Two bodies connected with an elastic thread moving with friction

  • #1
HighPhy
89
8
Homework Statement
Consider two bodies ##A## and ##B## connected as in the figure below. The horizontal surface where B is laid is rough with static friction coefficient ##\mu_s = 0.4## and kinetic friction coefficient ##\mu_k = 0.35##. The mass of B is ##m_B = 4 \ \mathrm{kg}##, the wire that connects ##A## and ##B## is an elastic of negligible mass and elastic constant ##k = 160 \ \mathrm{N/m}##; initially the system is at rest because ##A## is supported with an external force and the elastic has a length equal to that at rest. At instant ##t = 0##, the force ##m_A\boldsymbol{g}## that holds ##A## in equilibrium is eliminated.

##a)## What is the maximum value ##m^{\ast}## that the mass of ##A## can have if ##B## is to remain stationary?

##b)## If ##m_A = 3m^{\ast}##, with which acceleration does ##A## leave and at what instant, and with which acceleration does ##B## leave?
Relevant Equations
Too many to summarize. See "Attempt at a Solution".
(This is a homework assignment that my sister (younger than me) didn't manage to solve, but I am not sure about the attempt I thought of either. Especially in the last point. So I ask you to correct where I am wrong).

I solved ##a)## with the following:

1. Since ##B## must be stationary, I derived the maximum elongation of the spring from the equation of harmonic motion of ##A## (solution of a non-homogeneous ODE);

2.I imposed equilibrium on mass ##B## using the above elongation and then equated the corresponding elastic force to the static friction force between mass ##B## and the supporting plane. The relationship between ##m^{\ast}## and ##m_B## is thus obtained;


I am developing these two points.

1. Suppose we let the mass down gently so it comes to rest at the equilibrium length of the spring.
This equilibrium length is given by ##ky_0 = mg##, so: ##y_0 = \frac{m_Ag}{k}##.
Now we'll define a new variable ##z## as ##z = y - y_0##, so using this new variable the equilibrium position of the mass is ##z = 0##.
And then the EoM using the variable ##z## is:
$$\frac{\mathrm{d}^2z}{\mathrm{d}t^2} = -\frac{k}{m_A}z \ .$$
Suppose we displace the mass a distance $z$ from the equilibrium position, then the net force on the mass is $$F = kz \ .$$
And if we solve this equation we get:
$$z(t) = A \sin(\omega t) + B \cos(\omega t) \ ,$$
where ##\omega^2 = \dfrac{k}{m} \ .##

Now we know that at time ##t = 0## we have ##z = -y_0## and ##\frac{\mathrm{d}z}{\mathrm{d}t} = 0 \ .##
And if we substitute these values of ##z## and ##t## into our equation we find ##A = 0## and ##B = -y_0 \ .##
So our equation is:
$$z(t) = -y_0 \cos(\omega t) \ .$$
And we defined our variable ##z## as:
$$z = y - y_0 \ .$$
So now we can substitute for ##z## to get:
$$y - y_0 = -y_0 \cos(\omega t) \ .$$
And rearranging gives:
$$y = y_0 - y_0 \cos(\omega t)$$
or:
$$y = y_0(1 - \cos(\omega t)) \ .$$

Ultimately:
\begin{equation}
y = \frac{m_Ag}{k}\left(1 - \cos (\omega t) \right)
\end{equation}
Deriving with respect to time, we obtain the velocity:
$$\dot{y} = \frac{m_Ag}{k} \omega \sin (\omega t) \ ,$$
which is zero at ##\omega t= \pi \quad (1.1)##, instant at which the elongation (and thus the restoring force) is maximum. So:

$$\dot{y} = 0 \iff \omega t= \pi$$

Plugging ##(1.1)## in ##(1)##:

$$y_{\mathrm{max}} = \frac{m_Ag}{k}\left(1 - \cos (\pi) \right) = \frac{2m_A g}{k}$$

2. For balance of forces:

$$F_{\mathrm{friction}} = F_{\mathrm{elastic}} \implies F_{\mathrm{friction}} = \cancel{k} \frac{2m_A g}{\bcancel{k}} = 2 m_A g \ .$$

Using the fundamental inequality of friction ##F_{\mathrm{friction}} \leq F_{\mathrm{friction, \ max}} = \mu_s m_B g##:

$$2 m_A g \leq \mu_s m_B g \Rightarrow m_A \leq \frac{\mu_s}{2} m_B \ ,$$

from which

$$\color{Red}{\boxed{\color{Black}{m_{A, \ \mathrm{max}} = m^{\ast} = \frac{\mu_s}{2} m_B = 0.8 \ \mathrm{kg}}}}$$

##b)## I obtained the acceleration with which mass ##A## departs by deriving the EoM ##(1)## twice and setting ##t = 0##.

$$\color{blue}{\mathrm{First \ step)}} \ \color{Black}{\ddot{y} = \frac{mg}{k} \omega^2 \cos(\omega t) \stackrel{\omega^2 = k/m}{\boldsymbol{=\!=\!=\!=}} \frac{\cancel{m}}{\bcancel{k}} g \frac{\cancel{k}}{\bcancel{m}} \cos(\omega t) = g \cos(\omega t)}$$

$$\color{blue}{\mathrm{Second \ step)}}
\ \color{Black}{\ddot{y} = g \cos(\omega t) \stackrel{t = 0}{\boldsymbol{=\!=\!\Rightarrow}}\ddot{y} = g \cos 0} \Rightarrow \color{Red}{\boxed{\color{Black}{\ddot{y} = g}}}$$


I then obtained after how long ##B## starts by substituting the elongation of the wire that starts ##B## found previously (i.e. the one that wins the force of static friction) in the EoM of mass ##A## .

