Relationship between horizontal force and friction in an exercise

[Venturi365]

Homework Statement

A mass of $20\,\mathrm{N}$ lays on a horizontal surface. The static friction coefficient and the kinetic friction coefficient are, respectively: $\mu_{s}=0.8$ and $\mu_{k}=0.6$. A horizontal cable is tied to the mass with a constant tension $T$. What's the friction force that acts on the block if $T=15\,\mathrm{N}$ and $T=20\,\mathrm{N}$?

Relevant Equations

$f_{s}=\mu_{s}\cdot N$
$f_{k}=\mu_{k}\cdot N$

The thing with this exercise is that I don't think that the question makes sense at all (or, at least, is incomplete).

First of all, we don't know if the mass moves with any of those tensions, therefore I cannot know which coefficient apply. Second of all, even if we suppose that the mass is moving or not, the friction force has nothing to do with any horizontal force, but rather it is related to the movement and weight of the mass. Right?

Anyway, if we suppose that in both cases the mass doesn't move, then:

$$
\begin{align}
f_{s}=0.8\cdot 20 =16\,\mathrm{N}
\end{align}
$$

If we suppose that in both cases the mass moves, then:

$$
\begin{align}
f_{k}=0.6\cdot 20=12\,\mathrm{N}
\end{align}
$$

Am I missing something or is it the wording of the problem just uncomplete or misleading?

 

[erobz]

Homework Statement: A mass of $20\,\mathrm{N}$ lays on a horizontal surface. The static friction coefficient and the kinetic friction coefficient are, respectively: $\mu_{s}=0.8$ and $\mu_{k}=0.6$. A horizontal cable is tied to the mass with a constant tension $T$. What's the friction force that acts on the block if $T=15\,\mathrm{N}$ and $T=20\,\mathrm{N}$?
Relevant Equations: $f_{s}=\mu_{s}\cdot N$
$f_{k}=\mu_{k}\cdot N$

The thing with this exercise is that I don't think that the question makes sense at all (or, at least, is incomplete).

First of all, we don't know if the mass moves with any of those tensions, therefore I cannot know which coefficient apply. Second of all, even if we suppose that the mass is moving or not, the friction force has nothing to do with any horizontal force, but rather it is related to the movement and weight of the mass. Right?

Anyway, if we suppose that in both cases the mass doesn't move, then:

$$
\begin{align}
f_{s}=0.8\cdot 20 =16\,\mathrm{N}
\end{align}
$$

If we suppose that in both cases the mass moves, then:

$$
\begin{align}
f_{k}=0.6\cdot 20=12\,\mathrm{N}
\end{align}
$$

Am I missing something or is it the wording of the problem just uncomplete or misleading?

Notice that the coefficient of static friction is larger than the kinetic. Initially the block is "laying there". In order to get it to move you must overcome the force of static friction. Once moving the frictional force coefficient sharply drops off to $\mu_k$

Concept question: When the block is sitting there without any horizontal force applied. What is the force of friction?

 

[Venturi365]

Notice that the coefficient of static friction is larger than the kinetic. Initially the block is "laying there". In order to get it to move you must overcome the force of static friction. Once moving the frictional force coefficient sharply drops off to $\mu_k$

Ohhhh, so therefore when $T=15$ then $f\leq 16$ and when $T=20$ then $f=12$

 

[erobz]

Ohhhh, so therefore when $T=15$ then $f\leq 16$

No, you are on the right track but this statement is not correct.

When the block is sitting there without any horizontal tension force, what is the force of static friction acting on it?

 

[Venturi365]

When the block is sitting there without any horizontal tension force, what is the force of static friction acting on it?

Well, since the friction is a reaction force, it has to be $f=0\,\mathrm{N}$

I guess that technically, when $T=15\,\mathrm{N}$ then $f=15\,\mathrm{N}$ because there's no resulting movement and $\mu_{s}=0.8$ is just the maximum value that it could take.

 

[erobz]

Well, since the friction is a reaction force, it has to be $f=0\,\mathrm{N}$

I guess that technically, when $T=15\,\mathrm{N}$ then $f=15\,\mathrm{N}$ because there's no resulting movement and $\mu_{s}=0.8$ is just the maximum value that it could take.

There you go!

 

[Venturi365]

There you go!

Thank you for your help!

 

[kuruman]

Relevant Equations: $f_{s}=\mu_{s}\cdot N$

No. $$f_s\leq \mu_s~N##. This says that the force static friction is at most $\mu_s~N$. Otherwise, it's whatever is necessary to provide the observed acceleration. It's like a credit card with a maximum of $2,000. When you go to a store to buy something, you can use the card to buy anything under $2,000. What you charge to the card is the price of the object, not $2,000.