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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 3. The inclination θ of this inclined plane from the horizontal plane is gradually increased from 0o. Then

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Solution

The correct option is **B** the block will remain at rest on the plane up to certain θ and then it will topple

The maximum angle when the block will not slide should be equal to the angle of repose.

ϕ=tan−1μ=tan−1√3=60o

If the block does not topple, the line of action of weight should pass within the base of the block

And for toppling, maximum angle of inclination,

θ=tan−1b/2h/2

θ=tan−1(10/2)(15/2)=tan−123

θ=34o<60o

For 60o, mg is already outside the base.

∴ As the angle θ is increased first it will topple before it starts sliding.

The maximum angle when the block will not slide should be equal to the angle of repose.

ϕ=tan−1μ=tan−1√3=60o

If the block does not topple, the line of action of weight should pass within the base of the block

And for toppling, maximum angle of inclination,

θ=tan−1b/2h/2

θ=tan−1(10/2)(15/2)=tan−123

θ=34o<60o

For 60o, mg is already outside the base.

∴ As the angle θ is increased first it will topple before it starts sliding.

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