Back to the on-screen lesson ·
Forces along a line combine into one net force: same directions add, opposite directions subtract, and zero means balanced.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to combine forces along a line into a net force and predict whether the motion will change.
You can draw a free-body diagram with every force on an object, and find an unknown force by balancing when the object is at rest. This lesson combines forces that do not balance into one net force, which tells you how the motion will change.
| Term | What it means |
|---|---|
| Net force | The single force that has the same effect as all the forces on an object together. |
| Resultant force | Another name for the net force. |
| Balanced forces | Forces whose net force is zero. |
| Unbalanced forces | Forces with a nonzero net force, which change the motion. |
| Newton | The unit of force, N. |
| Meganewton | A million newtons, MN, used for rocket engines. |
Forces along one line combine into a single net force:
$$F_{\text{net}} = F_1 + F_2 + F_3 + \cdots$$
If the net force is zero, the forces are balanced and the motion does not change. If it is not zero, the object speeds up, slows down or changes direction, in the direction of the net force.
Another way: picture
Picture two friends pushing a stalled car. Pushing together from behind, their forces add. If one pushes from the front and one from behind, their forces work against each other and mostly cancel. What matters for the car is the net force, the sum with directions taken into account.
Another way: steps
When two forces point the same way, their effects add. Two students each pushing a car forward with $300$ N give a net forward force of $600$ N. Three horses each pulling a wagon with $1000$ N give $3000$ N.
This is why moving a heavy piano takes several people pushing together. Each person's force adds to the others', and together they overcome friction that no one could beat alone.
When forces point in opposite directions, they work against each other. A push of $80$ N forward against $30$ N of friction leaves a net force of $50$ N forward. The net force points the way of the larger force.
Choosing a positive direction makes this automatic: forces that way are positive, forces the other way are negative, and adding them does the subtracting for you.
When the forces in opposite directions are equal, the net force is zero. The forces are balanced. A book on a table, a car cruising at steady speed, a tug-of-war rope that is not moving: each has balanced forces.
Balanced forces do not stop motion. A car cruising on the highway has the engine's push balancing friction and air drag, and it keeps moving steadily. Balanced forces keep the motion the same, whatever it is.
When the net force is not zero, the motion changes. A net force in the direction of motion speeds the object up; against the motion, it slows it down; sideways, it turns it.
A soccer ball kicked forward speeds up while the foot pushes on it. Rolling afterward, it has a net backward force from friction, so it slows down. The next lesson finds how quickly the motion changes.
Vertical forces combine the same way. For a hot-air balloon, the lift of the hot air points up and the weight points down. If lift is larger, the net force is upward and the balloon rises faster; if weight is larger, it sinks.
Pilots control the net force by heating the air to increase lift or venting it to reduce lift. When lift equals weight, the balloon floats at a steady height.
Checking an answer. The net force can never be larger than all the forces added without signs. Its direction must match the side with the larger total. Equal opposite forces must give zero.
Forces along a line behave like signed numbers: a force of $+50$ N and one of $-50$ N have exactly opposite effects and cancel. So the combined effect of many forces is found by adding them with their signs.
The net force is useful because an object responds only to it. Two sets of forces with the same net force change an object's motion in exactly the same way, however different they look.
In a tug-of-war, both teams pull on the same rope in opposite directions. If the teams pull equally hard, the net force on the rope is zero and it does not move. If one team pulls harder, the rope speeds up toward that team.
Winning is about the net force, not about strength alone. A team that digs its feet in, getting more friction from the ground, can pull harder without sliding forward itself.
A car driving on a level road has four main forces along its motion: the forward push from the road on its tires, and the backward forces of rolling friction and air drag. Vertically, its weight balances the road's upward push.
At steady speed, the forward push balances the backward forces. Pressing the gas increases the push, the net force points forward, and the car speeds up. Lifting off the gas lets the backward forces win, and the car slows.
At liftoff, a rocket's engines push up with enormous force, while its weight pulls down. The net force is the difference. For a SpaceX Falcon 9, the thrust is only about forty percent more than the weight, so the net force is modest and the rocket rises slowly at first.
