# Equations

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Write an equation for the centripetal force F in terms of reaction force R and weight W for someone rotating on the Earth.

Considering you need the radius to find the centripetal force, I do not really see how that can be achieved.

Considering you need the radius to find the centripetal force, I do not really see how that can be achieved.

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(Original post by

Write an equation for the centripetal force F in terms of reaction force R and weight W for someone rotating on the Earth.

Considering you need the radius to find the centripetal force, I do not really see how that can be achieved.

**Mlopez14**)Write an equation for the centripetal force F in terms of reaction force R and weight W for someone rotating on the Earth.

Considering you need the radius to find the centripetal force, I do not really see how that can be achieved.

However , if there is more to this question than you have written here, please post again with the rest of it.

Last edited by Stonebridge; 8 months ago

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(Original post by

It looks like they don't want you to consider the radius of the Earth or its speed of rotation. Just those 2 forces, R and W. The question is, in this case, much simpler than you have imagined. Just a simple resultant force question.

However , if there is more to this question than you have written here, please post again with the rest of it.

**Stonebridge**)It looks like they don't want you to consider the radius of the Earth or its speed of rotation. Just those 2 forces, R and W. The question is, in this case, much simpler than you have imagined. Just a simple resultant force question.

However , if there is more to this question than you have written here, please post again with the rest of it.

Weighing scales don't measure your mass, just the reaction force R that the scales push back on you

I'd imagine that would help but I am still quite confused

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(Original post by

Indeed the following is mentioned earlier:

Weighing scales don't measure your mass, just the reaction force R that the scales push back on you

I'd imagine that would help but I am still quite confused

**Mlopez14**)Indeed the following is mentioned earlier:

Weighing scales don't measure your mass, just the reaction force R that the scales push back on you

I'd imagine that would help but I am still quite confused

When you stand on scales, there are 2 forces acting on you, W, your weight downwards and R, the force of the scales pushing up on you.

If the Earth was completely still - no rotation in this case - then you would be at rest on its surface.

If you are at rest then there is zero resultant force on you. (Newton's Law.)

This means that the resultant of those two forces is zero.

Taking

**down**as positive direction it means W - R = 0

(or W = R)

In other words, the measurement on the scale (R) would be equal to your weight (W).

If the Earth is rotating, however, you are moving in a very large circle around with the Earth's surface at that point.

For you to move in a circle there must be an accelerating force towards the centre of that circle, the so called centripetal force.

As there are only 2 forces acting on you still, W and R, then these 2 forces combined must produce that centripetal force.

This force acts in the downwards direction towards the centre of the circle you are moving in.

In the previous paragraph, I wrote down the equation for the

**resultant**downwards force when this force is zero. (When you were at rest)

This equation will

**always**give me the resultant downwards force, but now it isn't zero. It's equal to the centripetal force (F) on you, because you are moving.

It is this that connects with the first bit of the question. As a result of the inclusion of the force F now,

**W is not equal to R**.

In other words, the reaction of the scales on you pushing up, is not equal to your weight.

It is the reaction force of the scales on you that is actually measured by the scales as your 'weight'. (Through the spring system inside)

The conclusion is that now the scales are not (quite) measuring your weight. (Because R is not equal to W)

The difference is so small that it is insignificant. But it's a useful bit of physics.

BTW. At the north or south pole the scales would read correctly. Do you see why?

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(Original post by

Thanks. So it looks like it is as I thought.

When you stand on scales, there are 2 forces acting on you, W, your weight downwards and R, the force of the scales pushing up on you.

If the Earth was completely still - no rotation in this case - then you would be at rest on its surface.

If you are at rest then there is zero resultant force on you. (Newton's Law.)

This means that the resultant of those two forces is zero.

Taking

(or W = R)

In other words, the measurement on the scale (R) would be equal to your weight (W).

If the Earth is rotating, however, you are moving in a very large circle around with the Earth's surface at that point.

