A block rests on a smooth inclined plane making an angle of 30^∘ with the horizontal. The coeffi

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A block rests on a smooth inclined plane making an angle of...

A block rests on a smooth inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \mathrm{~N}$, the mass of the block (in $\mathrm{kg}$ ) is (Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$ ) (a) $2.0$ (b) $4.0$ (c) $1.6$ (d) $2.5$

Key Concepts

Equilibrium Conditions

In static equilibrium, the net force in all directions is zero. For an object on an inclined plane, this means that the static friction must exactly balance the component of gravity that is trying to pull the object down the slope. This balance is used to relate frictional force, gravitational force, and mass.

Normal Force

The normal force is the perpendicular contact force exerted by a surface on an object. On an inclined plane, it is calculated by the cosine component of the object's weight, and it plays a crucial role in determining the maximum possible static friction.

Static Friction

Static friction is the resistance that prevents an object from sliding along a surface. It adjusts to counteract the component of gravitational force along the inclined plane up to a maximum given by the product of the coefficient of static friction and the normal force.

Force Decomposition on an Inclined Plane

This is the process of breaking down the gravitational force acting on an object into two perpendicular components: one parallel to the surface of the incline and one perpendicular to it. Understanding this concept is key when analyzing the motion or equilibrium of objects on slopes.

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