Lesson 3 of 5Forces

Unit 2 · Lesson 3 of 5

Net Force in One Dimension

Adding forces with signs to get the net force, then feeding it into Newton's second law — including lifts, ropes and connected objects.

3 min read10 practice questionsAsk AI about this →

Key things to remember

1 $Σ F = F_{n e t} = F_{+} + F_{-}$ — add the forces with their signs, after declaring a positive direction.
2Balanced forces (net zero) mean equilibrium: at rest or moving at constant velocity.
3Unbalanced forces produce acceleration in the direction of the net force, with $F_{n e t} = m a$ .
4A scale reads the normal force, not the weight. They differ whenever there is vertical acceleration.
5Accelerating upward makes you feel heavier; accelerating downward makes you feel lighter; free fall gives $F_{N} = 0$ .
6Connected objects share one acceleration. Find it with the system approach, then get tension from one object alone.
7In an ideal rope the tension is the same throughout, and an ideal pulley only redirects it.

Almost every dynamics question reduces to the same two moves: add the forces to get $F_{n e t}$ , then apply $F_{n e t} = m a$ . In one dimension, adding forces means adding signed numbers, which is why choosing and stating a positive direction is the step that everything else depends on.

Σ F = F_{n e t} = F_{+} + F_{-}

The Greek sigma just means 'add them all up'. Forces in the positive direction count as positive; the rest count as negative.

Fig 1Forces laid out on a number line. The net force is where you land.

Draw the free-body diagram. Every force, labelled, from the object.
Choose a positive direction and write it down. Direction of motion is a good default.
Handle each dimension separately. Vertical forces balance against vertical; horizontal against horizontal.
Add the signed forces to get $F_{n e t}$ in that direction.
Apply Newton's second law, $F_{n e t} = m a$ , and solve for whatever is missing.
Interpret the sign of the answer using the convention you declared in step 2.

Balanced and unbalanced

Balanced forces sum to zero, so the object is either at rest or moving at constant velocity — this is called equilibrium. Unbalanced forces sum to something non-zero, so the object accelerates in the direction of that net force. 'Not moving' and 'moving steadily' are the same case as far as the forces are concerned.

Vertical problems: lifts and apparent weight

Fig 2The scale reads the normal force, not the weight. They differ whenever the lift accelerates.

A bathroom scale does not measure weight. It measures the normal force it has to push up with, and calls that your weight. Standing still, those are the same number. In an accelerating lift they are not — and the difference is what you actually feel.

Lift is…	Net force	Scale reads
At rest or at constant speed	Zero	$F_{N} = F_{g}$ — your true weight
Accelerating upward	Upward	$F_{N} > F_{g}$ — heavier, pressed into the floor
Accelerating downward	Downward	$F_{N} < F_{g}$ — lighter, that dropping sensation
In free fall (cable cut)	$F_{g}$ only	$F_{N} = 0$ — apparent weightlessness

Moving up is not the same as accelerating up

A lift moving upward at a constant

3 m/s

has zero net force and the scale reads your normal weight. The feeling of being heavier happens only while the speed is changing. That is why you feel it at the start and end of a lift ride, not in the middle.

Connected objects and tension

When two objects are joined by a rope, they must have the same acceleration — the rope guarantees it. That gives two ways to attack the problem, and using both is the best available check.

ASystem approach

Treat both objects as one combined mass
Internal tension cancels and disappears
Fastest route to the acceleration
Cannot give you the tension

BIndividual approach

Draw a separate free-body diagram for each object
Tension appears explicitly on both
Needed whenever the tension is asked for
Use the acceleration from the system approach

The standard two-step

Find the acceleration by treating everything as one system, then find the tension by applying

F_{n e t} = m a

to just one of the objects. Then check by applying it to the other object — if both give the same tension, the answer is right.

An ideal rope in this course is massless and does not stretch, and an ideal pulley is frictionless. That means the tension is the same everywhere along the rope, and the pulley only changes the direction of the pull without changing its size.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

20 kg

crate is pushed with

150 N

forward while friction of

50 N

opposes it. Find the net force and the acceleration.

Example 2

70 kg

person stands on a scale in a lift that accelerates upward at

2.0 m/s^{2}

. What does the scale read, in newtons?

Example 3

Two crates,

8.0 kg

and

12.0 kg

, are joined by a light rope on a frictionless surface. A

60 N

force pulls the

12.0 kg

crate. Find the acceleration and the tension.

Where marks get lost

Adding force magnitudes without signs, so a $150 N$ push against $50 N$ of friction comes out as $200 N$ .
Including the weight and normal force in a horizontal net-force calculation. Handle each direction separately.
Treating 'moving upward' as 'accelerating upward'. A lift at constant speed has zero net force and reads your true weight.
Including the rope's tension as an external force when using the system approach. Internal forces cancel.
Using only one object's mass to find the acceleration of a connected pair. The whole system accelerates together.
Forgetting that the scale reads the normal force, and so quoting the weight whatever the lift is doing.
Declaring up as positive and then writing the weight as a positive number.

Flashcards 9

Net force

Equilibrium

Unbalanced forces

What does a bathroom scale measure?

+ 5 more

Practice

Forces of

80 N

right and

30 N

left act on an object. What is the magnitude of the net force, in newtons?

15 kg

object is pushed with

90 N

against

30 N

of friction. What is its acceleration in

m/s^{2}

m/s^2

A lift moves upward at a constant

3.0 m/s

. What does a scale inside read compared with the passenger's true weight?

60 kg

person stands on a scale in a lift accelerating downward at

1.5 m/s^{2}

. What does the scale read, in newtons?

An object moves at constant velocity in a straight line. What can you conclude?

Two crates of

5.0 kg

and

10.0 kg

are joined by a rope on a frictionless floor and pulled with

45 N

. What is their shared acceleration, in

m/s^{2}

m/s^2

What is the term for the state in which all the forces on an object sum to zero?

When using the system approach on two crates joined by a rope, why is the tension left out?

Order the steps for solving a one-dimensional net-force problem.

1Interpret the sign of the answer using the stated convention
2Apply $F_{n e t} = m a$ and solve
3Add the forces in each direction with their signs
4Draw the free-body diagram of the object
5Choose and write down a positive direction

Match each lift situation to what the scale reads.

At rest

Accelerating upward

Accelerating downward

Cable cut, in free fall

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