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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 

 
  1. at θ=30o, the block will start sliding down the plane 
  2. the block will remain at rest on the plane up to certain θ and then it will topple 
  3. at θ=60o the block will start sliding down the plane and continue to do so at higher angles 
  4. 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=tan1μ=tan13=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=tan1b/2h/2
θ=tan1(10/2)(15/2)=tan123
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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