Inclined PlaneJEE Main

Forces, friction & the angle of repose — interactive Physics simulation for IIT-JEE.

Concept

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

Key formula

a=g (sin⁡θ−μcos⁡θ),θrepose=tan⁡−1μa = g\,(\sin\theta - \mu\cos\theta), \qquad \theta_{\text{repose}} = \tan^{-1}\mu

Derivation

Resolve the weight mgmg along and perpendicular to the surface. Perpendicular balance gives N=mgcos⁡θN = mg\cos\theta, so the maximum static friction is fmax=μN=μmgcos⁡θf_{max} = \mu N = \mu mg\cos\theta.

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

Once mgsin⁡θ>μmgcos⁡θmg\sin\theta > \mu mg\cos\theta, kinetic friction μN\mu N acts and a=g(sin⁡θ−μcos⁡θ)a = g(\sin\theta - \mu\cos\theta). The threshold tan⁡θ=μ\tan\theta = \mu defines the angle of repose.

Scenarios to explore

  • Inclined Plane — Block sliding on a frictional incline — forces & acceleration.
  • Frictionless slide: a = g sinθ — With μ = 0 the mass does not matter — only the angle sets the acceleration.
  • On the verge of sliding — Here μ ≈ tanθ — the block barely moves. The angle of repose, made visible.
  • Steep and slippery — A 60° incline with almost no friction — nearly free fall along the slope.

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=gsin⁡θa = g\sin\theta, independent of mass.

Common mistakes

  • Using mgmg for the normal force instead of mgcos⁡θmg\cos\theta.
  • Assuming friction is always μN\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.