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e2m2a

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I have been struggling with this problem for years and never have found an answer to it anywhere on the web or in textbooks. And I can’t derive the formula for it.

Suppose we have a mass designated as ##m_2##. This mass is constrained to move with one degree of freedom along a linear track. We assume no friction between ##m_2## and the track. The track is rigidly attached to the earth. We define the motion of ##m_2## to be along the y-axis.

There is a near-massless aluminum rod attached at one end to ##m_2##. At the other end of the rod is attached a magnet with mass ##m_1##. A person holds another magnet near ##m_1##, such that there is a magnetic force experienced by ##m_1## in the radially outward direction. We designate this force on ##m_1## as ##f_{mag}##. Because of this ##f_{mag}##, ##m_2## accelerates in the positive y-direction along the linear track.

The person hovers over the Earth in some kind of spacecraft or whatever, so there is no contact forces between the Earth and the bottom of the shoes of the person. And with this spacecraft , the person is able to keep the distance between the two magnets constant and keeps up with the accelerating ##m_2##, such that ##f_{mag}## remains constant.

We define the angle ##\theta## as the angle between the rod and the horizontal x-axis.

Here is my conundrum. I want to know the tension force, designated ##f_{tension}## acting on ##m_2## for all angles of the rod.

I know for the special case where ##\theta## is equal to 90 degrees, it would simply be:

$$f_{tension} = \frac {m_2} {m_1 + m_2} f_{mag}$$

And for ##\theta## equal to 0 degrees, it would be:

$$f_{tension} = \frac {m_{earth}} {m_1 + m_{earth}} f_{mag}$$

Where, ##m_{earth}## is the combined mass of the Earth and ##m_2##.

But I cannot figure out the formula for ##f_{tension}## for any angle ##\theta## between 0 and 90 degrees. Can someone help me out with this?

I have been struggling with this problem for years and never have found an answer to it anywhere on the web or in textbooks. And I can’t derive the formula for it.

Suppose we have a mass designated as ##m_2##. This mass is constrained to move with one degree of freedom along a linear track. We assume no friction between ##m_2## and the track. The track is rigidly attached to the earth. We define the motion of ##m_2## to be along the y-axis.

There is a near-massless aluminum rod attached at one end to ##m_2##. At the other end of the rod is attached a magnet with mass ##m_1##. A person holds another magnet near ##m_1##, such that there is a magnetic force experienced by ##m_1## in the radially outward direction. We designate this force on ##m_1## as ##f_{mag}##. Because of this ##f_{mag}##, ##m_2## accelerates in the positive y-direction along the linear track.

The person hovers over the Earth in some kind of spacecraft or whatever, so there is no contact forces between the Earth and the bottom of the shoes of the person. And with this spacecraft , the person is able to keep the distance between the two magnets constant and keeps up with the accelerating ##m_2##, such that ##f_{mag}## remains constant.

We define the angle ##\theta## as the angle between the rod and the horizontal x-axis.

Here is my conundrum. I want to know the tension force, designated ##f_{tension}## acting on ##m_2## for all angles of the rod.

I know for the special case where ##\theta## is equal to 90 degrees, it would simply be:

$$f_{tension} = \frac {m_2} {m_1 + m_2} f_{mag}$$

And for ##\theta## equal to 0 degrees, it would be:

$$f_{tension} = \frac {m_{earth}} {m_1 + m_{earth}} f_{mag}$$

Where, ##m_{earth}## is the combined mass of the Earth and ##m_2##.

But I cannot figure out the formula for ##f_{tension}## for any angle ##\theta## between 0 and 90 degrees. Can someone help me out with this?

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