## gr9277 #6 (edited post)

aby
Posts: 9
Joined: Fri Sep 30, 2011 6:41 am

### gr9277 #6 (edited post)

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

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 ?

diliu
Posts: 9
Joined: Sun Aug 29, 2010 2:05 pm

### Re: gr9277 #6 (edited post)

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
Posts: 1198
Joined: Sat Nov 07, 2009 11:44 am

### Re: gr9277 #6 (edited post)

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
Posts: 9
Joined: Sun Aug 29, 2010 2:05 pm

### Re: gr9277 #6 (edited post)

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