Free Body Diagrams: The Art of Seeing One Object at a Time

You are standing in a lift. The doors close. The lift begins to move upward.

For a brief moment — just as it starts — you feel heavier. Your legs press harder into the floor. The sensation is unmistakable.

Now the question: are you actually heavier?

Most students say yes. The lift is going up, the force must be greater, therefore you are heavier.

This is wrong. And understanding exactly why it is wrong — with precision, not just intuition — is what this article is about. By the end, you will have a method that solves this problem and every problem like it, automatically, without confusion.

The one shift that fixes everything

Newton's second law is F = ma. You know this. But there is a condition attached to it that most textbooks mention once and never emphasize again:

F = ma applies to one object at a time.

Not the lift and you together. Not the rope and the block together. One object, in isolation, with only the forces acting directly on that object counted.

This sounds obvious. It is not. The moment a problem has two objects touching each other, most students unconsciously start mixing forces from both. The equation becomes wrong not because F = ma is wrong, but because F is being computed for the wrong system.

The free body diagram is the technique that enforces this discipline. You draw one object. You draw every force acting on it. You draw nothing else. Then you write F = ma for that object alone.

That is the entire method. The rest is practice.

Back to the lift

Let us apply this method to you standing in the lift.

Step 1: Isolate the object.

The object is you. Not you-plus-lift. Not the lift. You, alone, drawn as a simple shape in space.

Step 2: Identify every force acting directly on you.

Two forces act on you:

Your weight, mg, pulling downward. This is Earth's gravity acting on your mass. It acts whether the lift moves or not, whether the lift exists or not. It is always mg downward.

The normal force, N, pushing upward. This is the lift floor pushing on your feet. It is a contact force — it exists only because you are in contact with the floor.

Nothing else. The lift's engine does not act on you directly. The cable does not act on you. The walls do not act on you (assuming you are not touching them). Only these two forces.

Step 3: Write F = ma.

Take upward as positive. The net force on you is N − mg. This equals your mass times your acceleration.

$N - mg = ma$

$N = mg + ma = m(g + a)$

When the lift accelerates upward, a is positive. So N > mg. The floor pushes you harder than your weight alone. Your feet feel more force. You feel heavier.

But your weight — mg — has not changed. Earth's gravitational pull on you is exactly what it always was. What changed is the normal force. The floor is pushing up harder because it needs to both support you and accelerate you upward.

The sensation of heaviness is not weight. It is normal force. These are different things, and confusing them is the source of most lift-problem errors.

The three cases, clearly

The equation N = m(g + a) handles every lift scenario:

Lift stationary or moving at constant velocity. Acceleration is zero. N = mg. The floor pushes exactly as hard as your weight. You feel normal.

Lift accelerating upward. Acceleration is positive. N = m(g + a) > mg. Floor pushes harder. You feel heavier. This is what you felt when the lift started moving up.

Lift accelerating downward — or decelerating while moving up. Acceleration is negative. N = m(g − a) < mg. Floor pushes less hard. You feel lighter. In free fall, a = g, N = 0. You feel weightless — the floor pushes you with zero force.

Notice that what matters is not the direction of motion but the direction of acceleration. A lift moving upward but slowing down has downward acceleration — and you feel lighter, even though you are moving up. This is the twist that most students get wrong on exams.

The misconception that kills marks

Here is the wrong approach, written out explicitly so you can recognize it in yourself:

"The lift is going up, so I will add an upward force for the lift's motion."

There is no such force. Motion is not a force. A lift moving upward at constant speed exerts exactly the same normal force as a stationary lift. Speed has no effect on the normal force. Only acceleration does.

The free body diagram prevents this error automatically. When you draw the diagram and ask "what forces act directly on this object," there is no entry for "the motion of the lift." Motion is not on the list. Only real interactions — gravity, contact, tension, friction — appear on the list.

Draw the diagram. The mistake cannot survive it.

A second example: the Atwood machine

Two blocks, masses m₁ and m₂, connected by a string over a frictionless pulley. m₁ > m₂. The system is released. Find the acceleration.

The wrong approach: treat the whole system as one object, write (m₁ - m₂)g = (m₁ + m₂)a, get the answer. This works for finding acceleration — but it hides the tension, which is what exams ask for.

The right approach: one free body diagram for each block.

Block 1 (heavier, going down): Weight m₁g downward, tension T upward. Taking downward as positive:

$m_1g - T = m_1a$

Block 2 (lighter, going up): Weight m₂g downward, tension T upward. Taking upward as positive:

$T - m_2g = m_2a$

Two equations, two unknowns (T and a). Add them:

$m_1g - m_2g = m_1a + m_2a$

$a = \frac{(m_1 - m_2)g}{m_1 + m_2}$

Substitute back to find T:

$T = \frac{2m_1m_2g}{m_1 + m_2}$

The tension is always less than m₁g and always greater than m₂g. This makes physical sense — the string must be pulling block 1 back (less than its weight) while pulling block 2 up (more than its weight).

Notice what made this possible: drawing two separate diagrams forced us to write two separate equations. The tension appeared in both — and that is exactly what allowed us to solve for it.

The twist variant

Same Atwood machine. Now the pulley has friction, and the string slips slightly. The tension on the two sides of the string is no longer equal — call them T₁ and T₂.

The free body diagram method still works, unchanged. Block 1 has T₁ upward. Block 2 has T₂ upward. You now have two equations and three unknowns (a, T₁, T₂) — and you need one more equation from the pulley's rotational dynamics to solve the system.

The method didn't break. It told you exactly how many equations you need. That is what a good method does.

What to remember

Newton's second law applies to one object at a time. Draw that object alone. List only the forces acting directly on it. Write F = ma.

The normal force and weight are different quantities. Weight is mg, fixed by gravity. Normal force is whatever the surface must provide to produce the observed acceleration. In a lift, they differ whenever the lift accelerates.

Motion is not a force. Never draw an arrow for "the direction of motion" on a free body diagram. Only real interactions appear.

When a problem has multiple objects, draw multiple diagrams. The connecting forces — tension, normal force between surfaces — appear in both diagrams, and that is what links the equations.

The next article applies this method to friction — where a hidden force appears that most students either forget entirely or apply incorrectly. The free body diagram makes friction visible.