A block of base 10 cm times 10 cm and heigh | vedclass

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(B) The condition for sliding is $	heta > \phi$,where $\phi$ is the angle of repose. Given $\mu = \sqrt{3}$,the angle of repose is $\phi = 	an^{-1}(

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the block will remain at rest on the plane up to certain $	heta$ and then it will topple

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(B) The condition for sliding is $	heta > \phi$,where $\phi$ is the angle of repose. Given $\mu = \sqrt{3}$,the angle of repose is $\phi = 	an^{-1}(\mu) = 	an^{-1}(\sqrt{3}) = 60^{\circ}$.The condition for toppling is that the line of action of the weight must pass outside the base of the block. This occurs when $	an 	heta > \frac{b}{h}$,where $b$ is the base width and $h$ is the height.For the given block,$b = 10 \ cm$ and $h = 15 \ cm$. The angle at which toppling begins is $	heta_{topple} = 	an^{-1}(\frac{b}{h}) = 	an^{-1}(\frac{10}{15}) = 	an^{-1}(\frac{2}{3}) \approx 33.7^{\circ}$.Since $	heta_{topple} \approx 33.7^{\circ}$ is less than the angle of repose $\phi = 60^{\circ}$,the block will start toppling before it can start sliding. Therefore,as $	heta$ increases from $0^{\circ}$,the block remains at rest until $	heta \approx 33.7^{\circ}$,at which point it will topple.

$A$ block of base $10 \ cm \times 10 \ cm$ and height $15 \ cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then

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