# M1 moments hard question

Watch
Announcements
#1
Two interlocking gears are in equilibrium. The gear on the right has a radius of 10 cm and has a loop 8 cm from the centre. The loop is to the right of, and level with the centre of the gear. A 10 kg mass hangs from the loop. The other gear has a radius of 5 cm and a loop 2 cm from the centre. The loop is to the left of, and level with the centre of the gear. A mass M kg hangs from the left loop.
Find the value of M.

A diagram can be seen on this word document(WB10)
https://gmaths28.files.wordpress.com/.../m201_1-workbook-moments-wb1_11.docx

I posted this a few days ago and got a response which I thought I'd understood but it turns out I haven't. I'm struggling to understand how the gears work tbh.
0
2 years ago
#2
(Original post by dont know it)
Two interlocking gears are in equilibrium. The gear on the right has a radius of 10 cm and has a loop 8 cm from the centre. The loop is to the right of, and level with the centre of the gear. A 10 kg mass hangs from the loop. The other gear has a radius of 5 cm and a loop 2 cm from the centre. The loop is to the left of, and level with the centre of the gear. A mass M kg hangs from the left loop.
Find the value of M.

A diagram can be seen on this word document(WB10)
https://gmaths28.files.wordpress.com/.../m201_1-workbook-moments-wb1_11.docx

I posted this a few days ago and got a response which I thought I'd understood but it turns out I haven't. I'm struggling to understand how the gears work tbh.
Where did this come from? Am I misunderstanding something, the 10kg mass would cause a clockwise rotation in the first gear and an anticlockwise rotation in the second gear. The M mass also produces an anticlockwise rotation, so I don't see how there is an equilibrium?

If the second mass was to the right, it could cause a clockwise torque which could maintain equilibrium?
0
2 years ago
#3
(Original post by mqb2766)
Where did this come from? Am I misunderstanding something, the 10kg mass would cause a clockwise rotation in the first gear and an anticlockwise rotation in the second gear. The M mass also produces an anticlockwise rotation, so I don't see how there is an equilibrium?

If the second mass was to the right, it could cause a clockwise torque which could maintain equilibrium?
Agreed, this is what was preventing me from replying earlier as I thought I was missing something obvious... maybe they did mean the M mass to be to the right of the centre.
0
2 years ago
#4
(Original post by RDKGames)
Agreed, this is what was preventing me from replying earlier as I thought I was missing something obvious... maybe they did mean the M mass to be to the right of the centre.
If so, it should be straight forward.
Torque for gear 1 is
10*g*0.08
the gears are in the ratio 2:1, so use that to balance the mass on the 2nd gear?
Last edited by mqb2766; 2 years ago
1
#5
(Original post by mqb2766)
Where did this come from? Am I misunderstanding something, the 10kg mass would cause a clockwise rotation in the first gear and an anticlockwise rotation in the second gear. The M mass also produces an anticlockwise rotation, so I don't see how there is an equilibrium?

If the second mass was to the right, it could cause a clockwise torque which could maintain equilibrium?
I was thinking the same tbh so you're probably right and the questions wrong.
0
#6
(Original post by mqb2766)
If so, it should be straight forward.
Torque for gear 1 is
10*g*0.08
the gears are in the ratio 2:1, so use that to balance the mass on the 2nd gear?
Thanks.
0
#7
(Original post by RDKGames)
Agreed, this is what was preventing me from replying earlier as I thought I was missing something obvious... maybe they did mean the M mass to be to the right of the centre.
Yeah I think they did.
0
#8
(Original post by mqb2766)
If so, it should be straight forward.
Torque for gear 1 is
10*g*0.08
the gears are in the ratio 2:1, so use that to balance the mass on the 2nd gear?
Sorry to bring this up again, but I don't think I get it still. This is what the solution says(slide 19): https://gmaths28.files.wordpress.com...ts-wb1_11.pptx

I don't get the force on left gear by right gear part and where it comes from? Could you explain it please?
0
#9
(Original post by mqb2766)
If so, it should be straight forward.
Torque for gear 1 is
10*g*0.08
the gears are in the ratio 2:1, so use that to balance the mass on the 2nd gear?
Sorry to bring this up again, but I don't think I get it still. This is what the solution says(slide 19): https://gmaths28.files.wordpress.com...ts-wb1_11.pptx

I don't get the force on left gear by right gear part and where it comes from? Could you explain it please?
0
2 years ago
#10
(Original post by dont know it)
Sorry to bring this up again, but I don't think I get it still. This is what the solution says(slide 19): https://gmaths28.files.wordpress.com...ts-wb1_11.pptx

