CalcOpenly
Deutsch

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.

Aktualisiert Geprüfte Beispiele: 6

Ausprobieren
Beschleunigung
m/s²
Beschleunigung: 2 m/s²
Signifikante Stellen: 6; Zum nächsten Wert, bei Gleichstand von null weg
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
So wird gerechnet 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}

Über 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.

Durchgerechnete Beispiele

3000 N on a 1500 kg car

Problem
F = ma
Solve for
Beschleunigung
Net force F
3000 N
Mass m
1500 kg
Ergebniseinheit
SI
Beschleunigung
2 m/s²

Prüfquelle: Python 3.8: a = F/m = 3000/1500

Force to give 2 kg an acceleration of 1 g

Problem
F = ma
Solve for
Kraft
Mass m
2 kg
Acceleration a
1 g
Ergebniseinheit
SI
Net force
19.6133 N

Prüfquelle: 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
Ergebniseinheit
SI
Masse
45.3592 kg

Prüfquelle: 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
Ergebniseinheit
SI
Beschleunigung
3.20476 m/s²
Normal force
84.9281 N
Friction force
16.9856 N
Block
Slides down

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

Fragen

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.

Wie genau arbeitet „F = ma calculator: force, friction and inclined plane“?

Die Genauigkeit hängt von Ihren Eingaben und den Annahmen der Methode ab. Die Dezimalrechnung nutzt 50 signifikante Stellen, doch Schätzungen, numerische Verfahren und Quelldaten können ungenauer sein. Die angezeigte Rundung beseitigt diese Grenzen nicht. Anhand unabhängiger Quellen geprüfte Rechenbeispiele: 6. Beispielsweise wird „3000 N on a 1500 kg car“ anhand von Python 3.8: a = F/m = 3000/1500 geprüft.

Woher stammt die Methode?

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

Über diesen Rechner

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

Quellen

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

Anhand von Quellen geprüft

Dieser Rechner enthält 6 Rechenbeispiele mit Ergebnissen aus unabhängigen Quellen. Sie laufen in der Testsuite und können auch hier ausgeführt werden.

Verwandte Rechner