Inclined PlaneJEE Main

Concept

A block on an incline feels gravity resolved into a component along the slope ($mg\sin\theta$) that drives motion and a component into the slope ($mg\cos\theta$) that sets the normal force. Friction opposes sliding up to a maximum of $\mu N$; the block moves only when the driving component beats that maximum.

Key formula

Derivation

Resolve the weight $mg$ along and perpendicular to the surface. Perpendicular balance gives $N = mg\cos\theta$, so the maximum static friction is $f_{max} = \mu N = \mu mg\cos\theta$.

Along the slope, Newton's second law reads $ma = mg\sin\theta - f$. While $mg\sin\theta \le f_{max}$ the block stays put ($a=0$) and friction self-adjusts to exactly $mg\sin\theta$.

Once $mg\sin\theta > \mu mg\cos\theta$, kinetic friction $\mu N$ acts and $a = g(\sin\theta - \mu\cos\theta)$. The threshold $\tan\theta = \mu$ defines the angle of repose.

Scenarios to explore

• Inclined Plane — Block sliding on a frictional incline — forces & acceleration.

Real-world applications

• Designing ramps, conveyor belts and chutes so loads do (or don't) slide.
• Measuring μ by tilting a surface until an object just begins to slide.
• Vehicle traction and parking on hills.

JEE exam tips

• Always draw the free-body diagram with axes along and perpendicular to the incline.
• On a frictionless incline $a = g\sin\theta$, independent of mass.

Common mistakes

• Using $mg$ for the normal force instead of $mg\cos\theta$.
• Assuming friction is always $\mu N$ — static friction is only up to that value.
• Forgetting that the block can be in equilibrium below the angle of repose.

Exam traps to avoid

• Adding an applied force changes both the net driving force and possibly the normal force.
• The angle of repose depends only on μ, never on the mass.