- 2067 Asoj · 6 marks
Force P to prevent block sliding down incline
Short answer
Resolve forces parallel and perpendicular to the incline. Set friction F = 0.4 N acting up the slope to prevent downward slide. Equilibrium gives P (cos 20° +0.4 sin 20°) =490.5 (sin 30° -0.4 cos 30°). Solving yields P ≈70 N.
Find the value of P to prevent the slide down of block as given below. [Figure: 50 kg block on a 30° incline, μ = 0.4, force P applied at 20° to the incline/horizontal as shown.]
Reading of the figure: a 50 kg block on a 30° incline, μ = 0.4. P pushes the block up the slope. Its line of action makes 20° with the incline surface, directed towards the surface (so P has a component along the slope, up, and a component pressing the block on to the plane). Take g = 9.81 m/s²: W = 50 × 9.81 = 490.5 N.
Does it slide without P? tan 30° = 0.577 > μ = 0.4, so (φ = 21.8° < 30°) the block slides down by itself. P is needed to hold it.
Impending motion down the slope (friction acts up the slope, F = μN)
Axes along and perpendicular to the incline.
Perpendicular: N = W cos30° + P sin20°
Along incline: P cos20° + F = W sin30°, F = 0.4 N
Substituting: P cos20° + 0.4 (W cos30° + P sin20°) = W sin30° P (cos20° + 0.4 sin20°) = W (sin30° − 0.4 cos30°) P (0.9397 + 0.1368) = 490.5 (0.5 − 0.3464) P × 1.0765 = 490.5 × 0.1536 = 75.35 P = 70.0 N
Check: N = 490.5 × 0.8660 + 70.0 × 0.3420 = 448.7 N, F = 0.4 × 448.7 = 179.5 N, and P cos20° + F = 65.8 + 179.5 = 245.3 N = W sin30° ✓.
Answer: P ≈ 70 N (minimum value to prevent the block from sliding down). If the 20° is instead taken from the horizontal, the same method gives P ≈ 71 N.
✓ Answer checked by Claude Opus 5.5 (high effort)
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