A block of base and height is kept on an inclined plane. The coefficient of friction between them is . The inclination of this inclined plane from the horizontal plane is gradually increased from . Then [2009]

Question

A block of base $10 cm \times 10 cm$ 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 [2009]

Smart Achievers · Accepted Answer

(2)

Maximum angle not to slide the block, angle of inclination = angle of repose,

i.e., $\tan^{- 1} μ = \tan^{- 1} \sqrt{3} = 60 °$

For the block to topple, the condition of the block has been shown in the figure.

In $∆ P O M$ , $\tan θ = \frac{P M}{O M} = \frac{10 / 2}{15 / 2} = \frac{5 cm}{7.5 cm} = \frac{2}{3}$

So, $θ < 60 °$ .

From this we can conclude that the block will topple at a lesser angle of inclination. Clearly the block will remain at rest on the plane up to a certain angle $θ$ and then it will topple.

Solution

Q.
A block of base $10 cm \times 10 cm$ 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 [2009]

1 at $θ = 30 °$ , 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 $θ = 60 °$ , the block will start sliding down the plane and continue to do so at higher angles

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

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Solution

Q. 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 0°. Then [2009]

1 at θ=30°, 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=60°, the block will start sliding down the plane and continue to do so at higher angles