# Normal reaction force when plane is not inclined or flat

Watch
#1
https://prnt.sc/n1oh8s .

For questions like these, is there a way to know the direction of the normal reaction force? When we have a flat surface it's just perpendicular to that surface in contact, but I don't know what we do here. I posted this question before and the response was the reaction acts tangential to the contact surface but I'm not sure what that means.
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11 months ago
#2
(Original post by dont know it)
https://prnt.sc/n1oh8s .

For questions like these, is there a way to know the direction of the normal reaction force? When we have a flat surface it's just perpendicular to that surface in contact, but I don't know what we do here. I posted this question before and the response was the reaction acts tangential to the contact surface but I'm not sure what that means.
Normal means perpendicular, at the point of contact. Here we have normal contact forces acting towards the centre of the circle. There are no frictional forces required for equilibrium, so they're zero.

Without friction, any force can only be perpendicular to the surface at the point of contact. With circles, that's always towards the centre, i.e. perpendicular to the surface.
Last edited by RogerOxon; 11 months ago
2
11 months ago
#3
(Original post by dont know it)
https://prnt.sc/n1oh8s .

For questions like these, is there a way to know the direction of the normal reaction force? When we have a flat surface it's just perpendicular to that surface in contact, but I don't know what we do here. I posted this question before and the response was the reaction acts tangential to the contact surface but I'm not sure what that means.
Take the point where the object touches the externals surface(s) and construct a line (or plane) of tangency that goes through that point.

These are precisely the blue lines on the image. Having that, you reduced your seemingly complicated issue to something you're already familiar with... you have a "fake" flat surface, and the reaction acts normal to this surface which counters the weight.

Due to symmetry, the two reactions are equivalent.

Last edited by RDKGames; 11 months ago
1
#4
(Original post by RDKGames)
Take the point where the object touches the externals surface(s) and construct a line (or plane) of tangency that goes through that point.

These are precisely the blue lines on the image. Having that, you reduced your seemingly complicated issue to something you're already familiar with... you have a "fake" flat surface, and the reaction acts normal to this surface which counters the weight.

Due to symmetry, the two reactions are equivalent.

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#5
(Original post by RogerOxon)
Normal means perpendicular, at the point of contact. Here we have normal contact forces acting towards the centre of the circle. There are no frictional forces required for equilibrium, so they're zero.

Without friction, any force can only be perpendicular to the surface at the point of contact. With circles, that's always towards the centre, i.e. perpendicular to the surface.
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11 months ago
#6
(Original post by dont know it)
You only care about the moment applied by the contact forces, so only need to consider the component tangential to the circle where the gears mesh.
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#7
(Original post by RogerOxon)
You only care about the moment applied by the contact forces, so only need to consider the component tangential to the circle where the gears mesh.
I'm a bit confused..so we wouldn't draw a tangent to the point of contact then draw the normal perpendicular to this tangent?
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11 months ago
#8
(Original post by dont know it)
I'm a bit confused..so we wouldn't draw a tangent to the point of contact then draw the normal perpendicular to this tangent?
With meshed gears, that gets complicated, and changes as they turn. What we care about is the nett component of the contact fores that applies a moment about the centre of each gear. Any component that goes through their centres doesn't apply a moment about it. We will also have reaction forces at the gear centres - they don't apply a moment either, so can also be ignored - draw them, but don't attemp to calculate their magnitudes.
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#9
(Original post by RogerOxon)
With meshed gears, that gets complicated, and changes as they turn. What we care about is the nett component of the contact fores that applies a moment about the centre of each gear. Any component that goes through their centres doesn't apply a moment about it. We will also have reaction forces at the gear centres - they don't apply a moment either, so can also be ignored - draw them, but don't attemp to calculate their magnitudes.
So with meshed gears, we take the reaction force to act tangential to the contact surface, but in other cases we take the reaction force to act perpendicular to the tangent? Is that correct? And thanks.
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11 months ago
#10
(Original post by dont know it)
So with meshed gears, we take the reaction force to act tangential to the contact surface, but in other cases we take the reaction force to act perpendicular to the tangent? Is that correct? And thanks.
The contact force may have components both tangential and perpendicular to the point of contact, or, in the case of meshed gears, we consider the circles that logically contact. You need to consider which component is important for the particular case. With meshed gears, their purpose is to transfer torque, so it's the force tangential to the circle of contact that you care matters. For equilibrium, there may not be any frictional forces, so it could be only the normal contact force, but you will have cases where both are non-zero, e.g. a ladder against a wall.
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#11
(Original post by RogerOxon)
The contact force may have components both tangential and perpendicular to the point of contact, or, in the case of meshed gears, we consider the circles that logically contact. You need to consider which component is important for the particular case. With meshed gears, their purpose is to transfer torque, so it's the force tangential to the circle of contact that you care matters. For equilibrium, there may not be any frictional forces, so it could be only the normal contact force, but you will have cases where both are non-zero, e.g. a ladder against a wall.
I think I get what you're saying, so does that mean for the initial question there was a contact force tangential to the point of contact as well, but we ignored it because it doesn't help us answer the question?
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11 months ago
#12
(Original post by dont know it)
I think I get what you're saying, so does that mean for the initial question there was a contact force tangential to the point of contact as well, but we ignored it because it doesn't help us answer the question?
You'd have to have an elastic cylinder to have frictional forces, which is beyond the level of these questions. You need to assume that there is no friction for that question, although I'm surprised that they didn't explicitly state it.
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#13
(Original post by RogerOxon)
You'd have to have an elastic cylinder to have frictional forces, which is beyond the level of these questions. You need to assume that there is no friction for that question, although I'm surprised that they didn't explicitly state it.
Ok thanks
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