A block with mass M rests on a frictionless surface and is connected to a horizontal spring of force constant k, the other end of which is attached 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 mu*s. Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.

I did the exact same question as this only with numbers and got the right answer subbing my numbers into
A=mu(coefficient of static friction*g/w^2

So, i thought the answer would be
A=(mu*g*m)/k
But I was told that the correct answer does not depend on mu (the coefficient of static friction). So I'm guessing it must get cancelled out of the equation? I'm confused because I needed it in the first question when I actually subbed numbers in?

Well, I can tell you the coefficent of friction does enter into the answer. Consider nofriction between the blocks, than any acceleration will make the top block slip. If the coefficent is infinite, as you nailed the two blocks together, then you could have a very large acceleration and they wouldn't slip.

My concern here is what does Mu*s mean? does that mean the static coefficent of friction? If so, I stick with my above answer. If it means the static coefficent times some value s, then I need to know what s is.

Let look at it.

x= Asinwt
v= Aw coswt
a= -A w^2 sinwt

So the max occurs at sinwt=1, or max acceleration is Aw^2

Now, the max force on the top block is mu*mg, so acceleration <mu*mg/m
acceleration<mu*g

or Aw^2<mu*g
but w is... sqrt (k/(M+m)
Ak/2M<mu*g
or A<mu*g*(M+m)/k

m is the mass of the top box, M the bottom.

Ah, coefficients of friction, it's a slippery slope of confusion! But fear not, I'm here to untangle this mess with a touch of humor!

In this scenario, where the top block is resting on the bottom one, we're dealing with the force required to keep the top block from slipping off due to the static friction. However, when we consider the oscillations of the system, the magic of physics cancels out the coefficient of static friction.

It's kind of like when you're calculating your expenses but realize that you've been double-counting the cost of your daily coffee fix. Just because you needed it for one part of the problem doesn't mean it's applicable to the entire situation.

So, how do we find the maximum oscillation amplitude without the coefficient of static friction? We turn to the other forces at play. A key player here is the force exerted by the spring, counteracting the gravitational force on the top block. By balancing these forces, we can find the maximum amplitude.

Rather than relying on the coefficient of static friction, the correct equation for the maximum amplitude of oscillation is:

A = (m*g)/(k + M/m)

So, no need to worry about the coefficient of static friction slipping into the equation. It has already taken its leave, vanishing into thin air, like a magician's assistant disappearing in a cloud of smoke. Poof!

Remember, in physics, sometimes things that seem important in one context may not be relevant in another. It's all about understanding the specific forces at play in the given scenario. Keep your sense of humor intact, and your physics journey will be full of laughter and knowledge!

To find the maximum amplitude of oscillation such that the top block will not slip on the bottom block, we need to consider the forces acting on the system.

When the system is at its maximum amplitude, the second block is just about to slip on the first block. At this point, the force of static friction between the two blocks is at its maximum. Let that maximum static friction force be denoted as F_max.

The system experiences three forces:
1. The force of gravity acting on the second block (mg).
2. The force exerted by the spring (kx), where x is the displacement of the first block from its equilibrium position.
3. The maximum static friction force between the two blocks (F_max).

To prevent slipping, the force of static friction (F_max) must be equal to or less than the maximum static friction force (mu_s * (m + M) * g), where mu_s is the coefficient of static friction.

Since the force of static friction can be represented as F_max = mu_s * (m + M) * g, we can set this equal to the net force acting on the top block:
F_max = m * x'' + k * x.

We can rearrange this equation to get:
k * x = F_max - m * x''.

Using the equation F = m * x'' (Newton's second law), we can replace m * x'' with F_max / m:
k * x = F_max - F_max / m.

Simplifying further, we get:
k * x = F_max * (1 - 1/m).

Since x = A, where A is the amplitude of oscillation, we have:
k * A = F_max * (1 - 1/m).

Now, let's substitute the expression for F_max:
k * A = mu_s * (m + M) * g * (1 - 1/m).

Notice that mu_s appears in the equation. Thus, the maximum amplitude of oscillation (A) does depend on the coefficient of static friction (mu_s). It seems there might be an error in the statement that the correct answer does not depend on mu_s.

However, without specific numbers or additional information, it is difficult to determine whether there was an error in your previous calculation or if there is a misunderstanding in the problem statement.

To find the maximum amplitude of oscillation such that the top block will not slip on the bottom block, you can start by analyzing the forces acting on the blocks.

Let's consider the forces acting on the top block first. There are three forces acting on it: the gravitational force (mg) pointing downwards, the force from the spring (kA) pointing upwards, and the force of static friction (μsN) which opposes any tendency for the top block to slip relative to the bottom block. Here, N represents the normal force exerted by the bottom block on the top block.

Since we want to find the maximum amplitude of oscillation without any slipping, the force of static friction should be at its maximum value, which is μsN. When the blocks are about to slip, the force of static friction reaches its maximum, thus preventing slipping.

To calculate the normal force N, we can consider the forces acting on the bottom block. The normal force exerted by the surface on the bottom block cancels out the gravitational force of the bottom block, so N = mg.

Now, we can equate the forces acting on the top block:

mg - kA - μsN = 0

Substituting N = mg, we have:

mg - kA - μsmg = 0

Rearranging the equation, we find:

kA = (m + μsm)g

Finally, solving for the maximum amplitude A:

A = [(m + μsm)g] / k

Therefore, the correct expression for the maximum amplitude of oscillation is A = [(m + μsm)g] / k.

Now, why does the coefficient of static friction (μs) seem to cancel out when you substitute numbers into the equation?

This is because when you substitute values into the equation, you also need to consider the relationship between the masses of the blocks and the coefficient of static friction. In this case, when you substituted your specific numbers, the μs term might have canceled out with the (m + μsm) term.

However, it's important to note that this cancellation of the coefficient of static friction term is specific to your example and the relationship between the masses and the coefficient of friction in that particular case. In general, the coefficient of static friction (μs) is an important factor that affects the maximum amplitude of oscillation and cannot be canceled out in the equation.