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Most students write N = mg cos θ for every incline. Add a horizontal push and that line is wrong. Here is why, and a worked example with real numbers.
Why is the normal force not mg cos θ any more?
The normal force is the push a surface gives back, at right angles to the surface (OpenStax University Physics, section 5.6). That same section gives N = mg cos θ for an object at rest or sliding on an incline, which holds only when weight is the only force with a part perpendicular to the slope.
A horizontal force is not parallel to the slope, so part of it presses into the surface. The surface pushes back harder, and N goes up by exactly that part. If you want to see how often N stops being mg, our post on cases where normal reaction is not equal to mg walks through five of them.
How do you resolve a horizontal force on an incline?
Use axes along the slope and perpendicular to it. Weight splits into mg sin θ down the slope and mg cos θ into the slope (OpenStax, section 5.6). Now do the same for the horizontal push F, which makes an angle θ with the slope:
Along the slope (up the slope, if F pushes the block towards the incline): F cos θ.
Perpendicular to the slope (into the surface): F sin θ.
Perpendicular balance, since the block does not lift off or sink in: N = mg cos θ + F sin θ.
Along the slope: compare F cos θ with mg sin θ to see which way the block tends to move, then let friction oppose that tendency.
The quick check for the sin-cos swap: when θ is 0 the slope is flat floor, and a horizontal push should add nothing to N. F sin 0 = 0, so the formula passes. Our guide to resolving forces on an inclined plane without the sin-cos swap has more checks of this kind.
How does the push change friction?
Static friction can grow only up to μs N, and kinetic friction equals μk N (OpenStax University Physics, section 6.2). So a bigger N raises the most friction the slope can supply. The push does two jobs at once: F cos θ drives the block up the slope, while F sin θ makes it harder to slide.
Worked example: 5 kg block on a 30° slope
Take m = 5 kg, θ = 30°, g = 9.8 m/s², a horizontal push F = 40 N towards the slope, and μs = 0.5. That is the static coefficient OpenStax lists for wood on wood (OpenStax, section 6.2).
Weight: mg = 49 N. mg sin 30° = 24.5 N down the slope; mg cos 30° = 42.4 N into it.
Push: F cos 30° = 34.6 N up the slope; F sin 30° = 20.0 N into it.
Normal force: N = 42.4 + 20.0 = 62.4 N. Without the push it would be 42.4 N.
Tendency: 34.6 N up against 24.5 N down leaves 10.1 N trying to move the block up the slope.
Friction limit: μs N = 0.5 × 62.4 = 31.2 N. That is more than 10.1 N, so the block stays still.
Actual friction: 10.1 N, pointing down the slope. Not 31.2 N, and not up the slope.
Step 6 is the easy one to get wrong. Static friction is only as big as it needs to be, and here it points down the slope because the push is winning. The same idea drives the problem of the minimum and maximum force to keep a block at rest on an incline.
How hard must you push to make it slide up?
At the point of slipping up the slope, friction is μs N, pointing down:
F cos θ = mg sin θ + μs (mg cos θ + F sin θ)
so F = mg (sin θ + μs cos θ) / (cos θ − μs sin θ). With the numbers above, F = 49 × 0.933 / 0.616 ≈ 74 N.
Look at the bottom of that fraction. If μs sin θ is equal to or bigger than cos θ, which means tan θ ≥ 1/μs, no horizontal push can slide the block up. Pushing harder adds at least as much friction as it adds drive. With μs = 0.5 that happens at about 63.4°, the angle whose tangent is 2.
What should you practise next?
Redo the example with the push pointing away from the slope, then with μs = 0.2. If your answer goes wrong and you cannot find where, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. If what is shaky is the idea of normal reaction itself, the Fermi Concept Library gives each physics concept its own lesson.
