Video narration
We have a 10 kg block on a 45° rough incline with μ = 0.6. Forces along the plane are: weight component down the plane, a 3 N force down the plane as shown, an applied force P up the plane, and friction acting upward as stated. For the block to be just at rest with the minimum P, friction must be at its limiting value: f = μN. First find the normal reaction. Perpendicular to the plane, only N balances the perpendicular component of weight, so N = mg cos 45°. Now balance forces along the plane. Taking up the plane as positive: P + f − (mg sin 45° + 3) = 0. Substitute f = μN = μ mg cos 45°: P = mg sin 45° + 3 − μ mg cos 45°. With m = 10 kg, g ≈ 10 m/s², sin 45° = cos 45° = 1/√2: P = 100/√2 + 3 − 0.6×100/√2 = 40/√2 + 3 ≈ 28.3 + 3 ≈ 31.3 N, so the minimum P is about 32 N.