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I'm stuck on getting the solutions the professor gave us for this problem.

First of all the figure:

enter image description here

He asks us for the maximum tension $T$ of the rope between $m_!$ and $m_2$.

His solution:

$$T = \frac{m_1 + 2m_2}{m_1+m_2} m_2 g$$

I cannot think about a way of getting it, and I'm thinking it's wrong.

From the point of view of the two masses I shall have

$$ \begin{cases} m_2 g - T = m_2 a \\ m_1 a = T \end{cases} $$

But this leads me to $$T = \frac{m_1 m_2 g}{m_1 + m_2}$$

Where am I wrong?

DETAILS ADDED

The first question of the problem was to find the law of dynamics. This is what I have done:

From the point of view of $m_1$ I wrote down Newton's equations:

$$P_2 - F_k = (m_1 + m_2) a$$

$$m_2 g - kx = (m_1 + m_2) \ddot x$$

This can be written as

$$(m_1 + m_2) \ddot x + kx - m_2 g = 0$$

Dividing by $m_1+m_2$ and the differential equation can be solved analytically and gives

$$x(t) = \frac{m_2 g}{k} + c_1 \cos\left(\sqrt{\frac{k}{m_1+m_2}} t\right) + c_2 \sin\left(\sqrt{\frac{k}{m_1+m_2}}t\right)$$

By the initial condition $x(0) = d$ (as the problems states, that is, at $t = 0$ the spring is in its rest length $d$) we can see $c_2 = 0$ and the final solution is

$$x(t) = \frac{gm_2}{k} + \left(d - \frac{gm_2}{k}\right)\cos\left(\sqrt{\frac{k}{m_1+m_2}}t\right)$$

Any help now? Is also this right?

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  • $\begingroup$ What about the force exerted by the spring? $\endgroup$
    – Farcher
    Aug 29 '18 at 16:15
  • $\begingroup$ @Farcher It seems it doesn't take it into account. Like if there were a wall at its place.. meh. $\endgroup$
    – Les Adieux
    Aug 29 '18 at 16:26
  • $\begingroup$ I think one should start with the spring force taken into account and maximize the tension, at which the mass m2 stops. $\endgroup$
    – drvrm
    Aug 29 '18 at 17:06
  • $\begingroup$ the force of the spring needs to be included in your first equation and your textbook must have some kind of expression for x(t) that gives the result. $\endgroup$ Aug 29 '18 at 17:10
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    $\begingroup$ Neither solution makes sense as the problem is stated! If this is statics problem, then $T= m_2g$. If it is a dynamics problem (and the figure has an $x(t)$) then initial conditions need to be given, and neither the prof's or your solution seems to include them. $\endgroup$
    – mike stone
    Aug 29 '18 at 22:20
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Your solution for $x(t)$ is almost there. The problem you are having is that you don't know what $d$ is, it is an arbitrarily defined constant and you need to get rid of it. I'm going to start from the beginning and rewrite your free-body equations and work through the problem. The two equations are

$$ m_1 a_1 = T - k x(t), \\ m_2 a_2 = m_2 g - T, $$

where I have made sure that the coordinate systems are such that the two tension forces are pulling against each other. Now, we assume the cable is rigid, and as a consequence, the two masses will have the same acceleration: $a_1 = a_2$. We can then subtract the two equations to find (after some manipulation)

$$ T = \frac{m_2 k x(t) + m_1 m_2 g}{m_1 + m_2}. $$

Note that $m_1$, $m_2$, $k$ and $g$ are positive quantities, so the maximum tension will occur when $x(t)$ is a maximum. We can add the two equations to find the differential equation you did and use the boundary conditions you give ($x(0) = d$ and $\dot{x}(0) = 0$) to find what you did

$$ x(t) = \frac{m_2 g}{k} + \left( d - \frac{m_2 g}{k} \right) \cos\left( \sqrt{\frac{k}{m_1 + m_2}} t \right). $$

But $d$ is still an arbitrary constant and we've used up our two force equations finding $T$ and $x(t)$, so we need another equation. Let's use conservation of energy:

$$ \frac{(m_1 + m_2) \dot{x}(t)^2}{2} + \frac{k x(t)^2}{2} - m_2 g x = 0. $$

Note the (-) sign in front of the gravitational potential term is because we chose a (+) sign for when the spring is expanded and that happens when the mass $m_2$ is lower in the Earth's gravity well, i.e. the cable is still rigid.

We can take the time derivative of $x(t)$, plug both into the conservation of energy equation and solve for $d$ to find $d = 2 m_2 g / k$. This leads to

$$ x(t) = m_2 g \left[1 + \cos\left( \sqrt{\frac{k}{m_1 + m_2}} t \right) \right]. $$

This is a maximum when $x(t) = 2 m_2 g$. If you plug this in to the above equation for $T$, you'll get your professor's solution.

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Using your 2nd law equations with the spring force included and solving for the tension, with $F_k$ directed toward the stationary vertical wall on the left and eliminating $a$, we get

$T = \frac{m_1m_2g +m_2kx}{(m_1 + m_2)}$

The tension $T$ will be maximum at $x = x_{max}$. The force due to the spring will be directed opposite the tension and will be maximum at this point. Also, both masses will not be moving at this point or the point $x=x_{min}$. There is no maximum or any extremum for $T$ in the interval $x_{min}<x<x_{max}$, so we cannot use the first derivative to find $x$ that will give $T_{max}$ in the interval. We need to find $x_{max}$ by some other method. Fortunately there is one available.

The spring mass system is executing simple harmonic motion with no friction or damping. Consequently, the total mechanical energy of the system is a constant, independent of time. At $x_{max}$, the total energy is all potential energy stored in the spring and is equal to $\frac{k*x_{max}^2}{2}$. At the minimum value of $x = x_{min}$, the total energy of the system is the potential energy of the height $x_{max}$ of mass $m_2$ above the datum at $y=y_{min}$ (that we can set to zero) and is equal to $m_2gx_{max}$. Equating these energies and solving for $x_{max}$ gives

$x_{max} = \frac {2m_{2}g}{k}$.

The spring force $F_k=kx_{max}$ is equal to $2m_2g$. Inserting this into the equation for $T$ will give you the answer for $T_{max}$ given by the professor.

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