and depending on the magnitude on D he may or may not succeed. eg. I did lots of trial questions in an attempt to do well on the SAC -> does this necessarily mean I did badly on the SAC? Idk, maybe english language is working against me ;_;

At the end of day, don't lose sleep over this question - it's weird wording, and you won't get something so ambiguous in a real exam (welp, unless VCAA want to pay a bunch of marks to people/have hundreds of teachers complaining to them)

[Kuroyuki]

Hey guys,
What are the requirements for an object to be moving with constant acceleration?

Eg: A box of mass 5 kg is on a horizontal plane. A force of magnitude F kg-wt acting at an
angle of 30 degrees to the horizontal is applied to the box. The coefficient of friction between the
box and the plane is 3 root 3. Find a range of values of F, so that the box moves with constant
acceleration.

I am thinking its like N>0 and a>0 but is that correct?

Also if someone could provide a worked answer to this question it would be greatly appreciated!
Thanks in advance

[lzxnl]

Hey guys,
What are the requirements for an object to be moving with constant acceleration?

Eg: A box of mass 5 kg is on a horizontal plane. A force of magnitude F kg-wt acting at an
angle of 30 degrees to the horizontal is applied to the box. The coefficient of friction between the
box and the plane is 3 root 3. Find a range of values of F, so that the box moves with constant
acceleration.

I am thinking its like N>0 and a>0 but is that correct?

Also if someone could provide a worked answer to this question it would be greatly appreciated!
Thanks in advance

Constant acceleration => F>0, F constant
So here, you really need net force > 0
There's friction which attains a max value of mu*N and there's the force you're pulling with
Use those.

[psyxwar]

In a tech active exam, where say we need to find a unit vector perpendicular to 2 different vectors, can I just set up 3 equations and write 'solving (1), (2), (3) simultaneously' then write down the answers (solving using CAS)? Or should I manually show some signs of trying to solve it by hand? Not sure if I'm showing enough working lol

[lzxnl]

In a tech active exam, where say we need to find a unit vector perpendicular to 2 different vectors, can I just set up 3 equations and write 'solving (1), (2), (3) simultaneously' then write down the answers (solving using CAS)? Or should I manually show some signs of trying to solve it by hand? Not sure if I'm showing enough working lol

If it's tech active, then solving it by CAS should be fine.
Although honestly, this is why a cross product was invented