The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

[M1] Connected Particles

Here's the question


I managed to get the correct answers for the first 3 parts which were:
a) 3.92ms-1
b) 41.2m
c) 7.84ms-1

Not even sure how to start on d) tbh. I don't even have the actual masses of each particle.
Original post by DarkEnergy
Here's the question

I managed to get the correct answers for the first 3 parts which were:
a) 3.92ms-1
b) 41.2m
c) 7.84ms-1

Not even sure how to start on d) tbh. I don't even have the actual masses of each particle.

I guess that they mean "nett force". in that case, given that the pulley is not accelerating, what is the nett force acting on it, by Newton I?

OTOH, if they mean the downward force on the pulley, draw a free body diagram showing all the forces acting on it, and the answer should be obvious.

(The question is very poorly worded BTW. The "system" does not have a single acceleration a or velocity - the two masses have different accelerations, with the same magnitude, for example)
Reply 2
Avatar for DarkEnergy
DarkEnergy
OP
20
Original post by atsruser
I guess that they mean "nett force". in that case, given that the pulley is not accelerating, what is the nett force acting on it, by Newton I?

OTOH, if they mean the downward force on the pulley, draw a free body diagram showing all the forces acting on it, and the answer should be obvious.

(The question is very poorly worded BTW. The "system" does not have a single acceleration a or velocity - the two masses have different accelerations, with the same magnitude, for example)

It says at the top it is released from rest so doesn't that mean it is accelerating?

I don't understand how I'm supposed to get the answer (81.2N) when I don't have T nor 'm'.

Also can you please explain why the actual mass of each particle doesn't matter in part a? I got the answer right but still not sure why you can suddenly ignore that the weight is 7mg & 3mg and instead use 7g and 3g. Very confusing question, sorry if these are stupid questions that I'm asking.
Original post by DarkEnergy
It says at the top it is released from rest so doesn't that mean it is accelerating?

I don't understand how I'm supposed to get the answer (81.2N) when I don't have T nor 'm'.

Also can you please explain why the actual mass of each particle doesn't matter in part a? I got the answer right but still not sure why you can suddenly ignore that the weight is 7mg & 3mg and instead use 7g and 3g. Very confusing question, sorry if these are stupid questions that I'm asking.


You can think of the m like a constant - when working out problems with connected particles (like in this example) you are essentially working with the ratio of the masses. Try working out (a) again using a value of m (e.g. make m=2 so that the masses are 14kg and 3kg...you should get the same answer!) When you worked out part (a) (by making 2 simultaneous equations for T) you should've found that you cancelled out m anyway, so that's why the value of m has no effect :smile:
Reply 4
Avatar for DarkEnergy
DarkEnergy
OP
20
Original post by Lauren-x-
You can think of the m like a constant - when working out problems with connected particles (like in this example) you are essentially working with the ratio of the masses. Try working out (a) again using a value of m (e.g. make m=2 so that the masses are 14kg and 3kg...you should get the same answer!) When you worked out part (a) (by making 2 simultaneous equations for T) you should've found that you cancelled out m anyway, so that's why the value of m has no effect :smile:

Oh yeah I see now, thanks! Any ideas what to do for d) though?
Original post by DarkEnergy
Oh yeah I see now, thanks! Any ideas what to do for d) though?


Hmm...the force on the pulley is usually T + T (sum of the tensions) but that makes it 82.4N
Reply 6
Avatar for DarkEnergy
DarkEnergy
OP
20
Original post by Lauren-x-
Hmm...the force on the pulley is usually T + T (sum of the tensions) but that makes it 82.4N

Oh right, thanks! Yeah that's the right answer.
Original post by DarkEnergy
Oh right, thanks! Yeah that's the right answer.


Yay! But thanks for the M1 revision too - it was useful for me to try and work it out/remember what to do :tongue: (I really need to look over it when I get the time haha)
Reply 8
Avatar for DarkEnergy
DarkEnergy
OP
20
Original post by Lauren-x-
Yay! But thanks for the M1 revision too - it was useful for me to try and work it out/remember what to do :tongue: (I really need to look over it when I get the time haha)

Again though, why doesn't the value of m matter here? Because if T = (41.2m)N and R = 2 * T, surely the answer should be (82.3m)N not 82.3N?
(edited 7 years ago)
Original post by DarkEnergy
Again though, why doesn't the value of m matter here? Because if T = (41.2m)N and R = 2 * T, surely the answer should be (82.3m)N not 82.3N?



A better way to look at it would be to draw a forces diagram, modelling the whole system as a particle.

If you have the two weights (7mg and 3mg) down, and the two tensions (41.2m) going up, all of the 'm's cancel out. It links back to the idea that m is a constant, so if all of the forces contain m then you can eliminate it.


Not sure if this is 100% correct for all forces diagrams, but it works for this topic :tongue:
(edited 7 years ago)
Original post by Lauren-x-
A better way to look at it would be to draw a forces diagram, modelling the whole system as a particle.

If you have the two weights (7mg and 3mg) down, and the two tensions (41.2m) going up, all of the 'm's cancel out. It links back to the idea that m is a constant, so if all of the forces contain m then you can eliminate it.


Not sure if this is 100% correct for all forces diagrams, but it works for this topic :tongue:

Ah okay, thanks again.
Original post by Lauren-x-
...


Original post by DarkEnergy
...


There are only two reasonable answers to part d), IMO:

1) (Unlikely) The total force on the pulley is zero, since it's "fixed". The two tensions in the rope are balanced by whatever is holding the pulley in place, or

2) (Probably what they meant) The force on the pulley due to the two masses is 2T = 82.4m N. The lack of "m" in their answer is a typo.
(edited 7 years ago)
Original post by ghostwalker
There are only two reasonable answers to part d), IMO:

1) (Unlikely) The total force on the pulley is zero, since it's "fixed". The two tensions in the rope are balanced by whatever is holding the pulley in place, or

2) (Probably what they meant) The force on the pulley due to the two masses is 2T = 82.4m N. The lack of "m" in their answer is a typo.

Thank you. Spoke to my Mechanics teacher today and indeed it was a typo (missing out the m)
Original post by DarkEnergy
Thank you. Spoke to my Mechanics teacher today and indeed it was a typo (missing out the m)


Good to know. Unfortunately typos seem to be endemic in mathematical material - doesn't help when you're trying to learn it.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you