Friction on a Rough Incline
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Direct answer
A block on a rough inclined plane stays in equilibrium until the incline angle reaches the angle of repose θ_r = tan⁻¹μ, because equilibrium needs tan θ ≤ μ. To haul the block up the incline, the minimum applied force along the slope is F = mg(sin θ + μ cos θ); to just stop it sliding down, only mg(sin θ − μ cos θ) is needed. When the applied force acts horizontally instead, it pushes the block into the plane, raising the normal to N = mg cos θ + F sin θ and enlarging friction — so a horizontal push must exceed the along-slope value to start the same motion. All these results are one free-body diagram with weight, normal, friction and the applied force resolved along and perpendicular to the incline.
What you must remember
- Angle of repose: θ_r = tan⁻¹μ; at this angle the block is on the verge of slipping, with mg sin θ = μ mg cos θ.
- Force to move up along the incline: F_min = mg(sin θ + μ cos θ); replacing the static μ by the smaller kinetic μk gives the force needed to keep the block moving up steadily.
- Force to just hold from sliding down: F = mg(sin θ − μ cos θ), positive only when tan θ > μ; below the angle of repose no force is needed at all.
- Horizontal push version: N = mg cos θ + F sin θ, and motion up the plane starts when F cos θ = mg sin θ + μ(mg cos θ + F sin θ) — the enlarged normal is the entire difference from the parallel case.
- Friction on the moving block: once sliding down, kinetic friction μk acts up the slope and the acceleration is g(sin θ − μk cos θ); sliding stops accelerating and becomes equilibrium exactly at tan θ = μk.
- Work by friction along an incline: W = −μ mg cos θ × L for a slide of length L along the slope — the bridge between height problems and energy audits.
A block, two directions of push
A 5 kg block sits on a 37-degree incline with μ = 0.25 (take g = 10 m/s², so sin 37° = 0.6, cos 37° = 0.8). Down-slope gravity is 30 N; the friction limit is 10 N. Since 30 > 10, the block slides unless held, and the least force along the incline to hold it is 20 N. To haul it up, the force must beat both gravity and friction: 40 N along the slope. One diagram, two answers, and every JEE Main version of this problem is a costume change on these two lines.
Now push horizontally instead. The push's along-slope component is F cos 37° = 0.8F up the plane, but its perpendicular component F sin 37° = 0.6F presses the block harder onto the incline, so the normal becomes 40 + 0.6F and the friction limit 0.25(40 + 0.6F). Motion up begins at 0.8F = 30 + 0.25(40 + 0.6F), which expands to 0.65F = 40, giving F = 61.5 N — some 50 percent more than the along-slope force for the identical job. The horizontal push fights its own side-effect, and this "why is it harder to push horizontally" question with the algebra shown is a JEE Advanced paragraph-level staple.
Where students slip
The vertical trap is the normal force: students carry N = mg cos θ into a problem where an applied force has a perpendicular component, and every subsequent number inherits the error — resolve everything first, then compute friction. The second slip is treating friction as μN always; static friction is self-adjusting up to μN, so a block held below the angle of repose may need only mg sin θ of it, and "find the friction acting" is not "find μN". In haul-up problems, friction reverses direction the moment impending motion flips from down to up. Two-block systems on one incline add the check that both blocks can supply their required friction shares. And keep sin/cos honest: at 37 degrees the sine is 0.6 (the 3-4-5 triangle), and swapping them inverts answers in ways the options are designed to catch.
Frequently asked questions
What is the angle of repose?
The incline angle θ_r = tan⁻¹μ at which a block just begins to slip; below it, static friction alone can hold the block, because tan θ < μ.
What force along the incline starts upward motion?
F = mg(sin θ + μ cos θ), since the pull must overcome both the down-slope gravity component and the limiting friction, which now acts down the plane.
Why must a horizontal push be larger than an along-slope one?
Its perpendicular component F sin θ increases the normal reaction and hence the friction, so the push fights both gravity and the friction it itself amplifies.
When is friction equal to μN?
Only at the limiting state — impending or actual sliding; below that, static friction self-adjusts to whatever value below the limit balances the other forces.
How does a slide down a rough incline end energetically?
Friction dissipates μ mg cos θ per metre of slope, so the block arrives at the bottom with mgh − μ mg cos θ × L of kinetic energy, zero when tan θ ≤ μ.