Solve reaction forces for blocks, ramps, and lifts. Find beam support reactions from multiple loads. Download clean CSV and PDF summaries for sharing easily.
| Scenario | Key inputs | Key outputs |
|---|---|---|
| Normal reaction on an incline | m=20 kg, g=9.81, θ=30°, P=0, μ=0.35 | N≈169.914, μN≈59.470 |
| Normal reaction with pull-away force | m=15 kg, θ=0°, P=80 N, φ=90°, away, μ=0.20 | N≈67.100, μN≈13.420 |
| Beam reactions with UDL | L=6 m, w=200 N/m, no point loads | RA=600.000, RB=600.000 |
| Beam reactions with point loads | L=5 m, P1=500 N at x=2 m, P2=300 N at x=4 m | RA=420.000, RB=380.000 |
For a simply supported 5.0 m beam with a 500 N load at x = 2.0 m, moments give RB = (500×2)/5 = 200 N and RA = 300 N. Adding a 300 N load at x = 4.0 m shifts reactions to RA = 420 N and RB = 380 N.
For a block on a plane, the perpendicular weight component governs contact: N = m g cosθ. With m = 20 kg and g = 9.81 m/s², N is about 196.2 N at θ = 0°, 169.9 N at 30°, and 98.1 N at 60°. As θ rises, contact pressure and friction capacity drop even if mass is unchanged.
External forces change N through the perpendicular component, P sinφ. With m = 15 kg on level ground, weight alone gives N ≈ 147.1 N. A pull-away P = 80 N at φ = 90° reduces N to about 67.1 N, while a push toward the surface raises N to about 227.1 N. If N becomes negative, lift-off occurs and contact is lost.
Maximum static friction is limited by Fmax = μN. Using μ = 0.35 with N ≈ 169.9 N yields Fmax ≈ 59.5 N. If tangential demand is 45 N, the margin is 59.5/45 ≈ 1.32. When μ is uncertain, check a range such as 0.20 to 0.45 to see sensitivity.
Uniformly distributed loads reduce to an equivalent force W = wL acting at midspan. For L = 6 m and w = 200 N/m, W = 1200 N at x = 3 m. Symmetry produces RA = RB = 600 N. With point loads, the calculator combines Σ(Pi xi) with W(L/2) to preserve moment balance.
Results are strongest when assumptions are explicit: geometry, load directions, units, and rounding. Exporting CSV supports quick checks, while PDF provides a stable record for reviews. Adjust one variable at a time, then compare reaction shifts to component ratings, anchor capacities, and allowable bearing pressures. For audits, note input sources and keep a versioned calculation date in reports.
It is the contact force perpendicular to the surface that prevents interpenetration. It balances the perpendicular components of weight and any applied forces when the body remains in contact.
If an applied pull-away component exceeds the perpendicular weight component, the computed normal reaction becomes negative. That indicates the surface cannot “pull” the object, so contact is lost and N is set to zero.
In this simplified model, friction is tangential and does not alter the perpendicular balance. The calculator uses N to compute the maximum static friction limit μN for quick feasibility checks.
Use it for statically determinate, simply supported beams where reactions at two supports are found from force balance and moments. It supports a full-span uniform load plus up to three point loads.
Use kilograms, meters, and seconds with forces in newtons. Keep locations between 0 and L. Consistent units ensure reactions and friction limits are reported correctly.
CSV and PDF exports reflect the computed tables and your selected rounding for downloads. Accuracy depends on correct inputs, geometry assumptions, and whether the real system matches the idealized static models.
Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.