What is Newton's second law? (article)

Review your understanding of Newton's second law in this free article aligned to NGSS standards.

What is Newton's second Law?

In the world of introductory physics, Newton's second law is one of the most important laws you'll learn. It's used in almost every chapter of every physics textbook, so it's important to master this law as soon as possible.

We know objects can only accelerate if there are forces on the object. Newton's second law tells us exactly how much an object will accelerate for a given net force.

$a = \frac{Σ F}{m}$ ‍

To be clear, $a$ ‍ is the acceleration of the object, $Σ F$ ‍ is the net force on the object, and $m$ ‍ is the mass of the object.

It is. This is the same formula as $F = m a$ ‍, except we've written the force more precisely as the net force $Σ F$ ‍, and we've divided both sides by the mass $m$ ‍ to get the acceleration $a$ ‍ by itself on one side of the equation.

$a = \frac{Σ F}{m}$ ‍.

One advantage of writing Newton's second law in this form is that it makes people less likely to think that $m a$ ‍—mass times acceleration—is a specific force on an object. The expression $m a$ ‍ is not a force, $m a$ ‍ is what the net force equals.

Looking at the form of Newton's second law shown above, we see that the acceleration is proportional to the net force, $Σ F$ ‍, and is inversely proportional to the mass, $m$ ‍. In other words, if the net force were doubled, the acceleration of the object would be twice as large. Similarly, if the mass of the object were doubled, its acceleration would be half as large.

What does net force mean?

A force is a push or a pull, and the net force $Σ F$ ‍ is the total force—or sum of the forces—exerted on an object. Adding vectors is a little different from adding regular numbers. When adding vectors, we must take their direction into account. The net force is the vector sum of all the forces exerted on an object.

We can find the sum of multiple force vectors graphically by using the tail-to-tip method. This means that we place the tail of each force vector that we want to add to the tip of the previous force vector. Then, after placing the last force vector, the total force vector would point from the first tail to the last tip as seen in the diagram below.

For instance, consider the two forces of magnitude 30 N and 20 N that are exerted to the right and left respectively on the sheep shown above. If we assume rightward is the positive direction, the net force on the sheep can be found by

$Σ F = 30 N - 20 N$ ‍

$Σ F = 10 N to the right$ ‍

If there were more horizontal forces, we could find the net force by adding up all the forces to the right and subtracting all the forces to the left.

Since force is a vector, we can write Newton's second law as $\vec{a} = \frac{Σ \vec{F}}{m}$ ‍. This shows that the direction of the total acceleration vector points in the same direction as the net force vector. In other words, if the net force $Σ F$ ‍ points right, the acceleration $a$ ‍ must point right.

How do we use Newton's second law?

If the problem you're analyzing has many forces in many directions, it's often easier to analyze each direction independently.

In other words, for the horizontal direction we can write

$a_{x} = \frac{Σ F_{x}}{m}$ ‍

This shows that the acceleration $a_{x}$ ‍ in the horizontal direction is equal to the net force in the horizontal direction, $Σ F_{x}$ ‍, divided by the mass.

Similarly, for the vertical direction we can write

$a_{y} = \frac{Σ F_{y}}{m}$ ‍

This shows that the acceleration $a_{y}$ ‍ in the vertical direction is equal to the net force in the vertical direction, $Σ F_{y}$ ‍, divided by the mass.

When using these equations we must be careful to only plug horizontal forces into the horizontal form of Newton's second law and to plug vertical forces into the vertical form of Newton's second law. We do this because horizontal forces only affect the horizontal acceleration and vertical forces only affect the vertical acceleration. For instance, consider a hen of mass $m$ ‍ that has forces of magnitude $F_{1}$ ‍, $F_{2}$ ‍, $F_{3}$ ‍, and $F_{4}$ ‍ exerted on it in the directions shown below.

The forces $F_{1}$ ‍ and $F_{3}$ ‍ affect the horizontal acceleration since they lie along the horizontal direction. Applying Newton's second law to the horizontal direction and assuming rightward is positive, we get

$a_{x} = \frac{Σ F_{x}}{m} = \frac{F_{1} - F_{3}}{m}$ ‍

Similarly, the forces $F_{2}$ ‍ and $F_{4}$ ‍ affect the vertical acceleration since they lie along the vertical direction. Applying Newton's second law to the vertical direction and assuming upward is positive, we get

$a_{y} = \frac{Σ F_{y}}{m} = \frac{F_{2} - F_{4}}{m}$ ‍

Warning: A common mistake people make is to plug a vertical force into a horizontal equation, or vice versa.

