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F = ma calculator: force, friction and inclined plane

Solve Newton's second law F = ma for force, mass or acceleration, or find acceleration, normal force and friction for a block on an inclined plane.

Mis à jour Exemples vérifiés : 6

Essayer
Accélération
m/s²
Accélération: 2 m/s²
Chiffres significatifs : 6 ; Au plus proche, égalités en s’éloignant de zéro
Net force
3,000N
Masse
1,500kg
Weight
14,710N

A net force of 3,000 N on 1,500 kg gives an acceleration of 2 m/s² (0.2039 g).

Net force and acceleration

1,500 kgF = 3,000 Na = 2 m/s²
Comment le calcul est effectué S

  1. Newton's second law

    a=Fm=3,0001,500=2 m/s2a = \frac{F}{m} = \frac{3{,}000}{1{,}500} = 2\ \mathrm{m/s^2}

  2. Weight on Earth for comparison

    W=mgn=(1,500)(9.80665)=14,710 NW = mg_n = (1{,}500)(9.80665) = 14{,}710\ \mathrm{N}

À propos de F = ma calculator: force, friction and inclined plane

Newton's second law says the net force on a body equals its mass times its acceleration, F = ma, so one newton gives 1 kg an acceleration of 1 m/s². Enter two of force, mass and acceleration to get the third. The inclined-plane mode splits the block's weight into mg sin θ down the slope and mg cos θ into it; the normal force balances the second part, and kinetic friction μk × N opposes the sliding.

The default, 3,000 N on a 1,500 kg car, gives 2 m/s², about 0.2 g. On the slope, a 10 kg block on a 30° incline with μk = 0.2 slides down at 3.20 m/s² against 16.99 N of friction.

Friction coefficients depend on both surfaces and their condition, so treat slope results as estimates. The block stays put while tan θ is at or below the static coefficient μs; leave μs blank and μk is used as that threshold.

Exemples détaillés

3000 N on a 1500 kg car

Problem
F = ma
Solve for
Accélération
Net force F
3000 N
Mass m
1500 kg
Unité des résultats
SI
Accélération
2 m/s²

Source de vérification : Python 3.8: a = F/m = 3000/1500

Force to give 2 kg an acceleration of 1 g

Problem
F = ma
Solve for
Force
Mass m
2 kg
Acceleration a
1 g
Unité des résultats
SI
Net force
19.6133 N

Source de vérification : Python 3.8: 2 × 9.80665 = 19.6133

100 lbf gives 1 g: mass is 100 lb (45.359237 kg)

Problem
F = ma
Solve for
Masse
Net force F
100 lbf
Acceleration a
1 g
Unité des résultats
SI
Masse
45.3592 kg

Source de vérification : Definition of the pound-force (NIST SP 811 B.8): 1 lbf = 0.45359237 kg × 9.80665 m/s²

10 kg block on a 30° slope, μk = 0.2

Problem
Inclined plane
Block mass
10 kg
Incline angle
30 °
Kinetic friction coefficient μk
0.2
Unité des résultats
SI
Accélération
3.20476 m/s²
Normal force
84.9281 N
Friction force
16.9856 N
Block
Slides down

Source de vérification : Python 3.8 math: N = mg cos30° = 84.92808…, f = 0.2N, a = g(sin30° − 0.2 cos30°) = 3.2047634

Questions

What is one newton of force?

One newton is the force that gives a 1 kg mass an acceleration of 1 m/s², so 1 N = 1 kg·m/s² in the SI. A 1 kg mass weighs 9.80665 N under standard gravity, and one pound-force is exactly 4.4482216152605 N (NIST SP 811). A 100 lbf push that accelerates an object at 1 g therefore means a mass of 100 lb, or 45.36 kg.

How do you calculate the friction force?

Friction equals the coefficient of friction times the normal force, f = μN. On level ground N = mg, so a 10 kg crate with μk = 0.2 needs a 19.61 N push to keep sliding at constant speed. Static friction is a limit, not a fixed value: it matches the applied force up to μs × N, and the object starts to move once that limit is passed.

At what angle does an object start to slide down a slope?

It starts to slide once the slope angle exceeds arctan μs, the point where the pull down the slope, mg sin θ, overtakes the most static friction can supply, μs mg cos θ. With μs = 0.7 that angle is 34.99°; with μs = 0.2 it is 11.31°. Tilting a surface until an object slips and reading the angle is a standard way to measure μs.

What are typical coefficients of friction?

OpenStax University Physics Table 6.1 gives approximate values, static then kinetic: rubber on dry concrete 1.0 and 0.7, wood on wood 0.5 and 0.3, ice on ice 0.1 and 0.03. Real values shift with surface finish, moisture and temperature, so a measured coefficient beats a table value. The kinetic coefficient is lower than the static one for each of these pairs.

Does a heavier block slide down a slope faster?

No. Mass cancels in a = g(sin θ − μk cos θ), because both the pull down the slope and the friction are proportional to the weight. A 5 kg block and a 50 kg block on a frictionless 45° slope both accelerate at 6.93 m/s². Differences seen in practice come from air resistance and from friction coefficients that change with load or speed.

Quelle est la précision de « F = ma calculator: force, friction and inclined plane » ?

La précision dépend de vos données et des hypothèses de la méthode. Le calcul décimal utilise 50 chiffres significatifs, mais les estimations, méthodes numériques et données sources peuvent être moins précises ; l’arrondi affiché ne supprime pas ces limites. Exemples résolus vérifiés à partir de sources indépendantes : 6. Par exemple, « 3000 N on a 1500 kg car » est vérifié à l’aide de Python 3.8: a = F/m = 3000/1500.

D’où vient cette méthode ?

OpenStax University Physics Volume 1, §5.3 Newton's second law and §6.2 Friction; HyperPhysics — Inclined plane with friction.

À propos de ce calculateur

F=ma;N=mgcos⁡θ,a=g(sin⁡θ−μkcos⁡θ) when tan⁡θ>μsF = ma;\qquad N = mg\cos\theta,\quad a = g(\sin\theta - \mu_k\cos\theta)\ \text{when } \tan\theta > \mu_s

Sources

  1. OpenStax University Physics Volume 1, §5.3 Newton's second law and §6.2 Friction
  2. HyperPhysics — Inclined plane with friction

Vérifié avec les références

Ce calculateur comprend 6 exemples résolus dont les réponses proviennent de sources indépendantes. Ils font partie de la suite de tests et peuvent aussi être exécutés ici.

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