Find the tensions acting in different ropes that conect three bodies

[Like Tony Stark]

Homework Statement

Three bodies are conected by ideal ropes. The mass of the smallest body is $10 kg$. The coefficient of static friction between the bodies and the surface is $0.3$ and they are pulled to the right by a $60 N$ force. Determine the tension acting on each rope.

Relevant Equations

Newton's equations

I called the smallest body $1$ and the biggest $3$. All of them are disturbed by horizontal forces. So the free body diagrams are
For $1$
$-T_1 -Fr_1 +F=10a$

For $2$
$T_1 -T_2 -Fr_2=20a$

For $3$
$T_2 -Fr_3=30a$

Then, if I consider $a=0$ and I solve for $1$ I get $T_1=30 N$, and then if I plug that value in $2$ I get $T_2=-30 N$. But why the signs of both tensions are different?
And then if I tried to find $T_2$ in $3$ I get $T_2=90$, and clearly that's different from the value calculated in $2$.

 

[haruspex]

Homework Statement: Three bodies are conected by ideal ropes. The mass of the smallest body is $10 kg$. The coefficient of static friction between the bodies and the surface is $0.3$ and they are pulled to the right by a $60 N$ force. Determine the tension acting on each rope.
Homework Equations: Newton's equations

I called the smallest body $1$ and the biggest $3$. All of them are disturbed by horizontal forces. So the free body diagrams are
For $1$
$-T_1 -Fr_1 +F=10a$

For $2$
$T_1 -T_2 -Fr_2=20a$

For $3$
$T_2 -Fr_3=30a$

Then, if I consider $a=0$ and I solve for $1$ I get $T_1=30 N$, and then if I plug that value in $2$ I get $T_2=-30 N$. But why the signs of both tensions are different?
And then if I tried to find $T_2$ in $3$ I get $T_2=90$, and clearly that's different from the value calculated in $2$.

At some point you must have plugged in values for the frictional forces. What is the "relevant equation " for static friction?
(Btw, I will demonstrate that the question is invalid, but leave that for now.)

 

[Like Tony Stark]

At some point you must have plugged in values for the frictional forces. What is the "relevant equation " for static friction?
(Btw, I will demonstrate that the question is invalid, but leave that for now.)

The "relevant equation" for static friction is $Fr_{s}=\mu_{s} . N$
What do you mean with "you must have plugged in values for the frictional forces"? I mean, I have to plug them because the friction is acting on the blocks

 

[haruspex]

The "relevant equation" for static friction is $Fr_{s}=\mu_{s} . N$

That equation is wrong in a subtle but crucial way.
E.g., consider a block resting on a horizontal rough floor with no lateral forces exerted on it. What would be the frictional force?

 

[Like Tony Stark]

That equation is wrong in a subtle but crucial way.
E.g., consider a block resting on a horizontal rough floor with no lateral forces exerted on it. What would be the frictional force?

Obviously, the static coefficient only applies a few moments before starting to move. But in this case we are assuming that the body is about to move, aren't we?

 

[haruspex]

Obviously, the static coefficient only applies a few moments before starting to move. But in this case we are assuming that the body is about to move, aren't we?

Ok, so now take my resting box and apply a horizontal force, but only half that needed to make it move. How great is the frictional force now?

 

[Like Tony Stark]

Ok, so now take my resting box and apply a horizontal force, but only half that needed to make it move. How great is the frictional force now?

It has the same magnitud that the force that you are applying

 

[haruspex]

It has the same magnitud that the force that you are applying

Right, so what is the correct form of the static friction equation?

 

[Like Tony Stark]

Right, so what is the correct form of the static friction equation?

The static friction has the same magnitude that the force applied to the body unless the force is greater than the maximum static friction

 

[haruspex]

The static friction has the same magnitude that the force applied to the body unless the force is greater than the maximum static friction

And as an equation, $F_s\leq\mu_sN$. Do try to remember that form.

So how does this alter your solution?

 

[Like Tony Stark]

And as an equation, $F_s\leq\mu_sN$. Do try to remember that form.

So how does this alter your solution?

Ok, I think I've noticed my mistake...
I have to calculate if the force applied to the blocks is bigger than the maximum static friction. If so, I have to plug the values for $N.\mu_s$ in Newton's equations. If not, the friction will compensate the force and they'll add up to zero. Right?

By the way, if the force is bigger, shouldn't I have to plug dynamic friction? Because it will be moving. But I'm not given the coefficient of dynamic friction.

 

[haruspex]

If not, the friction will compensate the force and they'll add up to zero.

Yes.

if the force is bigger, shouldn't I have to plug dynamic friction?

If the applied force exceeds the combined max static friction of the three blocks, yes - but it doesn't.
I think the question intends that you consider each block in turn, as you did. If the small block's friction is enough you can stop there; otherwise consider the next block ...