# A Spring + Two Masses [Force Analysis and Energy Analysis giving different answers]

1. Oct 21, 2005

### maverick280857

Hello
First of all, I am posting on PF after a LONG time so all these changes are new to me. As a result, I have, without reading prior announcements, posted this on the Advanced Physics forum instead of the other one. (To the moderator: Please shift it to the appropriate forum if you think its in the wrong place. Sorry for the inconvenience).

Okay..the problem:

Two bars of masses $m_{1}$ and $m_{2}$ connected by a light undeformed horizontal spring are lying on a uniformly rough horizontal tabletop, having a coefficient of friction $\mu$. The minimum force that has to be applied horizontally to the bar of mass $m_{1}$ along the length of the spring in order just to shift the other bar of mass $m_{2}$ is

(A) $F = \mu g ( m_{1} + \frac{1}{2}m_{2})$
(B) $F = \mu g (m_{1} + m_{2})g$
(C) $F = \mu g (\frac{m_{1} + m_{2}}{2})$
(D) $F = \mu g (\frac{1}{2}m_{1} + m_{2})$

Force Analysis

Suppose the spring has spring constant k and the elongation in it just when motion of $m_{2}$ is to begin is x, then

$$F = kx + \mu m_{1}g$$

$$kx = \mu m_{2}g[/itex] This gives part (B) as the answer. Energy Analysis The net force on the mass $m_{1}$ is $F - \mu m_{1} g$. The work of this force is $(F - \mu m_{1}g)x$. This work is stored as potential energy in the spring (since motion of $m_{1}$ has just started, its velocity is zero...to begin with). So, [tex]Fx = \mu m_{1}g x + \frac{1}{2}kx^{2}$$

Also,

$$kx = \mu m_{2} g$$

solving these two equations give (D) as the answer.

Which of these two methods is correct? Why is it correct? Which is wrong? Why is it wrong?

I'd be grateful if someone could help me with this...the two different answers are confusing me somewhat.

Thanks and cheers,
Vivek

2. Oct 21, 2005

### Staff: Mentor

This analysis is correct.

Mass $m_{2}$ will just begin to move when the spring is stretched a length $L$ such that $k L = \mu m_2 g$. To create this stretch, a force $F$ must be applied to $m_{1}$ such that $F = k L + \mu m_1 g = \mu g (m_1 + m_2)$.
This analysis is incorrect.

Realize that the applied force varies from an initial value of $\mu m_1 g$ to a final value of $F = \mu g (m_1 + m_2)$. The force is a function of x: $F(x) = \mu m_{1}g + kx$.

If you integrate this force over the range x = 0 to x = L, you will find that the work done by the applied force will equal $\mu m_{1}g x + \frac{1}{2}kx^{2}$. This just says that the work equals the energy stored in the spring plus the work done against friction. Nothing new here. Note: The work does not equal $F L$, as you assume in your analysis.

3. Oct 21, 2005

### maverick280857

Hello Doc

Thanks a lot for the explanation. [My answer (the one using forces) does not match with the answer given in the problem bank wherefrom this problem originates and they have given the (incorrect) energy method as the solution.] I am assuming that the "L" in your solution is the elongation of the spring when there is no relative motion between the two masses. Is this correct?

4. Oct 21, 2005

### maverick280857

HI again Doc,

I had a second look at the whole thing and it all fits in nicely. I see why the energy solution fails. But I have a few more queries.

Books on classical physics normally describe the ideal twobody oscillator consisting of two masses A and B with a spring between them which mediates the force. The mathematical analysis involves two coupled differential equations which when solved give the net elongation/compression of the spring. Suppose we now modify the problem like this: we exert a constant force F on A for all time $t \geq 0$ and we seek a solution to the differential equations one of which has a constant term F. (I am going to solve this in a while...just want to clarify some things). Some playing around with the conditions and constraints should--in principle at least--give a solution to this problem, because this is a twobody oscillator. Is this okay? (Of course, the present solution is much shorter).
If we insist on writing the differential equations of motion for the masses A and B given that their (possibly time varying) accelerations are $a_{1}$ and $a_{2}$ respectively, then they will be of the form

$$F_{net} - \mu m_{1}g - kx = m_{1}a_{1}$$
$$kx - \mu m_{2}g = m_{2}a_{2}$$

Only when the spring has reached maximum elongation L would the accelerations and velocities of the masses become equal and remain so subsequently. But how do we solve the above system of equations in general? Here, $x = x_{1} - x_{2} - l_{0}$ where x is the elongation/compression in the spring, $x_{1}$ and $x_{2}$ are the displacements of the masses with respect to an inertial frame. $l_{0}$ is the natural length of the spring. Further, $a_{1} = \ddot{x_{1}}$, $a_{2} = \ddot{x_{2}}$. Is all this okay?

Where can I find more detailed information about the general two body oscillator subjected to external forces?

Cheers
Vivek

Last edited: Oct 21, 2005
5. Oct 22, 2005

### daniel_i_l

I think that no analysis with the spring is needed to solve the problem. Since the spring is in between the to masses and we only want to know how moving m1 effects m2, the spring's energy and force will always cancel out. the total force needed is only the force that will overcome the total friction.

6. Oct 25, 2005

### maverick280857

Yes, thanks...I realize that. I am just wondering whether I will get an easily solvable system of de's...thats all. I am not thinking of solving this problem that way: I just want to work out a general solution to a twobody oscillator subject to forces in different ways.

7. Oct 25, 2005

### Staff: Mentor

You can make the general problem of coupled masses complicated by adding damping factors, etc. Here's a site that gives the math and a java simulation for such a problem: http://acoustics.me.uic.edu/Simulation/Mass_Spring_2DOF.htm [Broken]

Another interesting variation is to hang one mass from a second spring (fixed at one end) and explore the coupled oscillations under gravity. http://www.pcug.org.au/~apurdam/doublespring/doublespring.html [Broken] & http://www.unisanet.unisa.edu.au/10374/springs.html

As far as the original problem in this thread, one issue I have with the solution is that we tacitly assume that the coefficients of kinetic and static friction are equal. A more realistic model (with $\mu_k < \mu_s$) would reduce the force needed to just move the second mass.

Last edited by a moderator: May 2, 2017
8. Oct 29, 2005

### maverick280857

The links are great Doc. Thanks :-)