# Mechanics 4: Newton’s Laws

## Newton’s Laws of Motion

We’ve been going through the part of Physics called

*mechanics.*It is based on three laws, formalized by Isaac Newton. Let’s work through them.

The first law is: *A body in motion tends to stay in motion; a body at rest tends to stay at rest.* This is also referred to as the *Law of Inertia.* Inertia is defined as, “a tendency to do nothing or to remain unchanged.” This doesn’t mean that you *can’t* change what the object is doing — you can alter its movement (direction or speed) by applying a force.

This leads us the Newton’s second law: *The sum of all forces applied to an object is equal to its mass time its acceleration.* Written as a formula, the law looks like this

$$ \sum F_{app} = ma$$

Lastly, Newton’s third law is: *Any action produces an equal and opposite reaction.* For example, if a box is sitting on a table, then the box exerts a downward force on the table equal to its weight, and the table exerts an upward force on the box, also equal to its weight. This upward force (from a table, or platform, or the ground) is also called the *normal* force. Normal is a word that means “at right angles to a surface.” This is because the force is upwards, which is at right angles to the surface itself (table, ground, etc.).

The unit of force is, appropriately, called the newton. It is abbreviated “N” (upper-case letter, since it is named after a person).

### Summing Forces

To apply Newton’s Second Law, we must be able to calculate the force applied (\(F_{app}\)). Since there are typically 2 or more forces on an object (gravity and something else), we need to sum all the forces. Here’s where a bit more math comes in. Force is a vector quantity. A vector is something that has a magnitude and a direction. So we need to be able to sum up the force vectors and get what’s called the

*resultant*vector. To help with learning vectors, please see my

*Math Refresher*on vectors.

Let’s start with an example. Say that we have a block that is sitting on the ground. We are trying to push it horizontally with a 12 N force, and gravity is pulling down on it with a 10 N force. Here is how to write these vectors:

$$ \vec F_{push} = 12\,\hat\imath\ {\rm N} \qquad \qquad \vec F_g = 10\,\hat\jmath\ {\rm N}$$

Here we have used standard vector notation: An arrow above the variable letter indicates a vector quantity; the unit vector in the \(x\) direction is written \(\hat\imath,\) and in the \(y\) direction as \(\hat\jmath.\) Now, the resultant (applied) force is written as:

$$ \vec F_{app} = 12\,\hat\imath + 10\,\hat\jmath\ {\rm N}$$

The other standard way to indicate a vector is with a bold letter, like so:

$$ {\bf F}_{app} = 12\,\hat\imath + 10\,\hat\jmath\ {\rm N}$$

In order to sum the applied forces, we first must make sure that we have all of them listed. The standard way to do this is to use a “free-body” diagram. Such a diagram helps to analyze all the forces, and to not miss any. Let’s work through several examples to see how this works.

### Examples

1. Say we have a block of mass \(m,\) sitting on the ground, being pushed horizontally. This is the free-body diagram

Here, we’re calling the pushing force \(F_p,\) the force of gravity, \(F_g,\) and the normal force (from the ground) is \(F_n.\) So let’s sum these forces by direction

\(x\) direction | \(y\) direction |

\(\sum F_x = F_p\) | \( \sum F_y = F_n – F_g = 0 \) |

First, let’s apply Newton’s second law in the \(x\) direction and we get

$$\sum F_x = F_p = ma_x$$

This tells us that the acceleration in the \(x\) direction is \(a_x = F_p/m.\)

For the \(y\) direction, the forces balance (i.e., sum to 0), so there is no motion in that direction. We write it like this:

$$ \sum F_y = 0 = ma_y$$

which tells us that \(a_y = 0.\)

**(With Numbers)**

So, let mass, \(m=6\) kg and the pushing force, \(F_p = 12\)N. Calculate the normal force (\(F_n\)) and the acceleration in the \(x\) direction, \(a_x.\)

From the free-body diagram, \(F_n = F_g.\) The force of gravity equals mass \(\times\) the acceleration of gravity, or \(F_g = mg,\) so combining these we get \(F_n = mg.\) Recall that the acceleration of gravity is \(g = 9.81\,{\rm m/s}^2,\) so we have

$$F_n = mg = (6\,{\rm kg}) (9.81\, {\rm m/s^2}) = 68.9\,{\rm N}$$

Also from the free-body diagram, \(F_x = ma_x\) or (rearranging), \(a_x = F_x / m,\) giving

$$a_x = F_x / m = (12\,{\rm N}) / (6\,{\rm kg}) = 2\,{\rm m/s^2}$$

2. Say we have a block of mass \(m,\) sitting on the ground, being pushed horizontally by two opposing forces, \(F_1\) and \(F_2.\) We are told \(F_1 > F_2.\) This is the free-body diagram

Summing the forces by directions gives

\(x\) direction | \(y\) direction |

\(\sum F_x = F_1 – F_2 \) | \(\sum F_y = F_n – F_g = 0 \) |

Again here, the forces in the \(y\) direction sum to 0 (no movement in the \(y\) direction, so \(a_y = 0\). Then in the \(x\) direction, Newton’s second law gives us

$$\sum F_x = F_1 – F_2 = ma_x$$

And we can write

$$ ma_x = F_1 – F_2$$

or

$$ a_x = \frac{F_1 – F_2}{m} $$

**(With Numbers)**

Here, say that \(F_1 = 20\,\)N, \(F_2 = 15\,\)N, and mass \(m = 5\,\)kg. Again, calculate the normal force (\(F_n\)) and the acceleration in the \(x\) direction, \(a_x.\)

