All calculatorsNewton’s First Law Calculator
Analyze the net force on an object and see whether it’s in equilibrium. Enter mass and up to three forces; the calculator finds the resultant force vector, checks if ∑F = 0, and (if mass is given) reports the resulting acceleration.
Background
Newton’s First Law (the law of inertia) says that an object at rest or moving with constant velocity will continue in that state unless a net external force acts on it. If the vector sum of all forces is zero (∑F = 0), the object is in equilibrium. If ∑F ≠ 0, its motion changes according to Newton’s Second Law, ∑F = m a.
How to use this calculator
- Enter the object’s mass in kilograms. If you leave it blank, the tool still checks equilibrium but cannot compute acceleration.
- For each force, enter a magnitude in newtons and an angle in degrees from the +x axis (counterclockwise). Leave unused forces blank.
- Use the quick pick chips to auto-fill common setups like a book on a table or a horizontal pull on a box.
- Click Calculate to see the net force vector, check whether ∑F ≈ 0 (equilibrium), and (if mass is given) compute the acceleration vector.
How this calculator works
Each force is treated as a vector with a magnitude F and angle θ. The calculator resolves each one into components using Fx = F cosθ and Fy = F sinθ, then sums components to get the net force:
∑Fx = F1x + F2x + F3x,
∑Fy = F1y + F2y + F3y.
The magnitude of the net force is |∑F| = √(∑Fx² + ∑Fy²) and its direction is found with θnet = tan⁻¹(∑Fy / ∑Fx) using the correct quadrant.
If |∑F| is extremely small (within a numerical tolerance), the object is treated as being in equilibrium, consistent with Newton’s First Law. If mass m is provided and ∑F ≠ 0, the calculator uses Newton’s Second Law, ∑F = m a, to compute the acceleration vector: ax = ∑Fx/m, ay = ∑Fy/m.
Formula & Equation Used
1. Component form of forces
2. Net force from components
3. Newton’s First and Second Laws
When ∑F = 0, the object is in translational equilibrium. When ∑F ≠ 0, its acceleration is given by a = ∑F / m.
Example Problems & Step-by-Step Solutions
Example 1 — Book resting on a table
A 2.0 kg book rests on a horizontal table. The table exerts an
upward normal force of 19.6 N, while gravity pulls downward with
19.6 N. Treat upward as +y and downward as −y.
Forces: FN = 19.6 N at 90°, Fg = 19.6 N at −90°.
∑Fy = 19.6 + (−19.6) = 0 N, ∑Fx = 0 N.
|∑F| = 0 N ⇒ the book is in equilibrium and remains at rest
(Newton’s First Law).
Example 2 — Box pulled to the right
A 5.0 kg box is pulled with a 20 N horizontal force to the right.
Friction is negligible.
Forces: Fpull = 20 N at 0°, FN and Fg
cancel in the vertical direction.
∑Fx = 20 N, ∑Fy ≈ 0 N,
so |∑F| ≈ 20 N.
Using ∑F = m a: a = (20 N) / (5.0 kg) = 4.0 m·s⁻² to the right.
The box speeds up in the +x direction.
Frequently Asked Questions
Q: When is an object considered to be in equilibrium?
An object is in translational equilibrium when the vector sum of all external forces is zero (∑F = 0). In that case, its velocity is constant: it either remains at rest or moves with a constant speed in a straight line, consistent with Newton’s First Law.
Q: Why do we bother breaking forces into components?
Most forces act at angles, so adding them directly is tricky. By splitting each force into x and y components, we can add components separately (∑Fx, ∑Fy) and then reconstruct the net force vector using basic trigonometry.
Q: What happens if mass is zero or not given?
If mass is not provided, the calculator can still compute the net force and check equilibrium (∑F ≈ 0 or not). However, without a nonzero mass it cannot compute acceleration from ∑F = m a. If you enter mass = 0, the tool will warn you that acceleration can’t be determined.
