# Wire tension that might involve tensors

1. Jul 11, 2017

### e2m2a

I need help with this problem. This is not a homework assignment, so please don’t send it over to the homework forum.

It involves mechanical engineering dynamics that probably are more subtle and advanced then first year mechanical engineering dynamics. It might involve tensor analysis. I don’t know.

We let a mass, designated as $m_2$, slide along a straight track with one degree of freedom with no friction. The track is rigidly attached to the earth. There is a wire attached to $m_2$ so that it can be pulled at a constant acceleration. Attached on the other end of the wire is a second mass, designated as $m_1$. A known force of magnitude F is applied to $m_1$, such as a magnetic force, accelerating $m_1$ and $m_2$ along the track. We define the motion of $m_1$-$m_2$ to be in the positive y-direction. The wire makes an angle $\theta$ with respect to the x-axis, and for this problem we keep the angle $\theta$ constant. Thus, the givens are: $m_1$, $m_2$, F, and the angle $\theta$. I want to know the tension in the wire for any constant angle $\theta$ between 0 and 90 degrees with the above givens.

On the surface this might seem like a trivial problem, but when I do a deeper analysis, it seems to be more complicated than I can handle. For example, when $\theta$ is zero degrees, intuitively, the tension in the wire would be at a maximum, and there would be no acceleration of $m_1$-$m_2$ in the positive y-direction, and when $\theta$ is 90 degrees, the tension would be a minimum, and there would be a maximum acceleration of m1-m2. But what would the tension in the wire be at any angle between zero and 90 degrees? I can’t get my head around this. Could someone please help me with this?

2. Jul 11, 2017

### Dr.D

This does not sound too difficult, but I urge you to post a figure showing the situation and identifying the variables for clarity. If you do, I think you will have the help you need in short order.

3. Jul 11, 2017

### Staff: Mentor

4. Jul 11, 2017

### Dr.D

Thank you, Berkeman.

CWatters is definitely correct about the FBD, except that you need two FBDs, not just one. Draw them, write the equations of motion for each body, and solve for the tension. If you get stuck, but sure to post the work you have done.

I cannot imagine why you think tensors might be required in this simple problem. It really is not that hard at all.

5. Jul 11, 2017

### e2m2a

6. Jul 11, 2017

### e2m2a

So, I need to draw a FBD for $m_1$ and another FBD for $m_2$, and then solve a system of equations from the two diagrams? I will try.

7. Jul 11, 2017

### Nidum

The diagram link opens to a blank page for me ?

8. Jul 11, 2017

### e2m2a

9. Jul 11, 2017

### e2m2a

As an intuitive guess, I think the final expression for T would be as follows:

The tension force with respect to the x-axis would be:

$T_x = cos\theta F$

The tension force with respect to the y-axis would be:

$T_y = sin\theta \frac{m_2}{m_1 + m_2} F$

Thus, T would be:

$T = \sqrt{T_x^2 + T_y^2}$

Is this correct?

10. Jul 11, 2017

### Dr.D

Rather than guess, why not write the equations of motion for the particles and work it out mathematically?

11. Jul 11, 2017

### Dr.D

It opened OK for me using Libre Office.

12. Jul 11, 2017

### e2m2a

When I try to set up the equations of motion, all of the trigonometric expressions cancel out. I don't know how to set up the equations of motion correctly.

13. Jul 11, 2017

### Dr.D

Write the sum of forces on each body, once in the direction parallel to the guide and again in the direction perpendicular to the guide.

Be sure to post your work if you need more help.

14. Jul 11, 2017

### e2m2a

I think this is what I am suppose to do:

Forces on $m_1$:

$$a_y = \frac{sin\theta F – sin\theta T} {m_1}$$

$$a_x = \frac{cos\theta F – cos\theta T} {m_1}$$

Forces on $m_2$. (I have to introduce the constant $m_e$, the mass of the earth).

$$a_y = \frac{sin\theta T} {m_2}$$

$$a_x = \frac{cos\theta T} {m_e + m_2}$$

But I am not sure what to do next, for if I equate the terms for $a_y$ and $a_x$, all the trig expressions drop out, which leads to an answer that doesn't make sense. (The tension is constant for all angles of $\theta$).

15. Jul 11, 2017

### Dr.D

There is an error in your diagram that is probably leading you astray. The applied force F cannot be colinear with the wire.

In addition, you need to think through the kinematic relations between the motions of the two particles.

16. Jul 11, 2017

### e2m2a

Why is the applied F not colinear with the wire? I do not understand that. The tension force must be colinear with the wire, so why not F? I have defined the force F to be colinear with the wire, but if this is not physically possible, can you please explain why this is not possible? In addition, if F is not colinear, then this would cause $m_1$ to rotate and this is a complication I did not define to happen.

Last edited: Jul 11, 2017
17. Jul 11, 2017

### Dr.D

Both m1 and m2 must accelerate parallel to the guide. If F and T are colinear, there is no component of the sum of forces on m1 available to accelerate m1.

Rather than drawing F, draw simply two components, Fx, and Fy, and don't worry about exactly what direction they are in. This should clarify what happens considerably.

This is not a hard problem; do not make it one. But do not impose misunderstanding up on by insisting the F and T be colinear.