A block is placed on a rough inclined plane with inclination θ$$=45$$∘. The coefficient of static friction between the block and the incline is μ$$=0.75$$. Determine the maximum ratio $$h/b$$ for which the homogenous block will slide without tipping under the action of force F.

Question

A block is placed on a rough inclined plane with inclination θ$$=45$$∘. The coefficient of static friction between the block and the incline is μ$$=0.75$$. Determine the maximum ratio $$h/b$$ for which the homogenous block will slide without tipping under the action of force F.

Infinity Learn · Accepted Answer

Block has to slide without tipping, which implies that block is in translational as well as rotational equilibrium.When the block is on the verge of tipping about edge A, the normal reaction will pass through the edge of the block.From conditions of equilibrium along sloping surface:$$F+$$μ$$N=mgsin$$⁡θ              …(i)Perpendicular$$ to the sloping surface: $$N=mgcos$$⁡θ                      …$$(ii)The$$ block will not tip about A if ,F×$$h+(mgcos$$⁡θ$$)b2$$≥mg $$sin$$ θ ×$$b2$$ ⇒F ≥$$mg2($$$$sin$$ θ$$-$$ bhcos θ$$) ..............iii Solving we get , bh≥$$2($$μ$$-12$$ $$tan$$ θ$$)$$ $$=$$ $$2(0.75-0.5$$ ×$$1)$$                           ⇒hb≤$$2$$   Therefore $$hbmax=$$ $$2$$           …(iii)

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