# Mechanics, Static friction and springs

1. Nov 14, 2009

### Unto

1. The problem statement, all variables and given/known data
A block of mass m lies on a rough surface, inclined at angle theta to the horizontal. It is attached to a spring of force constant; the other end of the spring is fixed to a point on the table below the inclined block. The co-eff of static friction is mu-s and kinetic friction is mu-k. friction between the table and spring is negligible.

a) show that the friction force is 0 when the block is stationary and the spring is compressed by l sin theta where l = mg/k

2. Relevant equations
f = kl (spring force)
Fs = mu-s x Fn (friction force as a product of reaction force and co-eff of friction)

3. The attempt at a solution

i first drew a diagram and said that the system is in equilibrium. If the block is not moving, the the compression force from the spring + the friction force (acting against this compression) must = the wait of the block down the inclined plane.

So Fs = mu-s x Fn

Fn = mg sin theta, so Fs = mu-s x mg sin theta.

Therefore sum of forces = 0

kl (spring force) + mg sin theta + mu-s mg sin theta = 0

But I do not understand how the friction force can = 0.

The closest result I have been able to get (after finally assuming friction force = 0) is l = mg sin theta/k

I don't know what I'm doing wrong because I resolved the forces in the directions they are acting?/

2. Nov 14, 2009

### Unto

If I assume that the compression force works together with the friction force to oppose the motion of the b lock down the slope, I could assume that force from the spring >>>>> friction force? Therefore it is safe to assume that Friction force is 0 when the spring is in play, although I still can't work out how they got to their expression..

l sin theta, where l = mg/k. I only get l = mg sin theta / k..

It would help if I knew what l stood for, at first I thought it was a distance/length but I'm not so sure now...

3. Nov 14, 2009

### tiny-tim

Hi Unto!

(have a theta: θ )
You're making this very complicated!

You should have started "Let the distance be x".

Then you would have got x = mg sinθ / k (instead of your l = mg sinθ / k),

and therefore x = l sinθ, where l = mg/k.

4. Nov 14, 2009

### Unto

OMG!!!!!

I literally thought I was stupid and was going out of my head. I kept resolving, I even used F = kx and was like wtf am I doing wrong...

Thank you!!!!

There are other parts to this question, I will post them and my attempts if I get stuck if that is ok with you.

Thank you very much

5. Nov 14, 2009

### Unto

The block is now displaced by a distance A, compressing the spring further. For large displacements find the initial acceleration of the block.

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K so the block is going to get pushed up the inclined plane by the force from the spring due to it's compression. I gathered also that the friction is no longer static, but kinetic as it is opposing a now moving object.

So, SUM of Forces = ma

kx - (mu-k) mg cos θ - mg sin θ = ma.

Is this right? The compression force is being opposed by the weight of the block down the plane and the kinetic friction.

6. Nov 15, 2009

### tiny-tim

(just got up :zzz: …)
(have a mu: µ )

Yes, that's fine.

(but don't forget to write x in terms of A, and to use the first part of the question)

7. Nov 15, 2009

### Unto

Ok thank you. 1 Last question.

Since the acceleration of the block is kA - µk mg cos θ - mg sin θ = ma, I was asked to formulate an expression when acceleration was 0. I obtained:

x or A = l(µk cos θ + sin θ). This was from re-arranging the above equation once ma = 0.

I am now asked to find V at this point. If acceleration is 0 at this displacement, should I still use V^2 = U^2 + 2as? Actually no..

I can't use the others since they have time in them.

I cam up with V = sqrt(2l(µk cos θ + sin θ)) where l = mg/k from our previous expressions.

I'm not sure this is correct however.

When the spring is compressed to a certain displacement and released, the block is accelerated until the spring finishes relaxing (since the force from the spring is a resultant force), and even then, there is still the kinetic friction slowing it down until it stops. So where am I supposed to find V with so many things going on?

I assumed that S = x when there is no more acceleration. Since there is no more resultant force, and therefore no more acceleration when the spring has finished relaxing. But this would mean I have ignored Kinetic friction

But hasn't kinetic friction already been included in my above expression for when a = 0?

Man I'm confused :s. I really love Physics, I just always have to question what I do, I hate just following examples and not knowing why.

8. Nov 15, 2009

### Unto

This last question?

9. Nov 16, 2009

### tiny-tim

Hi Unto!
That's mA'' = kA + constant.