fgcse2018
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#1
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#1
How do you do this question? I got 40kg for M but not sure as it seems way too simple for a 10 marker. Name:  099D7ADE-4252-4AAA-B692-729B57F93C7F.jpg.jpeg
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mqb2766
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#2
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(Original post by fgcse2018)
How do you do this question? I got 40kg for M but not sure as it seems way too simple for a 10 marker. Name:  099D7ADE-4252-4AAA-B692-729B57F93C7F.jpg.jpeg
Views: 368
Size:  18.4 KB

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This came up fairly recently. One thing to notice is that with that arrangement, its actually impossible to be in equilibrium as both masses cause an anticlockwise moment on the smaller wheel. One of the masses must be on the other side of the relevant wheel for the moments to cancel.
Can you put your working down, I don't think 40kg is right.
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fgcse2018
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#3
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(Original post by mqb2766)
This came up fairly recently. One thing to notice is that with that arrangement, its actually impossible to be in equilibrium as both masses cause an anticlockwise moment on the smaller wheel. One of the masses must be on the other side of the relevant wheel for the moments to cancel.
Can you put your working down, I don't think 40kg is right.
I just did 10g x 0.08= Mg x 0.02 - but clearly this is wrong im just stuck as to what to do. Thanks
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mqb2766
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#4
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(Original post by fgcse2018)
I just did 10g x 0.08= Mg x 0.02 - but clearly this is wrong im just stuck as to what to do. Thanks
Where do the gears come into it?
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fgcse2018
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#5
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(Original post by mqb2766)
Where do the gears come into it?
No where, i didnt even consider them which was why i was sure i was probably wrong
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thekidwhogames
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#6
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It's not 40; it's half of that.
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mqb2766
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#7
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(Original post by fgcse2018)
No where, i didnt even consider them which was why i was sure i was probably wrong
So what do you understand about gears? Think about cycling, they change the speed and the torque. Have you covered this?
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fgcse2018
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#8
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(Original post by mqb2766)
So what do you understand about gears? Think about cycling, they change the speed and the torque. Have you covered this?
No i have not, i mean i know that torque is just the turning effect on something but thats about it, this is on stats year 2 moments edexcel unit test
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fgcse2018
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#9
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(Original post by thekidwhogames)
It's not 40; it's half of that.
How?
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mqb2766
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(Original post by fgcse2018)
No i have not, i mean i know that torque is just the turning effect on something but thats about it, this is on stats year 2 moments edexcel unit test
I'd do a bit reading about it. But for this question, think about how the force gets transferred from one wheel to the other.
In equilibrium, the forces(s) at the gear teeth must cause an equal and opposite moment to the mass on each wheel. What would be the force that the smaller wheel's teeth must exert on the larger wheel's teeth for the larger wheel to remain in equilibrium?
Then do the opposite, that force must generate an equal and opposite moment on the smaller wheel to the moment generated by mass M.
This is why its 10 marks.
https://www.youtube.com/watch?v=49DxlXs8tyk
isn't bad, but there are others

Edit: the answer given above is right, its because the gear ratio is 1:2 (or 5:10), but its important that you understand why it happens?
Last edited by mqb2766; 2 years ago
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fgcse2018
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#11
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(Original post by mqb2766)
I'd do a bit reading about it. But for this question, think about how the force gets transferred from one wheel to the other.
In equilibrium, the forces(s) at the gear teeth must cause an equal and opposite moment to the mass on each wheel. What would be the force that the smaller wheel's teeth must exert on the larger wheel's teeth for the larger wheel to remain in equilibrium?
Then do the opposite, that force must generate an equal and opposite moment on the smaller wheel to the moment generated by mass M.
This is why its 10 marks.
https://www.youtube.com/watch?v=49DxlXs8tyk
isn't bad, but there are others

Edit: the answer given above is right, its because the gear ratio is 1:2 (or 5:10), but its important that you understand why it happens?
Ahhh i see, thanks for the video and help, im sort of seeing it now but just need to work on it for a bit longer!
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mqb2766
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#12
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(Original post by fgcse2018)
Ahhh i see, thanks for the video and help, im sort of seeing it now but just need to work on it for a bit longer!
Sure, just post your working if you need any more help.
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fgcse2018
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#13
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#13
(Original post by mqb2766)
I'd do a bit reading about it. But for this question, think about how the force gets transferred from one wheel to the other.
In equilibrium, the forces(s) at the gear teeth must cause an equal and opposite moment to the mass on each wheel. What would be the force that the smaller wheel's teeth must exert on the larger wheel's teeth for the larger wheel to remain in equilibrium?
Then do the opposite, that force must generate an equal and opposite moment on the smaller wheel to the moment generated by mass M.
This is why its 10 marks.
https://www.youtube.com/watch?v=49DxlXs8tyk
isn't bad, but there are others

Edit: the answer given above is right, its because the gear ratio is 1:2 (or 5:10), but its important that you understand why it happens?
(Name:  BB9D09B0-F08C-46C2-B7CF-ECB9A47D9F53.jpg.jpeg
Views: 412
Size:  25.1 KB (RED PEN) so is this it? Or is the method wrong
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mqb2766
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#14
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#14
(Original post by fgcse2018)
(Name:  BB9D09B0-F08C-46C2-B7CF-ECB9A47D9F53.jpg.jpeg
Views: 412
Size:  25.1 KB (RED PEN) so is this it? Or is the method wrong
Looks about right. I'd probably be a bit more explicit about the forces that the gears are exerting on each other. On the larger wheel we must have
F*0.1 = 10*g*0.08
where F is the linear force exerted by the smaller gear's teeth on the larger wheel. So
F = 8g
In equilibrium, this is also the force (equal and opposite) exerted by the larger gear's teeth on the smaller wheel, so again in equilibrium
F*0.05 = M*g*0.02
so
40g = 2Mg
so M = 20. Again, I'll note that actually the masses are in the wrong position to get equilibrium, they'd both cause an anticlockwise motion of the smaller wheel.

The shortcut is the gear ratio as you mention, but I suspect that you'd need something like the above to get full marks for working.
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fgcse2018
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#15
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#15
(Original post by mqb2766)
Looks about right. I'd probably be a bit more explicit about the forces that the gears are exerting on each other. On the larger wheel we must have
F*0.1 = 10*g*0.08
where F is the linear force exerted by the smaller gear's teeth on the larger wheel. So
F = 8g
In equilibrium, this is also the force (equal and opposite) exerted by the larger gear's teeth on the smaller wheel, so again in equilibrium
F*0.05 = M*g*0.02
so
40g = 2Mg
so M = 20. Again, I'll note that actually the masses are in the wrong position to get equilibrium, they'd both cause an anticlockwise motion of the smaller wheel.

The shortcut is the gear ratio as you mention, but I suspect that you'd need something like the above to get full marks for working.
I see, thats very helpful actually, thanks alot once agin
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