A cube is placed on an inclined plane of inclination θ as shown in the figure. Coefficient of friction between the cube and the plane is μ. As the angle θ is gradually increased, the cube slides before toppling if
Question
A cube is placed on an inclined plane of inclination θ as shown in the figure. Coefficient of friction between the cube and the plane is μ. As the angle θ is gradually increased, the cube slides before toppling if
Solution
The cube will slide before toppling if the angle of inclination θ is such that the frictional force is less than the component of the weight of the cube acting down the slope.
Here's the step-by-step solution:
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The forces acting on the cube are its weight (mg, acting downwards), the normal force (N, perpendicular to the plane), and the frictional force (f, opposing motion).
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The weight of the cube can be resolved into two components: one perpendicular to the plane (mg cos θ) and one parallel to the plane (mg sin θ).
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The frictional force (f) is equal to the coefficient of friction (μ) times the normal force (N). Since the cube is not moving in the perpendicular direction, the normal force is equal to the perpendicular component of the weight, so f = μN = μmg cos θ.
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The cube will start to slide when the parallel component of the weight (mg sin θ) is greater than the frictional force (μmg cos θ). So, the condition for the cube to slide is mg sin θ > μmg cos θ.
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Dividing both sides by mg gives sin θ > μ cos θ. Dividing both sides by cos θ gives tan θ > μ.
So, the cube will slide before toppling if the tangent of the angle of inclination is greater than the coefficient of friction.
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