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This problem drove me crazy.

A small object is at rest on the edge of a horizontal table. It is pushed in such a way that it falls off the other side of the table, which is 1 m wide, after 2 s. Does the object have wheels?

Spoiler

Original post by tangotangopapa2
A small object is at rest on the edge of a horizontal table. It is pushed in such a way that it falls off the other side of the table, which is 1 m wide, after 2 s. Does the object have wheels?

Spoiler



might be on an air hockey table :unsure:

2 seconds seems like a long time for something to spend slowing down over 1m otherwise.
Original post by Joinedup
might be on an air hockey table :unsure:

2 seconds seems like a long time for something to spend slowing down over 1m otherwise.


Great.
Can you please prove mathematically that it is on an air hockey table? :tongue: What about wheels?
Alright, first time I'm answering a physics question on this forum, so bare with me :wink: (Oh, an I don't know if you have a formula editor built into this page, sry for not using it now if there is one.)

Start with the basing GCSE formula for distance cover while having constant acceleration:

s=(at^2)/2

We already have s and t given, what is a? Well, you have friction acting upon the body, which, in this case, is simply weight times the friction coefficient.

a = F/m = fmg/m = fg

Therefore:

s = (fgt^2)/2

Rearranging and plugging in the numbers will give you f = 0.05 [dimensionless]. Now ask yourself what actually happened. The object travelled exactly 1m and stopped, therefore if f = 0.05 then it will stop at the opposite side of the table, not fall. You can also find that the initial velocity of the object must have been exactly 1 m/s (you can calculated this yourself). So, if you want the object to fall, you have to decrease the coefficient of friction below 0.05 AND decrease the initial velocity to meet your conditions. The other extreme situation is when f=0 and v_0 = 0.5m/s. You could probably figure out the relationship between f and v_0, but I'll leave that to you.

Dunno, how this proves/disproves whether the object has wheels, but I hope this helps :smile:
assuming constant accelleration
s=0.5 (u+v) t

gives us initial speed of 1 m/s and therefore acceleration of 0.5 ms-2

since for the accelerating force on a horizontally sliding object F=ma and F=μmg
you can estimate the coefficient of friction
μ=a/g
which can be compared to a table of common materials http://www.engineersedge.com/coeffients_of_friction.htm
Original post by Joinedup
assuming constant accelleration
s=0.5 (u+v) t

gives us initial speed of 1 m/s and therefore acceleration of 0.5 ms-2

since for the accelerating force on a horizontally sliding object F=ma and F=μmg
you can estimate the coefficient of friction
μ=a/g
which can be compared to a table of common materials http://www.engineersedge.com/coeffients_of_friction.htm


Thank you so much. Since, the coefficient of friction is too low, whole body/ part of it should roll. So it has wheel. Very helpful.
Original post by Blank_Planet
Alright, first time I'm answering a physics question on this forum, so bare with me :wink: (Oh, an I don't know if you have a formula editor built into this page, sry for not using it now if there is one.)

Start with the basing GCSE formula for distance cover while having constant acceleration:

s=(at^2)/2

We already have s and t given, what is a? Well, you have friction acting upon the body, which, in this case, is simply weight times the friction coefficient.

a = F/m = fmg/m = fg

Therefore:

s = (fgt^2)/2

Rearranging and plugging in the numbers will give you f = 0.05 [dimensionless]. Now ask yourself what actually happened. The object travelled exactly 1m and stopped, therefore if f = 0.05 then it will stop at the opposite side of the table, not fall. You can also find that the initial velocity of the object must have been exactly 1 m/s (you can calculated this yourself). So, if you want the object to fall, you have to decrease the coefficient of friction below 0.05 AND decrease the initial velocity to meet your conditions. The other extreme situation is when f=0 and v_0 = 0.5m/s. You could probably figure out the relationship between f and v_0, but I'll leave that to you.

Dunno, how this proves/disproves whether the object has wheels, but I hope this helps :smile:


Thanks alot. So u cant be greater than 1ms-1 , can it? So coeff of friction < g/20.
Original post by tangotangopapa2
Thanks alot. So u cant be greater than 1ms-1 , can it? So coeff of friction < g/20.


yup :smile:
Reply 8
Original post by Blank_Planet
Alright, first time I'm answering a physics question on this forum, so bare with me :wink: (Oh, an I don't know if you have a formula editor built into this page, sry for not using it now if there is one.)

Start with the basing GCSE formula for distance cover while having constant acceleration:

s=(at^2)/2

We already have s and t given, what is a? Well, you have friction acting upon the body, which, in this case, is simply weight times the friction coefficient.

a = F/m = fmg/m = fg

Therefore:

s = (fgt^2)/2

Rearranging and plugging in the numbers will give you f = 0.05 [dimensionless]. Now ask yourself what actually happened. The object travelled exactly 1m and stopped, therefore if f = 0.05 then it will stop at the opposite side of the table, not fall. You can also find that the initial velocity of the object must have been exactly 1 m/s (you can calculated this yourself). So, if you want the object to fall, you have to decrease the coefficient of friction below 0.05 AND decrease the initial velocity to meet your conditions. The other extreme situation is when f=0 and v_0 = 0.5m/s. You could probably figure out the relationship between f and v_0, but I'll leave that to you.