What do we do when a force is directed at an angle?

When forces are directed in diagonal directions, we can still analyze the forces in each direction independently. But, diagonal forces will contribute to the acceleration in both the vertical and horizontal directions.

For instance, let's say the force $F_{3}$ ‍ on the hen is now directed at an angle $θ$ ‍ as seen below.

The force $F_{3}$ ‍ will affect both the horizontal and vertical accelerations, but only the horizontal component of $F_{3}$ ‍ will affect horizontal acceleration; only the vertical component of $F_{3}$ ‍ will affect the vertical acceleration. So we'll break the force $F_{3}$ ‍ into horizontal and vertical components as seen below.

Now we see that the force $F_{3}$ ‍ can be viewed as consisting of a horizontal force $F_{3 x}$ ‍ and a vertical force $F_{3 y}$ ‍.

Using trigonometry, we can find the magnitude of the horizontal component with $F_{3 x} = F_{3} cos θ$ ‍. Similarly, we can find the magnitude of the vertical component with $F_{3 y} = F_{3} sin θ$ ‍.

Since $sin θ = \frac{opposite}{hypotenuse}$ ‍, we can relate the magnitude of the vertical component $F_{3 y}$ ‍ to the magnitude of the force vector $F_{3}$ ‍ with

$sin θ = \frac{F_{3 y}}{F_{3}}$ ‍

And solving for $F_{3 y}$ ‍ we get

$F_{3 y} = F_{3} sin θ$ ‍

Similarly, since $cos θ = \frac{adjacent}{hypotenuse}$ ‍, we can relate the magnitude of the horizontal component $F_{3 x}$ ‍ to the magnitude of the force vector $F_{3}$ ‍ with

$cos θ = \frac{F_{3 x}}{F_{3}}$ ‍

And solving for $F_{3 x}$ ‍ we get

$F_{3 x} = F_{3} cos θ$ ‍

Now we can proceed as usual by plugging all horizontally directed forces into the horizontal form of Newton's second law.

$a_{x} = \frac{Σ F_{x}}{m} = \frac{F_{1} - F_{3 x}}{m} = \frac{F_{1} - F_{3} cos θ}{m}$ ‍

Similarly, we can plug all vertically directed forces into the vertical form of Newton's second law.

$a_{y} = \frac{Σ F_{y}}{m} = \frac{F_{2} - F_{4} + F_{3 y}}{m} = \frac{F_{2} - F_{4} + F_{3} sin θ}{m}$ ‍

We're assuming rightward and upward are the positive directions. And since the variables in this equation—e.g., $F_{2}$ ‍, $F_{3} cos θ$ ‍—represent the magnitudes of the forces exerted on the hen, we need to supply a negative sign for forces that point left or down.

What do solved examples involving Newton's second law look like?

Example 1: Newton the turtle

A 1.2 kg turtle named Newton has four forces exerted on it as shown in the diagram below.

What is the horizontal acceleration of Newton the turtle?
What is the vertical acceleration of Newton the turtle?

Example 2: String cheese

A wedge of cheese is suspended at rest by two strings which exert forces of magnitude $F_{1}$ ‍ and $F_{2}$ ‍, as seen below. There is also a downward force of gravity on the cheese of magnitude $20 N$ ‍.

What is the magnitude of the force $F_{1}$ ‍?
What is the magnitude of the force $F_{2}$ ‍?

We'll start by either using the horizontal or vertical version of Newton's second law. We don't know the value of any of the horizontal forces, but we do know the magnitude of one of the vertical forces— $20 N$ ‍. Since we know more information about the vertical direction, we'll analyze that direction first.

$a_{y} = \frac{Σ F_{y}}{m} (Start with Newton’s 2nd law for the vertical direction.)$ ‍

$a_{y} = \frac{F_{1} sin 60^{\circ} - 20 N}{m} (Plug in vertical forces with correct negative signs.)$ ‍

$0 = \frac{F_{1} sin 60^{\circ} - 20 N}{m} (Vertical acceleration is zero since the cheese is at rest.)$ ‍

$0 = F_{1} sin 60^{\circ} - 20 N (Multiply both sides by mass m .)$ ‍

$F_{1} = \frac{20 N}{sin 60^{\circ}} (Solve for F_{1} .)$ ‍

$F_{1} = 23 N (Calculate and celebrate!)$ ‍

Now to find the force $F_{2}$ ‍, we'll use Newton's second law for the horizontal direction.

