Solving Mechanics Problem with Friction Force

[voko]

It is still not obvious to me that the minimal force would be different from the rigid case. But I will have to postpone further analysis till tomorrow.

 

[kushan]

Ok this case is same as a vertical SHM in presence of gravity , like a spring tied to a tree with a mass m hanging , now if mass m is increased or in other words ( a force F is added) the SHM will still continue .
The relation to that with this question is that , For Force to be just minimum block M has to act like the stationary support and gravity as F.
Is this thing correct Ehild?

 

[ehild]

Not quite Kushan.

For this problem, Newton's second Law states for block m:

ma=F-μmg-kx.
For the mass hanging on spring:

ma=mg-kx.
You see the constant force is mg in the second case and F-μmg in the first one.

As the force of friction changes sign in every half period, it is not real SHM after the first maximum displacement.

For the time span from t=0 to the first maximum, the solution is

x=(F-μmg)/k[1-cos(wt)] (w=√(k/m)),
as initially x=0 and v=0.

How does x and v change with time? When is x maximum? What is the velocity when x is maximum?

I stop for today, and go to sleep. :zzz:

ehild

 

[voko]

To move the bigger mass, $\mu M g = k x$. On the other hand, $\frac {kx^2} {2} + \frac {m v^2} {2} = (F - \mu m g)x$. Since F is minimal, $\frac {m v^2} {2} = 0$, so $\frac {\mu Mg x} {2} = (F - \mu m g)x$. Indeed, we need a lesser force than in the rigid case.

 

[azizlwl]

I'll rephrase my question: does anything change with regard to the magnitude of the force if the link is rigid?

Is there different in force if we use different springs of different coefficient k.
What about if the value is so big where it is equivalent to rigid body.
Anyway no data given about the spring.

In calculating springs in series, the forces on each spring are equal.

 

[ehild]

To move the bigger mass, $\mu M g = k x$. On the other hand, $\frac {kx^2} {2} + \frac {m v^2} {2} = (F - \mu m g)x$. Since F is minimal, $\frac {m v^2} {2} = 0$, so $\frac {\mu Mg x} {2} = (F - \mu m g)x$. Indeed, we need a lesser force than in the rigid case.

Right!

ehild