Dunno, how this proves/disproves whether the object has wheels, but I hope this helps :smile:


You're so clever x
Reply 9
Original post by Joinedup
assuming constant accelleration
s=0.5 (u+v) t

gives us initial speed of 1 m/s and therefore acceleration of 0.5 ms-2

since for the accelerating force on a horizontally sliding object F=ma and F=μmg
you can estimate the coefficient of friction
μ=a/g
which can be compared to a table of common materials http://www.engineersedge.com/coeffients_of_friction.htm


What the final speed? I don't think you have done this correctly.
(edited 7 years ago)
Find the coefficient of friction assuming constant deceleration, as it is shown above it is 0.05. From experience you should know that this is too small, hence why it's probably using wheels.
Original post by B_9710
What the final speed? I don't think you have done this correctly.


Stops at the edge of a 1m table if I read it properly, so final speed is zero... In this case we really only care about the magnitude of the accn.
----
Fwiw
Suppose other explanations could be that you're on the Moon or in a lift that's accelerating downwards.
Reply 12
Original post by Joinedup
Stops at the edge of a 1m table if I read it properly, so final speed is zero... In this case we really only care about the magnitude of the accn.
----
Fwiw
Suppose other explanations could be that you're on the Moon or in a lift that's accelerating downwards.


Oh I see what you're saying, I misinterpreted it when I first read it.
You still cannot tell if it's on wheels though, there was no indication that the table is rough, and even if it is rough the coefficient of friction could infact be a small value. Plus how do wheels even affect the motion anyway?
(edited 7 years ago)
Original post by B_9710
Oh I see what you're saying, I misinterpreted it when I first read it.
You still cannot tell if it's on wheels though, there was no indication that the table is rough, and even if it is rough the coefficient of friction could infact be a small value. Plus how do wheels even affect the motion anyway?


Now that the coefficient of friction must be less than g/20. Though it does not prove that the object has wheel but highly indicates so. No known material used to make table has that small c.o.f. Had it stopped at the edge, then equality would have holded.
Note that 1 m/s is the maximum possible initial speed which gives c.o.f g/20. It could have been less. For instance, 0.5 m/s gives value of cof very very close to zero. Note that in all these calculations, final velocity is assumed to be zero, which may not be the case. It should have some speed in order to fall off the table. In that case, cof for each initial speed should be lower.

The way wheel works is by reducing friction. Only friction between wheel's bottom & ground and friction within internal system comes into account.
Original post by B_9710
Oh I see what you're saying, I misinterpreted it when I first read it.
You still cannot tell if it's on wheels though, there was no indication that the table is rough, and even if it is rough the coefficient of friction could infact be a small value. Plus how do wheels even affect the motion anyway?


Yeah I think it's more of a trick question than a fair question that you might meet in an exam...

wheels should reduce the resistance and therefore the acceleration though.
Reply 15
Original post by tangotangopapa2
Now that the coefficient of friction must be less than g/20. Though it does not prove that the object has wheel but highly indicates so. No known material used to make table has that small c.o.f. Had it stopped at the edge, then equality would have holded.
Note that 1 m/s is the maximum possible initial speed which gives c.o.f g/20. It could have been less. For instance, 0.5 m/s gives value of cof very very close to zero. Note that in all these calculations, final velocity is assumed to be zero, which may not be the case. It should have some speed in order to fall off the table. In that case, cof for each initial speed should be lower.

The way wheel works is by reducing friction. Only friction between wheel's bottom & ground and friction within internal system comes into account.


But we are actually modelling the object as a particle, so it's not like the object has any surface area so actually wheels make no difference to a particle in terms of the resistance acting on it as the only resistive force is friction. No obviously friction does increase as surface area in contact with the tbe increases, but as we are modelling it as a particle, it has no surface area and so you can really not say whether it has wheels or not. It's really a false question.
Original post by B_9710
But we are actually modelling the object as a particle, so it's not like the object has any surface area so actually wheels make no difference to a particle in terms of the resistance acting on it as the only resistive force is friction. No obviously friction does increase as surface area in contact with the tbe increases, but as we are modelling it as a particle, it has no surface area and so you can really not say whether it has wheels or not. It's really a false question.


Yes, you are right. As it has been mentioned earlier, it is not a fair question to meet in an exam. Do you think it makes sense to ask this question stating 'Is the object sliding or rolling?'
Reply 17
Original post by tangotangopapa2
Yes, you are right. As it has been mentioned earlier, it is not a fair question to meet in an exam. Do you think it makes sense to ask this question stating 'Is the object sliding or rolling?'


Again th same issues come up. I think that if you say that it can be assumed that no friction acts when the object is rolling it could then work.
Original post by tangotangopapa2
Now that the coefficient of friction must be less than g/20.


I actually made a tiny mistake: the coefficient of friction must be <1/20 not <g/20. The later is the max acceleration of the object. Coef. of friction is a dimensionless quantity. Small difference, but I thought its better to make things clear.
Original post by Blank_Planet
I actually made a tiny mistake: the coefficient of friction must be <1/20 not <g/20. The later is the max acceleration of the object. Coef. of friction is a dimensionless quantity. Small difference, but I thought its better to make things clear.


Thank you so much. Really appreciate it.

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