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# What is Newton's second law?

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.
$a=\frac{\mathrm{\Sigma }F}{m}$
To be clear, $a$ is the acceleration of the object, $\mathrm{\Sigma }F$ is the net force on the object, and $m$ is the mass of the object.
Looking at the form of Newton's second law shown above, we see that the acceleration is proportional to the net force, $\mathrm{\Sigma }F$, 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 $\mathrm{\Sigma }F$ 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.
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
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 $\stackrel{\to }{a}=\frac{\mathrm{\Sigma }\stackrel{\to }{F}}{m}$. 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 $\mathrm{\Sigma }F$ 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
${a}_{x}=\frac{\mathrm{\Sigma }{F}_{x}}{m}$
This shows that the acceleration ${a}_{x}$ in the horizontal direction is equal to the net force in the horizontal direction, $\mathrm{\Sigma }{F}_{x}$, divided by the mass.
Similarly, for the vertical direction we can write
${a}_{y}=\frac{\mathrm{\Sigma }{F}_{y}}{m}$
This shows that the acceleration ${a}_{y}$ in the vertical direction is equal to the net force in the vertical direction, $\mathrm{\Sigma }{F}_{y}$, 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
${a}_{x}=\frac{\mathrm{\Sigma }{F}_{x}}{m}=\frac{{F}_{1}-{F}_{3}}{m}$
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
${a}_{y}=\frac{\mathrm{\Sigma }{F}_{y}}{m}=\frac{{F}_{2}-{F}_{4}}{m}$
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 $\theta$ 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 ${F}_{3x}$ and a vertical force ${F}_{3y}$.
Using trigonometry, we can find the magnitude of the horizontal component with ${F}_{3x}={F}_{3}\text{cos}\theta$. Similarly, we can find the magnitude of the vertical component with ${F}_{3y}={F}_{3}\text{sin}\theta$.
Now we can proceed as usual by plugging all horizontally directed forces into the horizontal form of Newton's second law.
${a}_{x}=\frac{\mathrm{\Sigma }{F}_{x}}{m}=\frac{{F}_{1}-{F}_{3x}}{m}=\frac{{F}_{1}-{F}_{3}\text{cos}\theta }{m}$
Similarly, we can plug all vertically directed forces into the vertical form of Newton's second law.
${a}_{y}=\frac{\mathrm{\Sigma }{F}_{y}}{m}=\frac{{F}_{2}-{F}_{4}+{F}_{3y}}{m}=\frac{{F}_{2}-{F}_{4}+{F}_{3}\text{sin}\theta }{m}$

## 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?
To find the horizontal acceleration we'll use Newton's second law for the horizontal direction.
${a}_{x}=\frac{\mathrm{\Sigma }{F}_{x}}{m}\phantom{\rule{1em}{0ex}}\text{(Start with Newton’s 2nd law for the horizontal direction.)}$
${a}_{x}=3.3\frac{\text{m}}{{\text{s}}^{2}}\phantom{\rule{1em}{0ex}}\text{(Calculate and celebrate!)}$
To find the vertical acceleration, we'll use Newton's second law for the vertical direction.
${a}_{y}=\frac{\mathrm{\Sigma }{F}_{y}}{m}\phantom{\rule{1em}{0ex}}\text{(Start with Newton’s 2nd law for the vertical direction.)}$
${a}_{y}=-9.2\frac{\text{m}}{{\text{s}}^{2}}\phantom{\rule{1em}{0ex}}\text{(Calculate and celebrate!)}$

### 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 .
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—. Since we know more information about the vertical direction, we'll analyze that direction first.
${a}_{y}=\frac{\mathrm{\Sigma }{F}_{y}}{m}\phantom{\rule{1em}{0ex}}\text{(Start with Newton’s 2nd law for the vertical direction.)}$
Now to find the force ${F}_{2}$, we'll use Newton's second law for the horizontal direction.
${a}_{x}=\frac{\mathrm{\Sigma }{F}_{x}}{m}\phantom{\rule{1em}{0ex}}\text{(Use Newton’s 2nd law for the horizontal direction.)}$
${a}_{x}=\frac{{F}_{1}\text{cos}{60}^{\circ }-{F}_{2}}{m}\phantom{\rule{1em}{0ex}}\text{(Plug in horizontal forces with correct negative signs.)}$

## Want to join the conversation?

• 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?
• 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.
• For Newton's second law about acceleration, isn't their another way to calculate it by dividing the change in velocity by time?
• This means that p(or the Momentum) = F⋅∆t
• When do i know when to use cos and sin?
• 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.
• In Example 2, how does 20 get in the numerator and how did you get it to be divided by sin60?
• 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)
(1 vote)
• 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?
• The problem only asks for magnitude. Magnitude refers to a size or quantity ( disregards direction), it is always positive.
• 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 ?
• Well, you missed something. The numbers were both between modulus sign, which means that we are only going to work with the positive value.
• 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.
• As the second one is leftward it is -20N to the right. It is +20N to the left.
(1 vote)
• i need to know about newton's second law on variable mass systems.. in which playlist on khan academy i can find this?
(1 vote)
• All of the videos on Newton's second law is in Forces and Newton's laws of motion