What Is the Maximum Amplitude of Oscillation to Prevent Slipping?

• toesockshoe
In summary, 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.
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

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?

toesockshoe said:

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

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

Chestermiller said:
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...

toesockshoe said:

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

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 .

SammyS said:
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?

toesockshoe said:
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. FS = 2ma .

Last edited:
SammyS said:
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. FS = 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.

toesockshoe said:
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\ .##

What are oscillations and why are they important?

Oscillations refer to the repetitive back and forth motion of an object around its equilibrium position. They are important because they can help us understand and predict the behavior of various systems, such as pendulums, springs, and electrical circuits.

How do I solve an oscillations problem?

The first step in solving an oscillations problem is to identify the type of oscillation (e.g. simple harmonic, damped, forced) and determine the relevant equations or formulas to use. Then, plug in the given values and solve for the unknown variable. It can also be helpful to sketch a diagram or graph to visualize the problem.

What factors affect the frequency and amplitude of oscillations?

The frequency of oscillations is affected by the stiffness of the system (e.g. spring constant), the mass of the object, and any external forces acting on the system. The amplitude of oscillations is affected by the initial displacement and the energy of the system.

How do I calculate the period of oscillation?

The period of oscillation, or the time it takes for one complete cycle, can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant. If the system is not a simple harmonic oscillator, the period can be found by solving for the time it takes for the object to return to its initial position and direction of motion.

What are some real-life examples of oscillations?

Oscillations can be seen in many everyday objects and phenomena, such as a swinging pendulum, a bouncing basketball, a vibrating guitar string, and the motion of a rollercoaster. They are also used in various technologies, such as clocks, radios, and seismographs.

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