# Amplitude and friction

• abeltyukov

Hi,

## Homework Statement

There is a 2.0 kg mass riding on top of a 3.8 kg mass as it oscillates on a frictionless surface with a period of 3.0 s. The upper block just begins to slip when the amplitude is increased to 50 cm. What is the coefficient of static friction between the blocks? The spring constant is 70 N/m.

2. Homework Equations 3. The Attempt at a Solution

I am having a hard time incorporating friction into my equations. I believe that I am supposed to use F=uN, w = (k/m)^.5 as well as w = 2(pi)f and V = wA. Any ideas on how to start this one?

Thanks!

you may be making this more difficult than it need be. Forget the periodicity/ harmonic motion for a moment.

The lateral (restoring) force on the big block is at a maximum where?

What is the magnitude?

What is the normal force we need to know?

Under what condirions wiull slippage occur?

The restoring force is maximum when the amplitude is the highest, right? Here's what I tried doing:

A = (a(m1+m2)) / k

I solved for acceleration to get 6.03 m/s^2

I then used F = ma
I found F to be 35 N

I then used F = m1(u * g)
I got u to equal 0.892, which is still the wrong answer.

Am I still making it more complicated than it is?

Thanks!

heres how I see it. Peak force = 35N just as you did.

But remember the blocks are partly decoupled if slippage occurs--a common example is a golfball sitting on a tee. so its not m1 + m2 that are being accelerated when slippage occurs, its the bottom block.

Imagine instead that you have a pickup truck that has a box in the back. The pickup can accelerate up to a certain magnitude before the box begins to slide toward the tailgate? Does this help?

heres how I see it. Peak force = 35N just as you did.

But remember the blocks are partly decoupled if slippage occurs--a common example is a golfball sitting on a tee. so its not m1 + m2 that are being accelerated when slippage occurs, its the bottom block.

Imagine instead that you have a pickup truck that has a box in the back. The pickup can accelerate up to a certain magnitude before the box begins to slide toward the tailgate? Does this help?

So am I using the wrong acceleration? Should the equation be:

A = (a(mass of bottom block)) / k?

If so, I get u = 0.939, which still is the wrong answer. I think I misunderstood you.

I get the box in the pickup truck. So the bottom block is like the truck and the top block is like the box.

acceleration of big block= 35/38=.92

for small block to remain stationary,

acceleration of big block= 35/38=.92

for small block to remain stationary,

Did you mean 35/3.8? If so, I get friction to be 0.939, which is not the right answer.

Thanks!

i get an answer of 0.32

i get an answer of 0.32

How did you get that?

Thanks!

I think I have the wrong spring constant. The spring constant is actually unknown. I tried to calculate the k using the following equation:

k = (m1+m2)((2pi)(f))^2

I then used A = (u * g (m1+m2)) / k

Do I use m1 + m2 in both of the cases or do I just use one mass to solve for k?

Thanks!

ah I see. well if the two blocks are moving together the m1 plus m2 is the right mass to use in computing the k via the period. BTW This is why we ask for the whole problem exactly as stated. Too often a mistake made upstream of the actual post causes anyone attempting to help, a migraine headache. I think youre clear in this respect. But show all work in the future including derivation of k.

ah I see. well if the two blocks are moving together the m1 plus m2 is the right mass to use in computing the k via the period. BTW This is why we ask for the whole problem exactly as stated. Too often a mistake made upstream of the actual post causes anyone attempting to help, a migraine headache. I think youre clear in this respect. But show all work in the future including derivation of k.

Yes, sorry about that. Thank you very much for your help. I got the right answer of 0.223.

NP, no harm, no foul.