A force diagram shows every force on a single object as an arrow, drawn from the object. The resultant force follows from adding the arrows.
This lesson is part of SPM Physics force and motion I. Its main use is in applying Newton’s laws in context.
How do I draw one?
Follow the same order each time.
- Draw the object as a box or a dot.
- Draw the weight straight down from the centre.
- For each thing that touches the object, draw the force it exerts: surface, rope, hand, air.
- Label each arrow with a name and a value if given.
- Make arrow lengths roughly match the sizes.
Worked example: a box pulled along the floor
A 5.0 kg box is pulled along a horizontal floor by a rope with a 20 N horizontal force. Friction is 8.0 N. Use g = 9.8 m s⁻².
The forces are weight, 5.0 × 9.8 = 49 N down, the normal reaction, 49 N up, the pull of 20 N to the right and friction 8.0 N to the left.
The vertical forces balance, so the resultant is horizontal: 20 − 8.0 = 12 N to the right. The acceleration is a = F ÷ m = 12 ÷ 5.0 = 2.4 m s⁻².
The mistake that loses marks
A student adds an arrow labelled “force of motion” forward, and adds it to the 20 N pull, so the resultant comes out too large. No object exerts such a force.
Test every arrow with one question: what is pushing or pulling? If there is no answer, remove the arrow.
Check yourself
A ball has just reached the top of its path after being thrown straight up. Ignoring air resistance, name the forces on it, and state the resultant.
Answer
Only one force acts: the weight, downward. There is no “upward force from the throw”, because the hand is no longer touching the ball.
The resultant is the weight, mg, downward. The ball is momentarily at rest but still accelerating at g.
What to study next
Put the diagram to use in applying Newton’s laws in context. Then test yourself in the force and motion I practice set.
The units and significant figure checker checks the last line of a resultant calculation. If you want a teacher to watch you draw new diagrams, see online one-to-one Physics tuition.