# How do I find the minimum force done by a person pushing a box when two coefficients of frictions are given?

The problem is as follows:

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 with will let her to 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:

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

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

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

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

$$F=300\,N$$

Which would be the option four.

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:

$$F\approx 197.26\,N$$

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

• You may get a better answer over at Physics.SE. Nov 26 '19 at 15:17
• Can I ask you the title of the book from which this problem was taken? Dec 1 '19 at 7:40
• @Shootforthemoon I'm sorry the book has no author it comes from a collection of riddles in physics. Dec 2 '19 at 9:05