Forces have size and direction, so a correct answer has both: “30 N to the right”, not just “30”. Whenever you add or compare forces, direction decides whether they add up or cancel.
This page shows the fix with one original example. The reasoning board gives practice at choosing a relationship before calculating, and the bounds and rounding explainer helps you set a sensible number of significant figures.
Why is direction easy to forget?
In daily life “force” just means “push” or “strength”, and pushes are described by how big they are. In Physics the same word has a direction built in.
Numbers on a diagram invite you to add them. Nothing on the page stops you from writing 50 + 20, but the diagram is telling you one force goes each way.
A routine that keeps direction
- Draw the object as a box or dot, with every force as an arrow from it. Label each with name and size.
- Choose a positive direction and write it, for example “right is positive”.
- Add forces in that direction and subtract forces in the opposite direction. Forces at right angles to it are handled in their own line.
- State the resultant with direction, in words.
- Only then use F = ma, with the resultant, never one of the individual forces.
Worked example: pulling a box
A 6.0 kg box on a floor is pulled to the right with a force of 50 N. Friction acts on the box with a force of 20 N to the left. Take g = 10 N/kg. Find the resultant force and the acceleration.
Step 1, horizontal forces. Take right as positive. Pull = +50 N. Friction = −20 N.
Step 2, resultant. 50 − 20 = 30 N to the right.
Step 3, acceleration. a = F ÷ m = 30 ÷ 6.0 = 5.0 m/s² to the right.
Step 4, vertical forces. Weight = 6.0 × 10 = 60 N downwards. The floor pushes up with 60 N. These balance, so there is no vertical acceleration.
Check: with a = 5.0 m/s², the resultant should be m × a = 6.0 × 5.0 = 30 N. It matches.
The mistake to watch for
A student sees 50 N and 20 N on the diagram and writes:
Resultant = 50 + 20 = 70 N, so a = 70 ÷ 6.0 = 11.7 m/s².
The error is in step 2. The friction arrow points opposite to the pull, so it reduces the resultant. The student has treated “a force of 20 N” as a number to add, not as a push in a direction. The wrong acceleration is more than double the right one.
A second slip is leaving direction out of the answer. “5.0 m/s²” loses a mark that “5.0 m/s² to the right” keeps.
Check yourself
1. A drawer is pushed with 12 N to the right and a person pulls it with 18 N to the left. Find the resultant force.
Show answer
Take right as positive: +12 − 18 = −6. The resultant is 6 N to the left.
2. A car’s engine provides a forward force of 2000 N. Resistance forces total 2000 N backwards. State the resultant force and describe the car’s motion.
Show answer
Resultant = 2000 − 2000 = 0 N. There is no change in motion, so the car continues at its constant velocity. It has not stopped.
3. A 4.0 kg object hangs from a string that pulls up with 30 N. Take g = 10 N/kg. Find the resultant force and the acceleration, with directions.
Show answer
Weight = 4.0 × 10 = 40 N downwards. Take up as positive: +30 − 40 = −10 N, so the resultant is 10 N downwards. Acceleration = 10 ÷ 4.0 = 2.5 m/s² downwards.
Where this leads next
Start with drawing a force diagram for a stated situation, then calculating a resultant force and relating acceleration to resultant force. The idea of zero resultant is explained in balanced forces and the absence of forces. The forces and momentum module holds the full route and its mixed practice lets you test it.
If direction errors keep returning in your own working, a teacher can go through it with you in online one-to-one Physics tuition.