The problem is as follows:

The alternatives given in my book are as follows:

$\begin{array}{ll}

1.&40\,N\\

2.&200\,N\\

3.&240\,N\\

4.&300\,N\\

\end{array}$

In this problem, I'm confused at why they are giving two coefficients of friction. I'm assuming one is static friction and the other is kinetic friction. But exactly which is which? I think that the biggest one must be the static friction. I attempted my solution as included in the diagram from above.

If I were to use this, I attempted to use the equation:

$F\cos 37^{\circ}-\mu\left(w+\cdot F \sin 37^{\circ}\right)=0$

$F\left(\frac{4}{5}\right)-\frac{48}{100}\left(320+\frac{3}{5}F\right)=0$

$F\left(\frac{4}{5}\right)=\frac{48}{100}\left(320+\frac{3}{5}F\right)$

$\frac{4F}{5}=\frac{48}{100}\left(\frac{1600+3F}{5}\right)$

$F=300\,N$

But would this be the minimum force? Or instead?

$F\left(\frac{4}{5}\right)=\frac{36}{100}\left(320+\frac{3}{5}F\right)$

Which by doing all the calculations would give:

$F\approx 197.26\,N$

Btw, please excuse my drawing abilities; I did redraw the original sketch and I did my best.

Can somebody help me to establish exactly what's the meaning of those two coefficients of friction and are the vectors okay? :help:

A box of $320\,N$ is at rest over a horizontal terrain. The coefficients of friction between the box and the terrain are $0.36$ and $0.48$. A girl is pushing the box with her arms making an angle of $37^{\circ}$ with the horizontal. Find the magnitude of the minimum force that will let her begin the motion of the box.

The alternatives given in my book are as follows:

$\begin{array}{ll}

1.&40\,N\\

2.&200\,N\\

3.&240\,N\\

4.&300\,N\\

\end{array}$

In this problem, I'm confused at why they are giving two coefficients of friction. I'm assuming one is static friction and the other is kinetic friction. But exactly which is which? I think that the biggest one must be the static friction. I attempted my solution as included in the diagram from above.

If I were to use this, I attempted to use the equation:

$F\cos 37^{\circ}-\mu\left(w+\cdot F \sin 37^{\circ}\right)=0$

$F\left(\frac{4}{5}\right)-\frac{48}{100}\left(320+\frac{3}{5}F\right)=0$

$F\left(\frac{4}{5}\right)=\frac{48}{100}\left(320+\frac{3}{5}F\right)$

$\frac{4F}{5}=\frac{48}{100}\left(\frac{1600+3F}{5}\right)$

$F=300\,N$

But would this be the minimum force? Or instead?

$F\left(\frac{4}{5}\right)=\frac{36}{100}\left(320+\frac{3}{5}F\right)$

Which by doing all the calculations would give:

$F\approx 197.26\,N$

Btw, please excuse my drawing abilities; I did redraw the original sketch and I did my best.

Can somebody help me to establish exactly what's the meaning of those two coefficients of friction and are the vectors okay? :help:

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