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Net force calculator: find the resultant force and acceleration

The net force (resultant force) is all the forces on an object added together, taking their directions into account. For forces along a line, add the ones pushing one way and subtract the ones pushing the other: a car with 3000 N of driving force and 1000 N of resistance has a net force of 2000 N forwards. Newton's second law then gives the acceleration: F_net = ma. This net force calculator is part of our Force Calculator, which finds the resultant force and the acceleration it causes.

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Force Calculator

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How to find net force with the calculator

  1. Open the Force Calculator. Under Calculate, choose Resultant.
  2. To find the acceleration, choose Acceleration and type the driving force, the resistive force and the mass. The extra result is the resultant force.
  3. To find the push needed for a given acceleration, choose Driving Force and type the mass, the acceleration and the resistive force.
  4. Read the answer and the steps, and look at the force diagram: arrows for the driving and resistive forces, and the acceleration they give.
  5. If you already know the net force or want it from mass and acceleration, use the F = ma group instead, which finds force, mass or acceleration.

The Resultant modes handle two forces acting in opposite directions along one line, which covers most textbook questions. For forces at angles, use the method further down, or add them as vectors in the Vectors Calculator.

Net force formula

F_{\text{net}} = F_1 + F_2 + F_3 + \dots = ma

The net force is the vector sum of every external force on the object. Choose one direction as positive, give forces the other way a minus sign, and add. Force is in newtons (N), mass in kilograms and acceleration in m/s². For a driving force and a resistive force along one line, as in the calculator:

F_d - F_r = ma \qquad a = \frac{F_d - F_r}{m}

Worked examples you can check in the calculator

Example 1: net force and acceleration of a car

A 1200 kg car has a driving force of 3000 N and total resistive forces of 1000 N. These are the Resultant mode's starting values.

F_{\text{net}} = 3000 - 1000 = 2000 \qquad a = \frac{2000}{1200} = 1.67

The net force is 2000 N forwards and the car accelerates at 1.67 m/s².

Example 2: a tug of war

One team pulls left with 450 N and the other pulls right with 520 N. Taking right as positive, F_net = 520 − 450 = 70 N to the right, so the rope and both teams accelerate to the right.

Example 3: the driving force needed

A 1200 kg car needs to accelerate at 2 m/s² against 800 N of resistance. The net force must be ma = 1200 × 2 = 2400 N, so the driving force is 2400 + 800 = 3200 N. Choose Driving Force to check.

Example 4: balanced forces

A skydiver's weight is 700 N downwards and air resistance has grown to 700 N upwards. The net force is 700 − 700 = 0 N. Zero net force doesn't mean the skydiver stops: by Newton's first law they carry on at a constant velocity, called terminal velocity. In the calculator, equal driving and resistive forces give an acceleration of 0.

Try one, then press Mark to check your answer in the Laws of Motion course.

Find the force, mass or acceleration in a random F = ma problem.

Press Mark to have your answer marked in the Laws of Motion course.

Net force at an angle

When forces aren't along one line, split each into horizontal and vertical parts (components), add each set, then combine them with Pythagoras. A force F at angle θ to the horizontal has components F cos θ across and F sin θ up.

Example 5: two forces at right angles

A box is pulled with 30 N east and 40 N north. The net force is √(30² + 40²) = 50 N, at an angle of tan⁻¹(40 ÷ 30) = 53.1° north of east. In the Vectors Calculator, choose Two Vectors and type a = (30, 0) and b = (0, 40): the resultant |r| is 50 and its direction is 53.13°.

Example 6: one force at an angle

A sled is pulled with 50 N at 30° above the horizontal, while friction pulls back with 20 N. The pull's components are 50 cos 30° = 43.30 N across and 50 sin 30° = 25 N up.

DirectionForcesTotal
Horizontal43.30 − 2023.30 N
Vertical2525 N

Combining them, the net force from these two is √(23.30² + 25²) = 34.2 N at 47.0° above the horizontal. On the ground, weight and the normal force also act vertically, so in a full problem you would include them in the vertical total.

Try one

practice
Pick an answer, then check it.
A 5 kg trolley is pushed with 30 N while friction acts against it with 10 N. What is its acceleration?

Common mistakes

  • Adding forces that act in opposite directions. Give one direction a minus sign first.
  • Using one force instead of the net force in F = ma. The acceleration comes from the resultant, not from the driving force alone.
  • Thinking zero net force means not moving. It means constant velocity, which can be zero or not.
  • Adding forces at angles as plain numbers. 30 N and 40 N at right angles make 50 N, not 70 N.
  • Mixing up mass and weight. Mass is in kg. Weight is a force, W = mg, in newtons.

Common questions

How do you calculate net force?
Add all the forces on the object, with forces in opposite directions given opposite signs. If you know the mass and acceleration instead, F_net = ma.
Is net force the same as resultant force?
Yes. Both mean the single force that has the same effect as all the forces together.
What happens when the net force is zero?
The forces are balanced and the object doesn't accelerate. It stays still or keeps moving at a constant velocity.
How do you find net force with forces at angles?
Resolve each force into horizontal and vertical components, add the components in each direction, then use Pythagoras for the size and inverse tan for the direction.
What is the unit of net force?
The newton (N). One newton gives a 1 kg mass an acceleration of 1 m/s².

Revision cards

Read the front, say your answer, then flip to check.

Where to go next

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