Mass ##B## leaves when ##F_{\mathrm{friction}} = F_{\mathrm{friction, \ max}} = \mu_s m_B g##. From balance of forces:

$$F_{\mathrm{friction}} = F_{\mathrm{elastic}} \implies \mu_s m_B g = k \delta \ ,$$

where ##\delta## is the elongation of mass ##B##.

From that:

$$\delta = \frac{\mu_s m_B g}{k} \ ,$$

and plugging it in ##(1)## by placing ##\delta = y##, I got:

\begin{align*}
\frac{\mu_s m_B g}{k} = \frac{m_Ag}{k}\left(1 - \cos (\omega t) \right)
\end{align*}
\begin{align*}
\Downarrow
\end{align*}
\begin{align*}
\left(1 - \cos (\omega t)\right) = \mu_s \frac{m_B}{m_A}
\end{align*}
\begin{align*}
\Downarrow
\end{align*}
\begin{align*}
\cos (\omega t) = \left(1 - \mu_s \frac{m_B}{m_A}\right)
\end{align*}
\begin{align*}
\Bigg{\Downarrow} \ \omega = \sqrt{\frac{k}{m_A}}
\end{align*}
\begin{align*}
\sqrt{\frac{k}{m_A}} t = \arccos{\left(1 - \mu_s \frac{m_B}{m_A}\right)}
\end{align*}
\begin{align*}
\Downarrow
\end{align*}
\begin{align*}
t = \sqrt{\frac{m_A}{k}} \arccos{\left(1 - \mu_s \frac{m_B}{m_A}\right)}
\end{align*}
\begin{align*}
\Bigg{\Downarrow} \ m_A = 3 m^{\ast} = \frac{3}{2} \mu_s m_B
\end{align*}
\begin{align*}
t = \sqrt{\frac{3}{2} \frac{\mu_s m_B}{k}} \arccos{\left(1 - \cancel{\mu_s} \dfrac{\cancel{m_B}}{\frac{3}{2} \bcancel{\mu_s} \bcancel{m_B}}\right)}
\end{align*}
\begin{align*}
\Downarrow
\end{align*}
\begin{align*}
\color{Red}{\boxed{\color{Black} {t = \sqrt{\frac{3}{2} \frac{\mu_s m_B}{k}} \arccos \left({\frac{1}{3}}\right) \approx 0.15 \ \mathrm{s}}}}
\end{align*}

I now have to calculate with what acceleration ##B## starts, a point on which I found some conceptual difficulties; in particular, for a while I could not see how to introduce the kinetic friction coefficient (since body ##B## starts from a standstill) and which equations to use.

I made the following rationale:

The static friction for B is ##\mu_s m_B g##, so ##B## starts to move when the tension in the rope is equal to ##\mu_s m_B g##.
At the instant ##B## starts to move, the friction becomes kinetic so it decreases to ##\mu_k m_B g##, but the tension in the string is still ##\mu_s m_B g \ .##
So there is a net force on ##B## given by:
$$F = \mu_s m_B g - \mu_k m_B g = (\mu_s - \mu_k) m_B g \ .$$

Applying Newton's 2nd Law on mass ##B##:

$$F = m_B a \Rightarrow (\mu_s - \mu_k) m_B g = m_B a \ , $$

from which:

$$\color{Red}{\boxed{\color{Black} {a = (\mu_s - \mu_k) g}}}$$

About last point, this is the only physical explanation I could think of. But it is not convincing to me, it seems too forced. Could you provide the best physical explanation that effectively resolves this last point?






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  • #2
I agree with your answer for ##m*##. You could have got it more easily by considering energy. The GPE lost by ##m_a## must equal the EPE at its maximum extension, which corresponds to maximum force:
$$m_Ag\Delta y = \frac 1 2 k(\Delta y)^2$$$$\implies F_{max} = k(\Delta y) = 2m_Ag$$I'll take a look at part b).
 
Last edited:
  • #3
The initial acceleration of block B must be ##a_B = (\mu_s - \mu_k)g##, because the force increases continuously until it reaches ##\mu_s m_Bg##, at which point B moves.

I'm not sure whether the question is asking you to calculate the acceleration of ##A## at this point? That should be straightforward.

To get the time, then there's no option but to calculate the equation of SHM for A. It seems strange that they threw that it, when there was a neat solution for everything else.
 
  • #4
PeroK said:
the force increases continuously until it reaches ##\mu_s m_Bg##, at which point B moves.
Shouldn't it be ##\mu_k m_B g## (kinetic friction)?
But still, regarding this point, does the reasoning I wrote sound physically accurate to you? I don't know why, but I wasn't convinced when I thought of it...

PeroK said:
I'm not sure whether the question is asking you to calculate the acceleration of ##A## at this point? That should be straightforward.
I cannot be sure, but this is my interpretation from reading the text. If not, I wouldn't know which way to go.
 
  • #5
HighPhy said:
Shouldn't it be ##\mu_k m_B g## (kinetic friction)?
Mass B doesn't move when the force reaches ##\mu_km_Bg##. It only moves when the force reaches ##\mu_sm_Bg##. At which point there is a sudden instantaeous acceleration of ##(\mu_s - \mu_k)g##.
 
  • #6
I agree with ##t = 0.15s## as well.
 
  • #7
PeroK said:
I agree with ##t = 0.15s## as well.
So you agree with all my results?
 

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