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You need to drag a heavy box across a rough floor. Should you pull the rope flat or tilt it upward? Tilt too much and most of your effort lifts instead of drags. Somewhere in between is an angle that needs the least force. This is a favourite JEE problem because it combines a free-body diagram, the laws of friction and a maxima-minima step. This post derives the angle two ways and works three examples.
Is this on the syllabus?
The JEE (Main) 2026 syllabus lists "Static and Kinetic friction, laws of friction, rolling friction" under Laws of Motion, and the NEET (UG) 2026 syllabus has the same line. JEE Main mathematics also lists maxima and minima, which you need here.
NCERT Class 11 Chapter 4, Laws of Motion (reprint 2026-27), section 4.9.1, gives the rules: static friction satisfies fs ≤ μsN, kinetic friction fk = μkN, and both are empirical laws that are only approximately true. A worked example in the same section shows that a block on an incline starts to slide when tan θ = μs. You will see the same tan θ = μ appear below for a different reason.
Why does pulling upward help?
Two effects compete when you tilt the rope by θ:
The vertical part, F sin θ, lifts some weight off the floor. The normal force drops to N = mg − F sin θ, so limiting friction μN drops too.
The horizontal part, F cos θ, is the part that drags. It shrinks as θ grows.
At small angles the first effect wins. At large angles the second does.
How do you find the best angle?
Set the block on the point of sliding: F cos θ = μ(mg − F sin θ). Solve for F:
F = μmg / (cos θ + μ sin θ)
F is smallest when the denominator D = cos θ + μ sin θ is largest. Differentiate: dD/dθ = −sin θ + μ cos θ = 0, so tan θ = μ. Then cos θ = 1/√(1 + μ²) and sin θ = μ/√(1 + μ²), so D = √(1 + μ²) and
F_min = μmg / √(1 + μ²)
There is a second way that needs no calculus. Combine the floor's normal force and limiting friction into one contact force R. It always leans back from the vertical by the angle λ, where tan λ = μ. The block then feels three forces: mg, R in a fixed direction, and your pull. The shortest pull that closes the force triangle is perpendicular to R, which tilts it up by λ. That gives F_min = mg sin λ, the same answer.
Worked example 1: how much does the angle matter?
A 20 kg block, μs = 0.5, g = 9.8 m/s², so mg = 196 N.
Flat pull (0°): F = 0.5 × 196 = 98.0 N.
20°: 88.2 N.
26.6° (tan⁻¹ 0.5): F = 98/√1.25 = 87.7 N, the minimum.
30°: 87.8 N.
45°: 92.4 N.
60°: 105.0 N, worse than pulling flat.
Check the optimum: N = 196 − 87.7 × sin 26.6° = 156.8 N, so μN = 78.4 N, and F cos θ = 87.7 × 0.894 = 78.4 N. The forces balance. Note how flat the curve is near the minimum: anywhere from 20° to 30° is within 1 N.
Worked example 2: what if you push down instead?
Push the same block with the force pointing down at θ below the horizontal. Now N = mg + F sin θ, and F = μmg/(cos θ − μ sin θ).
20°: 127.5 N.
45°: 277.2 N.
When tan θ ≥ 1/μ = 2, that is θ ≥ 63.4°, the denominator is zero or negative. No push, however large, will move the block. Each extra newton adds more friction than drag.
Worked example 3: accelerating a moving block
The block is already sliding with μk = 0.3, and you want a = 0.5 m/s². Newton's second law gives F cos θ − μk(mg − F sin θ) = ma, so F = m(a + μkg)/(cos θ + μk sin θ).
The best angle is tan⁻¹ 0.3 = 16.7°. Then F = 20 × (0.5 + 2.94)/√1.09 = 65.9 N, against 68.8 N for a flat pull. The acceleration changes the force, not the best angle.
Which mistakes cost marks?
Taking N = mg when the force is angled.
Using μk to start the block moving. Starting needs μs.
Writing tan θ = 1/μ for the best angle. That is the jamming angle for a push.
Ignoring lift-off. The formula needs N > 0, so F sin θ must stay below mg.
Calculator in radians while you type degrees.
When the set-up is the hard part, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, including which forces go into each equation.
A practice routine
Draw the free-body diagram and derive F(θ) without notes.
Find the minimum by calculus, then again with the force triangle.
Redo example 1 with μ = 0.75. You should get 36.9° and F_min = 0.6mg.
Revise static versus kinetic friction with our friction problems guide, and the calculus with maxima and minima applications.
Before your next test, derive tan θ = μ both ways on one page.
