gr9277 #6 (edited post)

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gr9277 #6 (edited post)

Postby aby » Fri Sep 30, 2011 6:54 am

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

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Re: gr9277 #6 (edited post)

Postby diliu » Thu Nov 03, 2011 1:06 pm

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:

For static fraction we need:

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


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"

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Re: gr9277 #6 (edited post)

Postby bfollinprm » Thu Nov 03, 2011 1:16 pm

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.

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Re: gr9277 #6 (edited post)

Postby diliu » Thu Nov 03, 2011 1:49 pm

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
where, $\theta$ is the base angle.

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

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