How do I find the minimum force done by a person pushing a box when two coefficients?

Jun 2017
Lima, Peru
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 that will let her begin the motion of the box.​

The alternatives given in my book are as follows:


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:

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





But would this be the minimum force? Or instead?


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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Math Team
Jul 2011
$\mu_s > \mu_k$ ... there may be an instance where they are equal, but I haven't come across a situation where that has happened.

to get the box moving, the applied force's horizontal component must be at least $f_{s \, max} = \mu_s \cdot N$, where $N$ is the sum of the box's weight and the vertical component of the applied force acting downward.

$F\cos(37) \ge \mu_s[W + F\sin(37)] \implies F \ge \dfrac{\mu_s \cdot W}{\cos(37)-\mu_s \cdot \sin(37)} = 300 \text{ N}$
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