gr9277 #6 (edited post)

[aby]

pls help with this because i always do the same logic mistakes in this type of questions.



(correct answer : D)
the problem i'm having is with the given data of μ<0 . i understand it's given to prevent solutions C,D to explode, but the coefficient of friction could normally be very well above >1. and in such cases the derivation obviously fails and solutions C,D cannot be true. how could i solve this for every μ ?


edit: additional clarification

[diliu]

Considering the axial symmetry of this system, we firstly vertically divide it into two equal parts, and discuss the left one.

Clearly, there is a left-horizontal fore $$$f$$$ exerts on the half-cube, to keep the half-system balanced, $$f$$ is equal to the friction comes from the floor, which is right toward.

Let the force between the half-cube and the left-wedge be $$$N$$$, which is perpendicular to the interface, to be balanced we have:
$$$N\cos(45^\circ)=\frac{1}{2}Mg$$$.
$$$N\sin(45^\circ)=f$$$.

For static fraction we need:
$$$f\leq\mu(m+\frac{1}{2}M)g$$$

Note that $$$\cos(45^\circ)=\sin(45^\circ)$$$, combining the equations above, we have:

$$$\frac{1}{2}Mg\leq\mu(mg+\frac{1}{2}Mg)$$$.

Thus we get, if $$$\mu\geq1$$$, this inequality is naturally assured, there is no constraint on the value of $$$M$$$; when $$\mu<1$$, we get the answer "D"

[bfollinprm]

Or, in other words, if the static force is greater than the normal force, the mass would have an easier job boring through the triangular blocks than moving them across the floor.

[diliu]

Correct, that is the phenomena called "self-lock". You can put a triangular block under your door, and will never push the door to move if only
$$$\mu\geq\tan\theta$$$.
where, $$$\theta$$$ is the base angle.

In the problem above, $$$\theta=45^\circ$$$, so the critical value of $$$\mu$$$ is 1