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

- at θ=30o, 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=60o the block will start sliding down the plane and continue to do so at higher angles
- at θ=60o, the block will start sliding down the plane and on further increasing θ, it will topple at certain θ

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.

Suggest corrections

0 Upvotes

Similar questions

View More...

People also searched for

View More...

- About Us
- Contact Us
- Investors
- Careers
- BYJU'S in Media
- Students Stories - The Learning Tree
- Faces of BYJU'S – Life at BYJU'S
- Social Initiative - Education for All
- BYJU'S APP
- FAQ
- Support

© 2021, BYJU'S. All rights reserved.