I don't get the force on left gear by right gear part and where it comes from? Could you explain it please?
The moment or angular force or torque is simply
linear force * distance

1) the moment (angular force) on the right gear from the mass must be
10*g*0.08 (clockwise)
2) In equilibrium, this must be balanced by the force from the left gear's teeth, touching the right gear's teeth. Assume the linear force is F, then on the right gear wheel, we must have (balance angular forces)
F*0.1 = 10*g*0.08
F = 8g
3) On the left gear wheel, we also have equilbrium (balance angular forces)
F*0.05 = M*g*0.02
Assuming I've not done a typo mistake, that should give you M. Note the direction of the "rotation" of the left gear must be clockwise, so the mass should occur on the right. If this isn't clear, which bit isn't?

As a short cut, as the gears' radii are in the ratio 2:1, then the angular forces will also be in the same ratio (the gear ratio). The angular velocities would be in the ratio 1:2.

Edit see
http://furthermaths.org.uk/docs/Simp...udentFinal.pdf
The force part is on page 13.
Last edited by mqb2766; 2 years ago
0
#11
(Original post by mqb2766)
The moment or angular force or torque is simply
linear force * distance

1) the moment (angular force) on the right gear from the mass must be
10*g*0.08 (clockwise)
2) In equilibrium, this must be balanced by the force from the left gear's teeth, touching the right gear's teeth. Assume the linear force is F, then on the right gear wheel, we must have
F*0.1 = 10*g*0.08
F = 8g
3) On the left gear wheel, we also have equilbrium
F*0.05 = M*g*0.02
Assuming I've not done a typo mistake, that should give you M. Note the direction of the "rotation" of the left gear must be clockwise, so the mass should occur on the right. If this isn't clear, which bit isn't?

As a short cut, as the gears' radii are in the ratio 2:1, then the angular forces will also be in the same ratio (the gear ratio). The angular velocities would be in the ratio 1:2.

Edit see
http://furthermaths.org.uk/docs/Simp...udentFinal.pdf
I'm confused about the force F. Does it come from the weight of the left gear or is there a separate force at the teeth?
0
2 years ago
#12
(Original post by dont know it)
I'm confused about the force F. Does it come from the weight of the left gear or is there a separate force at the teeth?
The equilibrium is an angular force equilibrium about the gear wheels' centers - the gear wheels do not turn. The 10 kg mass means that the gears want to turn. The only thing stopping the right wheel turning is the "single" point of contact between the teeth of the two gears. At that point, they exert linear forces (on each other) which are "perpendicular" to the teeth (or the radii connecting that point of contact with the gears' centers) and because they are in equilibrium, the forces are equal and opposite, F. Even though the gears have different numbers of teeth, the linear force is the same (in equilibrium). However, beause the gears have different radii, the angular force at the centres is different because
angular force = linear force * radius
So the angular force at the center of a larger gear is larger than the angular force at a smaller gear. This is precisely what gears do. The gear ratio shortcuts the analysis about the linear force in the gear teeth and simply says the ratio of the angular forces at the gear wheels' centers is the same as the ratio of the radii (or gear teeth). Make sure you're clear about the difference between the linear force at the teeth and the angular force at the center.

The mass on the left gear wheel is simiply to ensure equilibirum, otherwise the gear wheels would turn.
Last edited by mqb2766; 2 years ago
1
#13
(Original post by mqb2766)
The equilibrium is an angular force equilibrium about the gear wheels' centers - the gear wheels do not turn. The 10 kg mass means that the gears want to turn. The only thing stopping the right wheel turning is the "single" point of contact between the teeth of the two gears. At that point, they exert linear forces (on each other) which are "perpendicular" to the teeth (or the radii connecting that point of contact with the gears' centers) and because they are in equilibrium, the forces are equal and opposite, F. Even though the gears have different numbers of teeth, the linear force is the same (in equilibrium). However, beause the gears have different radii, the angular force at the centres is different because
angular force = linear force * radius
So the angular force at the center of a larger gear is larger than the angular force at a smaller gear. This is precisely what gears do. The gear ratio shortcuts the analysis about the linear force in the gear teeth and simply says the ratio of the angular forces at the gear wheels' centers is the same as the ratio of the radii (or gear teeth). Make sure you're clear about the difference between the linear force at the teeth and the angular force at the center.

