# Force on a two block and a spring system

1. Jun 16, 2013

### Saitama

1. The problem statement, all variables and given/known data
Two bars connected by a weightless spring of stiffness $k$ and length (in the non-deformed state) $\ell_o$ rest on a horizontal plane. A constant horizontal force $F$ starts acting on one of the bars as shown in the figure. Find the maximum and minimum distance between the bars during the subsequent motion of the system, if the masses of the bars are:
(a)equal
(b)equal to $m_1$ and $m_2$ and the force $F$ is applied to the bar of mass $m_2$.

a)$l_{max}=l_o+F/k$ and $l_{min}=l_o$
b)$l_{max}=l_0+2m_1F/(k(m_1+m_2))$ and $l_{min}=l_o$

2. Relevant equations

3. The attempt at a solution
Case a):
From energy conservation,
$$Fx=\frac{1}{2}kx^2+\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2$$
where $x$ is the extension of the spring, $m$ is the mass of the blocks, $v_1$ and $v_2$ are the velocities of the blocks.
I still need one more equation to relate $v_1$ and $v_2$.

Thanks!

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2. Jun 16, 2013

### Tanya Sharma

Why should you go for energy conservation ?

Think for a while,how would the two blocks be moving with respect to each other when separation between them is minimum and when it is maximum .If you are able to figure out this,the problem becomes easy.

3. Jun 16, 2013

### voko

There could be a smarter way, but if everything else fails, just write down the diff. eq.'s of the system and solve them. Should not be too hard.

4. Jun 16, 2013

### Saitama

How should I write those diff eq.'s, can I have a few hints? :)

5. Jun 16, 2013

### voko

6. Jun 17, 2013

### Saitama

Forces (except the weight and normal reaction from ground) acting on $m_2$ are $F$ and force due to spring i.e $kx$ where $x$ is the extension of spring.
$$F-kx=m_2a_2$$

The only force acting on $m_1$ is $kx$
$$kx=m_1a_1$$

From the two equations, $$F=m_1a_1+m_2a_2$$
Should I replace $a_1$ with $dv_1/dt$ and $a_2$ with $dv_2/dt$? Is the above equation correct?

7. Jun 17, 2013

### voko

Yes, your equations are correct, assuming you can express $x$ via $x_1, \ x_2$ and $l_0$.

8. Jun 17, 2013

### Saitama

$x=x_1+x_2-l_0$?

So $a_1=d^2x_1/dt$ and $a_2=d^2x_2/dt$?

9. Jun 17, 2013

### voko

Let's say $x_1 = 0$ and $x_2 = l_0$. Then $x = 0$ as one would expect - the spring is relaxed. Now let's move $m_1$ toward $m_2$, say, by letting $x_1 = l_0/2$. Then $x = l_0/2$, i.e., greater than in the relaxed state, while it should be less.

10. Jun 17, 2013

### consciousness

I would prefer to do this problem by analyzing the motion of m2 w.r.t. m1. This removes the hassle of two variables x1,x2 leaving only one whose maximum we have to find. Remember to make the FBD of m2 carefully though.

11. Jun 17, 2013

### Saitama

I thought more about it. The best I could come up with is $x=x_2-x_1-l_0$. Is this correct?

Also, where should I use it?

I have $F=m_1a_1+m_2a_2$ but I don't have $x$ here.

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Last edited: Jun 17, 2013
12. Jun 17, 2013

### voko

Check if your equations make sense. Keep one variable fixed, then another one. Is that consistent with the simple "one mass" case? Pay attention to the signs.

13. Jun 17, 2013

### Saitama

You mean to say my equation for x is incorrect? I can't think of anything else. :(

If I consider your cases (post #9), when $x_1=0$ and $x_2=l_0$, $x=0$ i.e for this case, my equation is correct. When $m_2$ is fixed and $m_1$ is moved a distance $l_0/2$ i.e $x_2=l_0$ and $x_1=l_0/2$, $x=-l_0/2$, this is also correct, I can't see where my equation is wrong.

14. Jun 17, 2013

### voko

I am not saying anything is incorrect. I am saying that I would like you to check whether your equations in #6 are consistent with the your definition of x.

15. Jun 17, 2013

### Saitama

But still, where should I use this equation for $x$?

16. Jun 17, 2013

### voko

Did I not say #6?

17. Jun 17, 2013

### Saitama

The final equation I had was $F=m_1a_1+m_2a_2$ but there is no x in this.

18. Jun 17, 2013

### voko

Does that make it automatically correct or even useful to begin with?

19. Jun 17, 2013

### Saitama

I am still not sure what to do. I will begin with using $kx=m_1a_1$ as it involves $x$.
$$k(x_2-x_1-l_0)=m_1\frac{d^2x_1}{dt^2}$$

I can replace $x$ and $a_2$ in the equation, $F-kx=m_2a_2$ but what I am supposed to do with these equations?

20. Jun 17, 2013

### voko

Hmm. I think I outlined the general approach in #3, no?