Question 277: Laws of Motion

Question

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

Accepted Answer

The block slides on the incline when
$$mg\sin 	heta  \ge \mu mg\cos 	heta $$
$$ \Rightarrow 	an 	heta  \ge \mu $$
$$ \Rightarrow 	an 	heta  \ge \sqrt 3  = 1.732$$
For the block to topple, we have
$$f\left( {{{15} \over 2}} ight) \ge N(5)$$ (about centre of mass)
$$ \Rightarrow \mu N\left( {{{15} \over 2}} ight) \ge N(5)$$
$$ \Rightarrow \mu  \ge {2 \over 3} = 0.67$$
or $$	an 	heta  \ge 0.67$$
That is, the block neither slides nor topples till $$	heta  \le {	an ^{ - 1}}(0.67)$$ (i.e. it remains at rest) and on exceeding this, it topples first before sliding.

Answer

at θ = $$60^\circ $$, the block will start sliding down the plane

Answer

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

Answer

at θ = $$60^\circ $$, the block will start sliding down the plane and continue to do so at higher angles

Answer

at θ = $$60^\circ $$, the block will start sliding down the plane and on further increasing θ, it will topple at certain θ

Laws of Motion

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

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