Physics · Laws Of Motion · NEET
The surface only pushes back against the part of the weight pressing straight into it. On a slope, gravity mg points vertically down, but the surface is tilted. Only the component perpendicular to the surface, mg cosθ, presses on it, so the surface pushes back with N = mg cosθ. The other component, mg sinθ, runs along the slope and does not add to N. That is why N is always less than mg on an incline (cosθ < 1), and N shrinks as the incline gets steeper.
The component along the slope, mg sinθ, is the 'driving' force that pulls the block down the incline. The steeper the angle, the bigger mg sinθ, so the stronger the pull. This is why a block sits still on a gentle slope but slides on a steep one. Remember: mg sinθ acts DOWN the slope, mg cosθ acts INTO the slope.
Friction always opposes the tendency of motion. If the block is sliding down (or about to slide down), friction acts UP the slope to resist it. If you push the block up the slope, friction acts DOWN. For a block simply resting on a rough incline, the natural tendency is to slide down, so static friction points up the slope and balances part of mg sinθ.
When the block is already moving down, kinetic friction f = μ N = μ mg cosθ acts up the slope. Net force down the slope = mg sinθ − μ mg cosθ. Divide by mass m: a = g(sinθ − μ cosθ). If sinθ ≤ μ cosθ (that is, tanθ ≤ μ), the acceleration is zero or negative and the block will not slide on its own.
Zero. If the block is at rest, it is in equilibrium — the weight, the normal force, and the static friction all cancel out. Students often calculate mg sinθ and pick that as the answer, but static friction is exactly balancing it. So the correct net force on a stationary block is 0, no matter what μ or θ is (as long as it stays at rest). NEET 2023 tested exactly this.
A block of mass 2 kg is placed on a rough inclined surface AC of coefficient of friction μ. It remains at rest. Taking g = 10 m/s², the net force on the block is:
Two inclined surfaces have equal length L and the same inclination 45° with the horizontal; one is rough and the other perfectly smooth. A body takes 2 times as much time to slide down the rough surface as the smooth surface. The coefficient of kinetic friction μk between the body and the rough surface is close to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The block stays at rest as long as the required static friction does not exceed the maximum available. That is, mg sinθ ≤ μs mg cosθ, which simplifies to tanθ ≤ μs. If tanθ is larger than μs, the block slides.
a = g(sinθ − μk cosθ), directed down the slope. For a smooth incline (μk = 0) this reduces to the familiar a = g sinθ.
Because N = mg cosθ. As θ increases, cosθ decreases, so N decreases. Less of the weight presses into the surface, which also makes the maximum friction μ mg cosθ smaller.
No. The sliding condition tanθ > μ has no mass in it — mass cancels out. A heavy and a light block start to slide at the same angle on the same surface. Mass affects the actual forces but not the slide/no-slide decision.
The angle of repose θr, where tanθr = μs. In NCERT Example 4.8, a mass just begins to slide at θ = 15°, giving μs = tan15° ≈ 0.27. At this angle mg sinθ exactly equals the maximum static friction.
