Please help me get started with this mechanics question:
A cube of side 20cm and mass 2kg rests on a rough horizontal surface.
Thank you Maths Help.
This sort of question is best tackled in two stages:
If the first value of X is lower, the block will slide before it topples,
But if the second value is lower, the block will topple over without sliding.
The diagram shows all the external forces that act on the block.
For sliding to occur, the force X must exceed the friction force F.
Vertical equilibrium gives R = 2g ; and
limiting friction gives F = µR.
So for sliding, X > 0.6g.
The diagram shows the forces acting on the block just as the block topples.
Note that as the block is about to topple, the reaction force R and the friction force F are both zero.
Taking moments about A:
For toppling, 0.2X > 0.1 × 2g.
So for toppling, X > g.
The answer to the question is that
Equlibrium is broken by sliding,|
with a force F of 0.6g Newtons
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