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Laws of Motion

Section: Physics

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Question : 17 of 20

Marks: +1, -0

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°. Then

[JEE Adv 2009 P1]

at θ = 30°, the block will start sliding down the plane.
the block will remain at rest on the plane up to certain θ and then it will topple.
at θ = 60°, the block will start sliding down the plane and continue to do so at higher angles.
at θ = 60°, the block will start sliding down the plane and on further increasing θ , it will topple at certain θ.

Solution:

The block slides on the incline when

mgsinθ ≥ μmgcosθ

⇒ tan θ ≥ μ

⇒ tanθ ≥ 3 =1.732

For the block to topple, we have

f

\left(\frac{15}{2}\right)

≥ N (5) (about centre of masss)

⇒ µN

\left(\frac{15}{2}\right)

≥ N (5)

⇒ µ ≥

\frac{2}{3}

= 0.67

or tan θ ≥ 0.67

That is, the block neither slides nor topples till θ ≤

\tan^{-1}

(0.67) (i.e. it remains at rest) and on exceeding this, it topples first before sliding.

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