A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane
By Problem: 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 degree. Then, (IIT JEE 2009)
- at $\theta=30$ degree, the block will start sliding down the plane.
- the block will remain at rest on the plane up to certain $\theta$ and then it will topple.
- at $\theta=60$ degree, the block will start sliding down the plane and continue to do so at higher angles.
- at $\theta=60$ degree, the block will start sliding down the plane and on further increasing $\theta$, it will topple at certain $\theta$.
Solution: The forces acting on the block are its weight $mg$, frictional force $f$, and normal reaction $N$. When the block starts toppling, the contact point between the inclined plane and the block is A and at that instant $N$ acts on A.
The frictional force $f$ opposes downward motion of the block and attains its maximum value \begin{align} f_\text{max}=\mu N.\nonumber \end{align} Resolve $mg$ along and normal to the inclined plane. Newton's second law gives \begin{align} N=mg\cos\theta.\nonumber \end{align} The block starts sliding down when $mg\sin\theta > f_\text{max}$. Substitute the values of $f_\text{max}$ and $N$ to get \begin{align} \tan\theta>\mu=\sqrt{3}=1.73.\nonumber \end{align} The block starts toppling when net torque about A is anticlockwise. The torques about A due to $N$ and $f$ are zero because these forces pass through A. The torque due to $mg$ is \begin{align} \tau=mg\sin\theta\times 7.5-mg\cos\theta\times 5.\nonumber \end{align} Thus, the block will topple if $\tau > 0$ i.e., $\tan\theta > 2/3 = 0.66$. Thus, $\theta_\text{topple} < \theta_\text{slide}$ and hence toppling occurs before sliding.
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