$a_{x} = \frac{Σ F_{x}}{m} (Use Newton’s 2nd law for the horizontal direction.)$ ‍

$a_{x} = \frac{F_{1} cos 60^{\circ} - F_{2}}{m} (Plug in horizontal forces with correct negative signs.)$ ‍

$a_{x} = \frac{(23 N) cos 60^{\circ} - F_{2}}{m} (Plug in value of F_{1} = 23 N obtained in the vertical calculation.)$ ‍

$0 = \frac{(23 N) cos 60^{\circ} - F_{2}}{m} (Horizontal acceleration is zero since the cheese is at rest.)$ ‍

$0 = (23 N) cos 60^{\circ} - F_{2} (Multiply both sides by mass m .)$ ‍

$F_{2} = (23 N) cos 60^{\circ} (Solve for F_{2} .)$ ‍

$F_{2} = 11.5 N (Calculate and celebrate!)$ ‍

Want to join the conversation?

Sort by:

thawryluk11
6 years agoPosted 6 years ago. Direct link to thawryluk11's post “I don't understand how an...”
I don't understand how an object with an acceleration of 0 could have a Force. If F=ma and a=0 [so F=m(0)] then why doesn't the Force end up as 0?
Button navigates to signup page•Comment on thawryluk11's post “I don't understand how an...”
(11 votes)
- Davin Peng
  2 years agoPosted 2 years ago. Direct link to Davin Peng's post “Yes, the force would be z...”
  Yes, the force would be zero, but that is the Net Force. So the forces acting on the object can cancel each other out and the object would have 0 acceleration. Using the example of hanging cheese, the vertical forces cancel each other out, as sin60 times 23 is approximately equal to 20, so the net force would end up zero, but there are still these forces acting on it.
  Comment on Davin Peng's post “Yes, the force would be z...”
  (16 votes)
Suraj Gangaram
8 years agoPosted 8 years ago. Direct link to Suraj Gangaram's post “For Newton's second law a...”
For Newton's second law about acceleration, isn't their another way to calculate it by dividing the change in velocity by time?
Button navigates to signup page•Button navigates to signup page
(9 votes)
- 🍕⚡ ViςhαL Πaudel⚡🍕
  6 years agoPosted 6 years ago. Direct link to 🍕⚡ ViςhαL Πaudel⚡🍕's post “This means that p(or the ...”
  This means that p(or the Momentum) = F⋅∆t
  Button navigates to signup page
  (0 votes)
Suany Chavez
3 years agoPosted 3 years ago. Direct link to Suany Chavez's post “When do i know when to us...”
When do i know when to use cos and sin?
Button navigates to signup page•Button navigates to signup page
(6 votes)
- John Calvin
  3 years agoPosted 3 years ago. Direct link to John Calvin's post “There is a common saying ...”
  There is a common saying in math called Soh Cah Toa. For Sine, Cosine, and Tangent. If the problem gives you the opposite and hypotenuse sides then you will you Sine, because sine is Soh which contains o and h for opposite and hypotenuse. You will use cosine is you are given adjacent and hypotenuse.
  Button navigates to signup page
  (6 votes)
Kesia Otieno
7 years agoPosted 7 years ago. Direct link to Kesia Otieno's post “In Example 2, how does 20...”
In Example 2, how does 20 get in the numerator and how did you get it to be divided by sin60?
Button navigates to signup page•Comment on Kesia Otieno's post “In Example 2, how does 20...”
(7 votes)
- deka
  2 years agoPosted 2 years ago. Direct link to deka's post “1. "cheese is suspended a...”
  1. "cheese is suspended at rest" means a_total=0 and F_total=0, which means all components of F_y as well as F_x must cancel each other, respectively.
  F_y_total = F_y_down + F_y_up = 0N
  2. then 20N downward must be offset by 20N upward
  F_y_down + F_y_up = -20N + 20N = 0N
  F_y_up = 20N
  3. and 20N upward must be applied by y component of the diagonal Force as it is the only to offset the downward Force.
  F_y_up = 20N = F_y_dia*sin60
  F_y_dia = 20N/sin60 = 20N/~0.87 = ~23
  (~ means around, you can use a calculator if you want)
  in fact, i prefer this path to that starting from Newton's second law like above cause it's faster and irrelevant to mass (in fact, they are same but i simply don't follow the strict steps from a=net_F/m to the starting point of mine)
  Button navigates to signup page
  (1 vote)
Kayla Sims
7 years agoPosted 7 years ago. Direct link to Kayla Sims's post “Hey guys! My question is ...”
