Physics · Laws Of Motion · NEET
μmg is the friction only when the force is horizontal, because then the normal reaction N = mg stays full. But if you apply the force at an upward angle θ, the vertical part F sinθ shares the weight, so N = mg − F sinθ becomes smaller. Smaller N means smaller limiting friction μN. So the force needed to overcome friction also drops. Minimising F over all angles gives F(min) = μmg/√(1+μ²), which is always LESS than μmg.
The best angle is θ = tan⁻¹(μ), which is exactly the angle of friction λ. So you pull the body along the direction that makes an angle equal to the angle of friction with the horizontal. At this special angle the required force is smallest. If you pull flatter (0°) or steeper, you need more force.
When you PULL at an upward angle, F sinθ acts up and reduces the normal reaction, so friction falls. When you PUSH at a downward angle, F sinθ acts down and INCREASES the normal reaction, so friction rises and you need more force. That is why a heavy box is easier to drag with an upward-slanting rope than to shove down-and-forward.
Friction depends on normal reaction: f = μN. The horizontal component F cosθ tries to move the body; the vertical component F sinθ changes N. Pull up → N = mg − F sinθ (less friction). Push down → N = mg + F sinθ (more friction). So the same-sized force gives different friction just by changing its direction.
F(min) = μmg/√(1+μ²) uses the coefficient of static (limiting) friction, so it is the smallest force to just START motion. Once moving, kinetic friction μ(k) is usually smaller, so keeping it moving needs even less force. NEET problems almost always ask for the force to just begin motion using μ.
A body of mass m is kept on a rough horizontal surface (coefficient of friction μ). A horizontal force is applied but the body does not move. The magnitude of the resultant F of the normal reaction and the frictional force satisfies:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
F(min) = μmg / √(1 + μ²), where μ is the coefficient of friction, m the mass and g = 9.8 or 10 m/s². It must be applied at angle θ = tan⁻¹(μ) above the horizontal.
At θ = tan⁻¹(μ), which equals the angle of friction. At this angle both the horizontal pull and the reduction of normal reaction are balanced for least effort.
Yes. Since √(1+μ²) > 1, F(min) = μmg/√(1+μ²) is always smaller than the horizontal force μmg. Tilting the pull upward always helps.
No, mass stays in F(min) = μmg/√(1+μ²). Mass only cancels when the question asks for maximum acceleration (a = μg), not for force.
NEET regularly tests friction with forces applied at an angle, pull-vs-push, and the resultant of normal and friction (asked in NEET 2024). Knowing the √(1+μ²) factor saves time and avoids the μmg trap.
