- #1

Venturi365

- 12

- 3

- 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?

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?