# Help with oscillations problem

1. Jun 20, 2015

### toesockshoe

1. The problem statement, all variables and given/known data
A block with mass m rests on a frictionjless surfae and is connected to a horizontal spring of force constant k. The other end of the spring is attahced to a wall. A second block with mass m rests on top of the first block. The coefficient of static friction between the blocks is given. Find the maximum amplitude of the oscialltion such that the top block will not slip on the bottom block

2. Relevant equations

F=ma
3. The attempt at a solution

so i did 2 F=ma problems... and BOTH THE MASSES ARE EQUAL.... atleast that's what im assuming from the problem becuase they both have mass, m. (although having different masses wouldnt make it much more difficult... just a bit more algebra)

1.
F=ma ... system m on bottom..

$F_{el} - F_s = ma$
$-kA - \mu _s mg = ma$

F = ma... system m on the top

$F_s = ma$

$\mu mg = ma$

let the 2 ma's equal each other to get... but this poses a problem because when I set them equal... i get

$kA = 0$
because the two $\mu _s mg$ 's cancel out....

where am i going wrong?

2. Jun 20, 2015

### Staff: Mentor

What do you get if you add your following two equations together:

$F_{el} - F_s = ma$

$F_s = ma$

The static friction force is not equal to $\mu _s mg$ until the to block is just about to slip. You evaluate this in the very last step of your analysis.

Chet

3. Jun 20, 2015

### toesockshoe

ahh... i see. i think finals prep is penetrating into my 5th grade math knowledge....

4. Jun 20, 2015

### SammyS

Staff Emeritus
Yes, each block has mass, m, so their masses are the same.

Step #1. Draw a free body diagram for each block.

By the way, the spring must accelerate both blocks, so most of a correct solution will involve a mass of 2m .

5. Jun 20, 2015

### toesockshoe

i did do that... didn't the poster above me say that im correct? i just made a really dumb mistake when setting the 2 F=ma's equal... is there something worng with my F=ma equations?

6. Jun 20, 2015

### SammyS

Staff Emeritus
You didn't mention a free body diagram nor any direct results you got from them.

I don't see where Chet said you were correct.

You don't define much in the way of what your variables/symbols mean.

The problem states that each block has mass, m. I expect that means that the force exerted by the spring must = 2ma.

I.e. FS = 2ma .

Last edited: Jun 20, 2015
7. Jun 20, 2015

### toesockshoe

on the bottom mass, there is a spring attached to it. ill describe by free body diagram to you: the spring is pulling the mass to one side (lets say to the right)... and there is another mass on top that has a static friction which is to the left (becuase it causes the top mass to move to the right.... so it MUst act to the left on the bottom mass through NTL).... thus for the bottom system ( in the horizontal direction) the equation I got was: $F_{el} - F_s = ma$.... el stands for the elastic force by the spring, and s stands for static friction (the friction force). On the top mass, there is only one force acting in the horizontal direction.... the friction force. so for the top mass, i have $F_s = ma$ ... again s stand for static... NOT spring. Isn't this correct?

furthermore... if you go on to set both my F=ma's equal.... you will see that there will be a 2m.

8. Jun 20, 2015

### SammyS

Staff Emeritus
Yes. It really helps to define those forces.

So, I suppose that you did as Chet suggested and found that $\ F_{el}=2ma\ .$