An effigy of cyclohexane

Classical Mechanics Level 5

When i was studying it, I encountered "cyclohexane" and did not like it at all.

So I made a solid hexagonal prism as a model of cyclohexane, put it on floor and applied force on the top left vertex to knock it over. What is the minimal value of the coefficient of friction between the model and the floor, $\mu_\textrm{s}^\textrm{min}$ , required to topple the "cyclohexane" model in this fashion?

The answer is 0.288675.

1 solution

A Former Brilliant Member
Sep 25, 2016

I think it should be mentioned that force is to be applied horizontally , otherwise minimum value will be 0 as it can be toppled just by applying vertical force.

Sahil Bansal - 4 years, 7 months ago

what makes you even think that it can be toppled by applying a vertical force ! Think Again bro !

A Former Brilliant Member - 4 years, 7 months ago

@shubham dhull - Please allow the answer to more decimal places. I entered answer as 0.28867 and it marked me incorrect

Anubhav Tyagi - 4 years, 6 months ago

o, i do't believe it , the answer given is 0.288675. which is long enough , i think !

A Former Brilliant Member - 4 years, 6 months ago

I have changed the answer . So if someone else in future enters answer to more decimal places he will be marked correct

Anubhav Tyagi - 4 years, 6 months ago

@Anubhav Tyagi – how can you do that?

A Former Brilliant Member - 4 years, 6 months ago

@A Former Brilliant Member – are you a moderator ?

A Former Brilliant Member - 4 years, 6 months ago

An effigy of cyclohexane

The answer is 0.288675.

1 solution

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