
An inclined plane is a flat surface that forms an angle θ\thetaθ with the horizontal.
Inclined-plane problems are important because they require forces to be analyzed in components, making them a fundamental application of Newton’s Laws.
On a horizontal surface, gravity acts entirely perpendicular to the surface.
On an incline, gravity has:
The parallel component causes the object to accelerate down the slope.
For inclined-plane problems, it is usually easiest to choose axes:
This simplifies Newton’s Second Law.
For an object of mass mmm on an incline:
Weight acts vertically downward.
\(\vec{V}=m\vec{g}\)
The surface exerts a force perpendicular to the plane.
\(vec{N}\)
Friction acts parallel to the surface and opposes relative motion.
\(\vec{f}\)
The weight vector is decomposed into two components.
Acts along the incline:
\(F_{\parallel}=mg\sin\theta\)
This component tends to pull the object down the slope.
Acts into the surface:
\(F_{\perp}=mg\cos\theta\)
This component determines the normal force.
If there is no acceleration perpendicular to the plane:
\(\sum F_{\perp}=0\)
Therefore:
\(N=mg\cos\theta\)
As the incline becomes steeper, the normal force decreases.
If friction is absent:
Along the slope:
\(\sum F_{\parallel}=ma\)
\(mg\sin\theta=ma\)
Thus:
\(a=g\sin\theta\)
Important result:
The acceleration depends only on:
not on mass.
Kinetic friction:
\(f_{k}=\mu_{k}N\)
Substituting:
\(f_{k}=\mu_{k}mg\cos\theta\)
Newton’s Second Law along the slope:
\(mg\sin\theta-f_{k}=ma\)
Therefore:
\(mg\sin\theta-\mu_{k}mg\cos\theta=ma\)
Acceleration:
\(a=g(\sin\theta-\mu_{k}\cos\theta)\)
If the object remains at rest:
\(f_{s}=mg\sin\theta\)
provided that:
\(mg\sin\theta\les\mu_{s}N\)
If the downslope component exceeds the maximum static friction, the object begins to slide.
For every incline problem:
Draw all forces.
Choose axes parallel and perpendicular to the slope.
Resolve weight into:
\(mg\sin\theta\)
and
\(mg\cos\theta\)
Apply Newton’s Second Law along each axis.
As the angle increases:
\(mg\sin\theta\)
increases.
This means the tendency to slide becomes larger.
Meanwhile:
\(mg\cos\theta\)
decreases.
This reduces the normal force and therefore reduces friction.
For an object on an incline:
Weight:
\(W=mg\)
Parallel component:
\(mg\sin\theta\)
Perpendicular component:
\(mg\cos\theta\)
Normal force:
\(N=mg\cos\theta\)
Frictionless acceleration:
\(a=g\sin\theta\)
Inclined-plane problems are a fundamental application of:
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