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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

Find the value of

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.

*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.

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#2

(Original post by

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

Find the value of

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.

**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.

If the second mass was to the right, it could cause a clockwise torque which could maintain equilibrium?

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#3

(Original post by

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?

**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?

**did**mean the M mass to be to the right of the centre.

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#4

(Original post by

Agreed, this is what was preventing me from replying earlier as I thought I was missing something obvious... maybe they

**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.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; 1 year ago

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**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?

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(Original post by

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?

**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?

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**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.

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**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?

I don't get the force on left gear by right gear part and where it comes from? Could you explain it please?

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**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?

I don't get the force on left gear by right gear part and where it comes from? Could you explain it please?

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#10

(Original post by

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?

**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?

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; 1 year ago

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(Original post by

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

**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

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#12

(Original post by

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?

**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?

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; 1 year ago

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(Original post by

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.

**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.

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#14

(Original post by

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?

**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?

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(Original post by

Yes and because of the teeth, the linear force is applied at the perimeter of the wheels, tangential to the perimeter

**mqb2766**)Yes and because of the teeth, the linear force is applied at the perimeter of the wheels, tangential to the perimeter

Last edited by dont know it; 1 year ago

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#16

(Original post by

Nice. 1 more thing, I know this might sound stupid but since the mass is being hung, why is there no tension?

**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?

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.

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(Original post by

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.

**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.

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#18

(Original post by

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?

**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?

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.

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(Original post by

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.

**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.

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#20

(Original post by

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.

**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.

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.

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