Hey guys! My question is in reference to the second example and has to do with direction. We typically assign the left/down as negative and up/right as positive. Is it because the question is asking for the magnitude that the direction of the force not important? Clearly, F1 is pointing up and to the right, so I can see why that vector is positive, but F2 is pointing left, yet the magnitude was still positive. Why?
Button navigates to signup page•Button navigates to signup page
(4 votes)
- valeria
  7 years agoPosted 7 years ago. Direct link to valeria's post “The problem only asks for...”
  The problem only asks for magnitude. Magnitude refers to a size or quantity ( disregards direction), it is always positive.
  Comment on valeria's post “The problem only asks for...”
  (5 votes)
FoFo S
6 years agoPosted 6 years ago. Direct link to FoFo S's post “In example 1, the tan = ...”
In example 1, the tan = -9/3.3=2.7 why is positive not negative . is tan always(+), also the theta was positive it should be negative ! please explain this ?
Button navigates to signup page•Button navigates to signup page
(2 votes)
- Sajjad Bin Samad
  6 years agoPosted 6 years ago. Direct link to Sajjad Bin Samad's post “Well, you missed somethin...”
  Well, you missed something. The numbers were both between modulus sign, which means that we are only going to work with the positive value.
  Comment on Sajjad Bin Samad's post “Well, you missed somethin...”
  (5 votes)
wee
9 months agoPosted 9 months ago. Direct link to wee's post “In net force, it is also ...”
In net force, it is also a vector right. I am confused because in subtracting vectors, dont we invert the direction of the second vector. Hence in rightward 30N and leftward 20N, wouldnt the equation become 30N+20N? or is the sign inverted already? I am sorry if my question seems dumb. I am confused because I can not understand how subtracting horizontal vectors geometrically and not just in formulas.
Button navigates to signup page•Button navigates to signup page
(3 votes)
- ajay.madappat.93
  6 months agoPosted 6 months ago. Direct link to ajay.madappat.93's post “As the second one is left...”
  As the second one is leftward it is -20N to the right. It is +20N to the left.
  Button navigates to signup page
  (1 vote)
Abdelrahman Elaraby
7 years agoPosted 7 years ago. Direct link to Abdelrahman Elaraby's post “i need to know about newt...”
i need to know about newton's second law on variable mass systems.. in which playlist on khan academy i can find this?
Button navigates to signup page•Button navigates to signup page
(1 vote)
- Leafarian
  7 years agoPosted 7 years ago. Direct link to Leafarian's post “All of the videos on Newt...”
  All of the videos on Newton's second law is in Forces and Newton's laws of motion
  Button navigates to signup page
  (7 votes)
soap
6 years agoPosted 6 years ago. Direct link to soap's post “In example 1 (newton the ...”
In example 1 (newton the turtle) why is there tanθ = absolute values of acceleration vector a_y / a_x. In physics can't you have negative angles?
Button navigates to signup page•Button navigates to signup page
(2 votes)
- Sajjad Bin Samad
  6 years agoPosted 6 years ago. Direct link to Sajjad Bin Samad's post “I'm not sure why they did...”
  I'm not sure why they did that but maybe for simplicity, although they added a line The total acceleration vector points right and down it's says everything about the direction, there is no need to show that by using a negative sign too. This would've been counted as a repetitive error.
  You can have negative angles in physics though saying the direction of something is downward is the same as putting a negative sign followed by the magnitude or absolute value .
  Comment on Sajjad Bin Samad's post “I'm not sure why they did...”
  (3 votes)
Mohamed Nasr
6 years agoPosted 6 years ago. Direct link to Mohamed Nasr's post “I need to know how to cal...”
I need to know how to calculate acceleration with cooficient and friction I need to know the laws of cooficient
Button navigates to signup page•Button navigates to signup page
(3 votes)
- Sajjad Bin Samad
  6 years agoPosted 6 years ago. Direct link to Sajjad Bin Samad's post “Head to this playlist htt...”
  Head to this playlist https://www.khanacademy.org/science/physics/forces-newtons-laws/inclined-planes-friction/v/inclined-plane-force-components
  Comment on Sajjad Bin Samad's post “Head to this playlist htt...”
  (1 vote)

What is Newton's second law? (article) | Khan Academy (2024)