For you to move in a circle there must be an accelerating force towards the centre of that circle, the so called centripetal force.

As there are only 2 forces acting on you still, W and R, then these 2 forces combined must produce that centripetal force.

This force acts in the downwards direction towards the centre of the circle you are moving in.

In the previous paragraph, I wrote down the equation for the

This equation will

It is this that connects with the first bit of the question. As a result of the inclusion of the force F now,

In other words, the reaction of the scales on you pushing up, is not equal to your weight.

It is the reaction force of the scales on you that is actually measured by the scales as your 'weight'. (Through the spring system inside)

The conclusion is that now the scales are not (quite) measuring your weight. (Because R is not equal to W)

The difference is so small that it is insignificant. But it's a useful bit of physics.

BTW. At the north or south pole the scales would read correctly. Do you see why?

**Stonebridge**)Thanks. So it looks like it is as I thought.

When you stand on scales, there are 2 forces acting on you, W, your weight downwards and R, the force of the scales pushing up on you.

If the Earth was completely still - no rotation in this case - then you would be at rest on its surface.

If you are at rest then there is zero resultant force on you. (Newton's Law.)

This means that the resultant of those two forces is zero.

Taking

**down**as positive direction it means W - R = 0(or W = R)

In other words, the measurement on the scale (R) would be equal to your weight (W).

If the Earth is rotating, however, you are moving in a very large circle around with the Earth's surface at that point.

For you to move in a circle there must be an accelerating force towards the centre of that circle, the so called centripetal force.

As there are only 2 forces acting on you still, W and R, then these 2 forces combined must produce that centripetal force.

This force acts in the downwards direction towards the centre of the circle you are moving in.

In the previous paragraph, I wrote down the equation for the

**resultant**downwards force when this force is zero. (When you were at rest)This equation will

**always**give me the resultant downwards force, but now it isn't zero. It's equal to the centripetal force (F) on you, because you are moving.It is this that connects with the first bit of the question. As a result of the inclusion of the force F now,

**W is not equal to R**.In other words, the reaction of the scales on you pushing up, is not equal to your weight.

It is the reaction force of the scales on you that is actually measured by the scales as your 'weight'. (Through the spring system inside)

The conclusion is that now the scales are not (quite) measuring your weight. (Because R is not equal to W)

The difference is so small that it is insignificant. But it's a useful bit of physics.

BTW. At the north or south pole the scales would read correctly. Do you see why?

However, I am still confused about what you meant F not being equal to R

I would assume the scales would read correctly at poles since you are essentially at rest because of the poles being the axis of rotation of planet Earth?

Also if I understand correctly, the equation I am looking for is W - R = F (At rest), though I am still confused on what the missing force is when the person is moving?

Last edited by Mlopez14; 8 months ago

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(Original post by

I would assume the scales would read correctly at poles since you are essentially at rest because of the poles being the axis of rotation of planet Earth?

**Mlopez14**)I would assume the scales would read correctly at poles since you are essentially at rest because of the poles being the axis of rotation of planet Earth?

(Original post by

Also if I understand correctly, the equation I am looking for is W - R = F (At rest), though I am still confused on what the missing force is when the person is moving?

**Mlopez14**)Also if I understand correctly, the equation I am looking for is W - R = F (At rest), though I am still confused on what the missing force is when the person is moving?

It's a poor question, IMO, because it only really makes sense at the equator (or on inclined land), as the reaction force is perpendicular to the ground, your weight is towards the centre of mass of the Earth, and your acceleration perpendicular to its axis of rotation. If we model the Earth as a perfect sphere, then, in most places, you'd need friction too. Consider standing at a point close to the North pole - your weight acts downwards, the reaction upwards, but you need friction to provide the force to accelerate you around the Earth's axis of rotation. Of course, the net forces are small, so almost always ignored.

Last edited by RogerOxon; 8 months ago

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