As the rocket burns fuel, its weight drops while its thrust stays nearly the same, so the net force grows and the rocket speeds up faster and faster.
The most common error is adding all the forces without paying attention to their directions. Two opposite forces of $300$ N and $280$ N do not give $580$ N; they give $20$ N toward the larger.
Another is to forget the weight in vertical problems, or to use mass in kilograms instead of weight in newtons. Weight must be converted with $W = mg$ before it can be added to other forces.
The net force is often much smaller than the individual forces. A rocket may have tens of meganewtons of thrust and weight, yet a net force of just a few meganewtons. An object at rest may have large forces on it and a net force of zero.
When predicting motion, always find the net force first. Looking at one large force alone can be very misleading.
Everyday net forces are often small. A shopping cart rolling at steady speed has a net force of zero. Giving it a gentle shove might add a net force of twenty or thirty newtons for a moment, enough to speed it up noticeably.
Comparing forces to familiar weights helps: one newton is about the weight of a small apple, and a hundred newtons is about the weight of a ten-kilogram bag of dog food.
When forces act both horizontally and vertically, find the net force in each direction separately. A sled pulled forward across snow has its forward pull against friction horizontally, and its weight against the snow's push vertically.
Usually the vertical forces balance, since the sled neither sinks nor rises, and the interesting net force is the horizontal one that speeds it up or slows it down.
The net force does not have to point the way an object is moving. A car coasting to a stop at a red light moves forward while its net force points backward, which is exactly why it slows. A ball thrown upward moves up while its net force, its weight, points down.
The net force tells you which way the motion is changing, not which way the object is going. Keeping these two ideas apart is one of the most useful habits in all of physics, and it prepares you for the next lesson, where the size of the change is worked out.
In a classroom, forces are measured with spring scales, which stretch in proportion to the pull. Two scales pulling a cart in opposite directions let you read both forces and find the net force by subtracting.
If the scales read the same, the cart does not speed up: the forces balance. If one reads more, the cart starts moving toward that scale. Trying this with a partner is a quick way to feel what a net force means.
When a rocket lifts off from Cape Canaveral or Kennedy Space Center, two forces matter most: the engines' thrust pushing up and the rocket's weight pulling down. The rocket can rise only if the thrust is larger. The difference, the net force, decides how quickly it speeds up.
For a SpaceX Falcon 9, the thrust at liftoff is about seven and a half meganewtons and the weight about five and a half, a net force of about two meganewtons. That is why the rocket seems to rise slowly off the pad. NASA's Space Launch System has a much larger net force for its weight and climbs faster. As fuel burns away, every rocket's weight falls and its net force grows.
Every October, hundreds of hot-air balloons rise over New Mexico at the Albuquerque International Balloon Fiesta. Each pilot controls the balloon's height with a single burner. Heating the air inside the envelope increases the lift; letting it cool reduces it.
When the lift is larger than the weight of the balloon, basket and riders, the net force points up and the balloon rises faster. When the pilot lets the air cool, the lift drops below the weight and the balloon descends. To float steadily at one height, the pilot fires the burner in short bursts, keeping the lift just equal to the weight so the net force stays close to zero.
It is tempting to add every force on an object to find the total. But forces in opposite directions work against each other. A push of $80$ N against friction of $30$ N does not give $110$ N; it gives a net force of $50$ N in the direction of the push.
A related error is to think balanced forces mean an object must be at rest. Balanced forces keep the motion the same: an object at rest stays at rest, and a moving object keeps moving at a steady speed.
Two friends push a car forward with $250$ N and $300$ N, against $400$ N of friction. Choose forward as positive.
$+250, \ +300, \ -400$
Signs for direction.
Add the forward forces.
$250 + 300 = 550\ \text{N}$
Same direction.
Subtract the friction.
$550 - 400 = 150\ \text{N}$
Opposite direction.
State the net force.
$150\ \text{N forward}$
Positive.
Predict the motion.
$\text{the car speeds up}$
Unbalanced forward.