The mass on the left gear wheel is simiply to ensure equilibirum, otherwise the gear wheels would turn.
So the left gear exerts a force F on the right gear so it maintains equilibrium, and the reaction force from the right gear which is equal and opposite to force F is exerted on the left gear for it to also maintain equilibrium?
0
2 years ago
#14
(Original post by dont know it)
So the left gear exerts a force F on the right gear so it maintains equilibrium, and the reaction force from the right gear which is equal and opposite to force F is exerted on the left gear for it to also maintain equilibrium?
Yes and because of the teeth, the linear force is applied at the perimeter of the wheels, tangential to the perimeter
0
#15
(Original post by mqb2766)
Yes and because of the teeth, the linear force is applied at the perimeter of the wheels, tangential to the perimeter
Nice. 1 more thing, I know this might sound stupid but since the mass is being hung, why is there no tension? Is it because it's not acting on the gear and that's what our focus is?
Last edited by dont know it; 2 years ago
0
2 years ago
#16
(Original post by dont know it)
Nice. 1 more thing, I know this might sound stupid but since the mass is being hung, why is there no tension?
There would be in the string. However, it would be equal to
mg
as the forces are equal and opposite so the length of the string (and the tension in the string) is irrelevant. It would be equivalent to having the mass at the point of attachment of the string on the wheel.
0
#17
(Original post by mqb2766)
There would be in the string. However, it would be equal to
mg
as the forces are equal and opposite so the length of the string (and the tension in the string) is irrelevant. It would be equivalent to having the mass at the point of attachment of the string on the wheel.
I'm confused tbh. So are you saying we ignore it because it acts on the mass which is a different body to the one we are focusing on i.e. the left gear? Therefore there are only 2 forces acting on the left gear which is the normal reaction exerted by the right gear onto the left, and the weight of the mass. Or am I just completely wrong and missing the point?
0
2 years ago
#18
(Original post by dont know it)
I'm confused tbh. So are you saying we ignore it because it acts on the mass which is a different body to the one we are focusing on i.e. the left gear? Therefore there are only 2 forces acting on the left gear which is the normal reaction exerted by the right gear onto the left, and the weight of the mass. Or am I just completely wrong and missing the point?
I may have misunderstood what you were asking or you may be overthinking it a bit or ...

Just consider the right gear wheel. The 10kg mass is on a string. The downwards force this generates does not depend on the length of the string and would be the same if the 10 mass were attached to the gear wheel at the point where the string is attached to the gear wheel.

Imagine you have two 10kg masses. One has no string and one is attached to a string 1m long. You outstretch your arms (horizontally) and hold the mass with no string attached in your left hand, and hold the end of the string of the other mass in your right hand. Your shoulders would have the same angular force acting on them. The string simply transmits the linear force to the point of contact, because of the tension in the string.

This would be the same for each gear wheel.
0
#19
(Original post by mqb2766)
I may have misunderstood what you were asking or you may be overthinking it a bit or ...

Just consider the right gear wheel. The 10kg mass is on a string. The downwards force this generates does not depend on the length of the string and would be the same if the 10 mass were attached to the gear wheel at the point where the string is attached to the gear wheel.

Imagine you have two 10kg masses. One has no string and one is attached to a string 1m long. You outstretch your arms (horizontally) and hold the mass with no string attached in your left hand, and hold the end of the string of the other mass in your right hand. Your shoulders would have the same angular force acting on them. The string simply transmits the linear force to the point of contact, because of the tension in the string.

This would be the same for each gear wheel.
Hmm I'm still confused. I think I get your point about the tension but since we are taking moments about the centre of the wheel, do we ignore the tension because it's not actually acting on the gear itself, it's acting on the mass like you say. So like questions with horizontal beams in equilibrium, we would draw forces acting on the beam itself and then take moments about a point on the beam. So because the gears are in equilibrium, we only consider the forces acting on the gears and not the mass.
0
2 years ago
#20
(Original post by dont know it)
Hmm I'm still confused. I think I get your point about the tension but since we are taking moments about the centre of the wheel, do we ignore the tension because it's not actually acting on the gear itself, it's acting on the mass like you say. So like questions with horizontal beams in equilibrium, we would draw forces acting on the beam itself and then take moments about a point on the beam. So because the gears are in equilibrium, we only consider the forces acting on the gears and not the mass.
The 10 kg mass is attached (by a string) to the right gear wheel, where the point of attachment is horizontal to the center. It is that point (attachment) which the 10g force acts on the gear wheel's center to provide a rotational force.

If you had a beam in equilibrium with a 10 kg point mass or a 10 kg mass suspended by a string from the same point, the equations would be the same. It is the point of attachment to the beam where the force acts.
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Would you give consent for uni's to contact your parent/trusted person in a mental health crisis?

Yes - my parent/carer (51)
31.1%
Yes - a trusted person (42)
25.61%
No (48)
29.27%
I'm not sure (23)
14.02%