What is the tension along wire 1 after wire 2 is cut?

[Bestfrog]

Considering the figure, (a) what is the tension along the wire 1?
(b) What is the tension along wire 1 immediately after the wire 2 is cut?

Now, if I put myself in a reference frame with y-axis along $\vec{g}$ (with inverse direction) the part (a) is $$\begin{cases} T_1\cdot cos\theta -mg=0 \\ -T_1 \cdot sin\theta + T_2 =0 \end{cases}$$

So $T_1=\frac{mg}{cos\theta}$.
For part (b), because I consider the infinitesimal time after $T_2=0$ I can approximate $T_1=\frac{mg}{cos\theta}$, so the same in part (a).

Now, if I use a different reference frame, with y-axis along the wire 1(with same direction of $T_1$), I have $$\begin{cases} T_1 -mg cos\theta -T_2 sin\theta=0 \\ -T_2 cos\theta + mg sin\theta=0 \end{cases}$$

For part (b) I can approximate $T_1 \text{as} T_1= mg cos\theta$ (since the centripetal acceleration is still 0 for the first infinitesimal times).

My problem is that I have two different tensions, and it is not correct(?)

 

[Orodruin]

For part (b), because I consider the infinitesimal time after T2=0T2=0T_2=0 I can approximate T1=mgcosθT1=mgcosθT_1=\frac{mg}{cos\theta}, so the same in part (a).

This is not correct. The acceleration changes and therefore so must the force balance. Of course, this is in the ideal approximation where the system responds instantly.

 

[ehild]

Considering the figure, (a) what is the tension along the wire 1?
(b) What is the tension along wire 1 immediately after the wire 2 is cut?

Now, if I put myself in a reference frame with y-axis along $\vec{g}$ (with inverse direction) the part (a) is $$\begin{cases} T_1\cdot cos\theta -mg=0 \\ -T_1 \cdot sin\theta + T_2 =0 \end{cases}$$

So $T_1=\frac{mg}{cos\theta}$.
For part (b), because I consider the infinitesimal time after $T_2=0$ I can approximate $T_1=\frac{mg}{cos\theta}$, so the same in part (a).

As Orudruin pointed out, the equations set up for equilibrium are no longer valid if wire 2 is cut. The mass will accelerate.So you should write $$\begin{cases} T_1\cdot cos\theta -mg=ma_x \\ -T_1 \cdot sin\theta =ma_y \end{cases}$$

 

[Bestfrog]

Well, I show you the original problem and the point where I stuck up.

1. The problem.
Referring to the picture below, a mass M stays on a surface with coefficient of static friction $\mu$. This object is connected by a massless rope to an another mass m, that is in quiet thanks to a second massless rope (as shown in picture). What is the minimum $\theta$ for whom the mass M stays at rest immediately after the second rope is cut?

3. The attempt to a solution.
I wrote the free body diagrams for M and m.
Considering M $$\begin{cases} T_1 - f=0 \\ N- Mg=0 \end{cases}$$

Considering m (after rope 2 is cut) $$\begin{cases} -T_1 sin\theta = -m a_x \\ T_1 cos\theta -mg = -m a_y \end{cases}$$

Now, how can I continue? There are too much unknowns.

(If I knew what is $a_{tot}$ of object m I could conclude!)

 

[Bestfrog]

I used the fact that $a_{tot} = g sin\theta = \sqrt{a_x^2 + a_y^2}$ and I solved it.

 

[Orodruin]

There are too much unknowns.

No there are not. You have to find some relations that must be satisfied when the mass $M$ does not move. In particular, what is the relation between the tension in the string and the coefficient of friction and how is $a_x$ related to $a_y$?

 

[Bestfrog]

No there are not. You have to find some relations that must be satisfied when the mass $M$ does not move. In particular, what is the relation between the tension in the string and the coefficient of friction and how is $a_x$ related to $a_y$?

Is it correct my assumption in post #5?

 

[Orodruin]

Is it correct my assumption in post #5?

It should be, but please show your work and what you arrive at in the end.

 

[Bestfrog]

Using that assumption I have that $T=mgcos\theta$.

For the equilibrium of M I have $T \le f_{max}=Mg \mu$, so $\theta \ge arcos(\frac{M\cdot \mu}{m})$ that should be correct.
(My only doubt was only about finding the tension)

 

[Orodruin]

Using that assumption I have that $T=mgcos\theta$.

For the equilibrium of M I have $T \le f_{max}=Mg \mu$, so $\theta \ge arcos(\frac{M\cdot \mu}{m})$ that should be correct.
(My only doubt was only about finding the tension)

Yes, this is correct.

Just to point out that there is an easier way to arrive at the $T = mg\cos(\theta)$. If you instead put your coordinate system at the ball such that one of the directions is along the string, then you know that the acceleration in that direction must be zero (no radial acceleration). The force equilibrium in that direction directly becomes $T = mg\cos(\theta)$ after projecting the gravitational force onto that direction.

 

[Orodruin]

Also note that cutting the string actually results in a smaller tension than before the string is cut. In other words, if it was not moving before, it will not move when the string is cut.

Bonus assignment: What is the maximal $\theta$ for which the mass will never move?