Physics

Physics

Problem

A block of base 10 cm×10 cm10 \text{ cm} \times 10 \text{ cm} and height 15 cm15 \text{ cm} is kept on an inclined plane. The coefficient of friction between them is 3\sqrt{3} . The inclination θ\theta of this inclined plane from the horizontal plane is gradually increased from 0∘0^\circ . Then a. at θ=30∘\theta = 30^\circ , the block will start sliding down the plane b. the block will remain at rest on the plane up to certain θ\theta and then it will topple c. at θ=60∘\theta = 60^\circ , the block will start sliding down the plane and continue to do so at higher angles d. at θ=60∘\theta = 60^\circ , the block will start sliding down the plane and on further increasing θ\theta , it will topple at certain θ\theta .
Intuitive explanation

The key is recognizing that a block on an incline can fail in two ways: it can slide down, or it can tip over—and which happens first depends on the geometry and friction. As the angle increases, the friction force initially prevents sliding, but the torque from gravity about the lower edge grows faster. You need to find the critical angle where tipping becomes inevitable, comparing the condition for sliding (tan⁡θ=μ\tan\theta = \mu) against the condition for tipping (when the center of mass's horizontal offset exceeds the base's half-width). The block's tall, narrow shape makes tipping the likely failure mode.

Intuitive explanation

The key is recognizing that a block on an incline can fail in two ways: it can slide down, or it can tip over—and which happens first depends on the geometry and friction. As the angle increases, the friction force initially prevents sliding, but the torque from gravity about the lower edge grows faster. You need to find the critical angle where tipping becomes inevitable, comparing the condition for sliding (tan⁡θ=μ\tan\theta = \mu) against the condition for tipping (when the center of mass's horizontal offset exceeds the base's half-width). The block's tall, narrow shape makes tipping the likely failure mode.

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