A box on a slope has $60$ N pulling it down the slope, $45$ N of friction up the slope and a $20$ N rope pulling up. Choose down the slope as positive.
$+60, \ -45, \ -20$
Signs for direction.
Add the signed forces.
$60 - 45 - 20 = -5\ \text{N}$
Net force.
Read the sign.
$5\ \text{N up the slope}$
Negative means up.
Describe the box sliding down.
$\text{it slows down}$
Net force against its motion.
Find the rope pull that balances.
$60 - 45 = 15\ \text{N}$
Net force zero.
Describe the motion then.
$\text{steady speed}$
Balanced.
A $2$ kg weather balloon has a lift of $30$ N. Find its weight.
$W = 2 \times 9.8 = 19.6\ \text{N}$
Earth's pull.
Choose up as positive and give signs.
$+30, \ -19.6$
Lift up, weight down.
Add the forces.
$30 - 19.6 = 10.4\ \text{N}$
Net force.
Read the direction.
$\text{upward}$
Positive.
Add $8$ N of air drag as it rises.
$30 - 19.6 - 8 = 2.4\ \text{N}$
Drag opposes rising.
Find the drag that balances.
$30 - 19.6 = 10.4\ \text{N}$
Steady rise.
Describe the balanced motion.
$\text{rising at a steady speed}$
Balanced, not stopped.
Give the forces signs.
$+90, \ -65$
Push positive.
Add the signed forces.
$90 - 65$
Opposite directions subtract.
Evaluate the net force.
In a tug-of-war, one team pulls the rope to the right with $260$ N and the other pulls to the left with $175$ N. With right as positive, what is the net force on the rope, in N?
Complete the worked solution: in a school tug-of-war, team A has $3$ students each pulling with $180$ N, and team B has $3$ students each pulling with $170$ N. Find team A's total pull in N, team B's total pull in N, and the net force toward team A in N.
Find team A's total.
$F_A = n_A \times p_A =$ s
Same direction.
Find team B's total.
$F_B = n_B \times p_B =$ t
Same direction.
Find the net force.
$F_{\text{net}} = F_A - F_B =$ n
Opposite teams subtract.
Read the sign.
$\text{positive: team A wins}$
Toward team A.
Match each situation to its result.
| the forces add | the forces subtract | the forces balance: net force zero | the object speeds up, slows down or turns | |
|---|---|---|---|---|
| two people pushing a car the same way | ||||
| a push against friction | ||||
| two equal teams in a tug-of-war | ||||
| a nonzero net force |
A box on a loading dock feels a $25$ N push and a $10$ N gust of wind to the right, and $20$ N of friction and a $15$ N rope pull to the left. With right as positive, fill in the total force to the right in N, the total force to the left in N, and the net force in N.
| force | |
|---|---|
| total to the right (N) | |
| total to the left (N) | |
| net force (N) |
A shopper pushes a cart forward with a force $P$, in newtons, across a parking lot, against $30$ N of friction and $18$ N of wind pushing back. With forward as positive, write the net force on the cart, in N, as a function of $P$.
Answer:
A hot-air balloon with its basket and riders has a mass of $450$ kg. The hot air gives an upward lift of $4600$ N. With $g = 9.8$ N/kg and up as positive, what is the net force on the balloon, in N?
Answer: N
At liftoff from Florida, a SpaceX Falcon 9 has engines pushing up with about $7.6$ MN and a mass of about $549$ metric tons, a weight of $5.38$ MN. With up as positive, what is the net force on the rocket, in MN?
Answer: MN
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A shopper pushes a cart forward with a force $P$, in newtons, across a parking lot, against $20$ N of friction and $12$ N of wind pushing back. With forward as positive, write the net force on the cart, in N, as a function of $P$.
Answer:
You can find a net force. Explain to someone why pushing a box with $80$ N against $30$ N of friction does not give a net force of $110$ N.
27. Your turn: a $90$ N push acts on a crate against $65$ N of friction. What is the net force?, step 3
$25\ \text{N in the push direction}$
Unbalanced.