What Is the Maximum Amplitude of Oscillation to Prevent Slipping?

[toesockshoe]

Homework Statement

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

Homework Equations

F=ma

The Attempt at a Solution

so i did 2 F=ma problems... and BOTH THE MASSES ARE EQUAL... atleast that's what I am assuming from the problem becuase they both have mass, m. (although having different masses wouldn't 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

$F_s = ma$ [/B]
$\mu mg = ma$

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?

 

[Chestermiller]

Homework Statement

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

Homework Equations

F=ma

The Attempt at a Solution

so i did 2 F=ma problems... and BOTH THE MASSES ARE EQUAL... atleast that's what I am assuming from the problem becuase they both have mass, m. (although having different masses wouldn't 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

$F_s = ma$ [/B]
$\mu mg = ma$

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?

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

 

[toesockshoe]

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

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

 

[SammyS]

Homework Statement

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

Homework Equations

F=ma

The Attempt at a Solution

so i did 2 F=ma problems... and BOTH THE MASSES ARE EQUAL... atleast that's what I am assuming from the problem becuase they both have mass, m. (although having different masses wouldn't 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

$F_s = ma$ [/B]
$\mu mg = ma$

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?

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 .

 

[toesockshoe]

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 .

i did do that... didn't the poster above me say that I am 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?

 

[SammyS]

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

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. F_S = 2ma .

 

[toesockshoe]

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 forced exerted by the spring must = 2ma.

I.e. F_S = 2ma .

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.

 

[SammyS]

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.

Yes. It really helps to define those forces.

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