As before, \(F_n = mg,\) giving

$$F_n = mg = (5\,{\rm kg}) (9.81\, {\rm m/s^2}) = 49\,{\rm N}$$

From the free-body diagram, \(F_x = F_1 – F_2,\) which equals \(F_x = (20 – 15) = 5\,\)N, giving

$$a_x = F_x / m = 5/5 = 1\,{\rm m/s^2}$$

3. We have a block sitting on an inclined plane. We’ve shown the same 4 vectors as before, but now they are rotated by an angle theta (\(\theta\)) to be along the plane or perpendicular to the plane. Since it’s all rotated, the easiest way to do the math is to rotate the coordinate system — define \(x\) as along the plane and \(y\) as into/out of the plane. Also, some of the forces have new names. As in example 1, \(F_p\) is the pushing force and \(F_n\) is the normal force. But now, the force of gravity is split into \(F_{g_1}\) and \(F_{g_2}.\)

Intuitively, you know that on an incline, gravity will pull the block into the plane and down the plane. So we need to calculate how gravity splits into these two parts. How do we do this? You can probably guess this — we’ll use trig. In a right triangle, the two angles other than the right angle have to add up to \(90^\circ\) (they’re *complementary*). So let’s call the upper angle alpha (\(\alpha\)); its formula is \(\alpha = 90 – \theta.\) Next, we’ll draw in the force due to gravity (called \(F_g\)). It acts straight down, so it’s parallel to the right-hand-side of the big triangle (labelled side \(c\)). This means that it makes an angle \(\alpha\) with the inclined plane (thick line, side \(a\)). We also re-draw the vectors \(F_{g_1}\) and \(F_{g_2}\) to show how they combine to get \(F_g.\) Since \(F_{g_1}\) is perpendicular to the plane, that means that the angle between it and \(F_g\) must be \(\theta,\) the complement of \(\alpha.\)

Now, flip and rotate the small triangle of vectors to draw it in the “standard” trig way, and we have

And we can see that

$$ F_{g_1} = F_g \cos\theta \qquad {\rm and} \qquad F_{g_2} = F_g \sin\theta$$

**(With Numbers)**

Say we have a 15 kg block sitting on an incline of \(\theta = 30^\circ.\) (a) How much pushing force does it take to keep it from sliding down the plane, and (b) what is the normal force from the plane?

Start by looking at the diagram at the beginning of example 3. For part (a), we want to just hold the block in place. In terms of an equation, this is

$$\sum F_x = F_p – F_{g_2} = 0$$

for no acceleration. This tells us that \(F_p = F_{g_2}.\) We can solve for this using the above equation,

$$ F_{g_2} = F_g \sin\theta$$

where the force of gravity is \(F_g = mg.\) We are given that \(m = 15\,\)kg, and we know that the acceleration of gravity is \(g = 9.81\,{\rm m/s}^2.\) Putting it all together,

$$ F_g \sin\theta = (15\,{\rm kg})(9.81\,{\rm m/s^2}) \sin (30^\circ) = 73.6\,{\rm N}$$

For part (b), we use the forces in the \(y\) direction, which give us

$$\sum F_y = F_n – F_{g_1} = 0$$

again, for no acceleration. This tells us that \(F_n = F_{g_1}.\) As before, this equals

$$ F_{g_1} = F_g \cos\theta = (15\,{\rm kg}) (9.81\,{\rm m/s^2}) \cos(30^\circ)$$

which equals about 127 N of force.

4. Our final example is with two blocks, connected by a rope. We define a new force, \(T,\) the tension in the rope.

As before, we draw a free-body diagram, but now we need two diagrams — one for each block. These two diagrams are connected by the tension in the rope.

Notice that for \(m_a,\) \(T\) is horizontal; for \(m_b,\) \(T\) is vertical. Again, we can link the two free-body diagrams with the tension in the rope. Also, in the first diagram, the \(T\)s point at each other. This is because the rope is pulling \(m_a\) up the slope, and is preventing \(m_b\) from falling.

As long as block \(m_a\) stays in contact with the plane, there is no motion in the \(y\) direction, and, as before, \(F_{g_1} = F_n.\) Next, also assume that there’s no motion in the \(x\) direction, so \(F_{g_2} = T.\) No motion means that \(m_b\) does not move either, and \(F_g = T\) for that block. Let’s use what we derived in Example 3, and recall that \(F_g = mg.\) So for mass \(m_a\)

$$F_{g_1} = F_n \qquad \qquad F_{g_2} = T $$

$$m_ag \cos\theta = F_n \qquad m_ag \sin\theta = T $$

and for mass \(m_b\)

$$ F_g = T$$

$$ m_bg = T$$

then, setting the two \(T\) equations equal, we get:

$$ m_ag \sin\theta = m_bg$$

Dividing through by \(g\) gives us

$$ m_a \sin\theta = m_b$$

Remember that this is for no motion. So for a given mass, \(m_a,\) and incline angle, \(\theta,\) we can calculate mass \(m_b\) needed for no motion (\(m_b\) just holds \(m_a\) in place).

**(With Numbers)**

In this example, let the large mass be \(m_a = 8\,\)kg and the incline is at \(\theta = 20^\circ.\) Calculate the normal force, \(F_n\) and what must be the mass of \(m_b\) for no motion?

From above, the normal force \(F_n = m_ag \cos\theta,\) which gives us

$$F_n = (8\,{\rm kg}) (9.81\, {\rm m/s^2}) \cos 20^\circ = 73.7\,{\rm N}$$

And also from above, for no motion, \(m_b = m_a \sin\theta,\) giving

$$m_b = (8\,{\rm kg}) \sin 20^\circ = 2.74\,{\rm kg}$$

So, if the second mass (\(m_b\)) is this much, there is no motion. If it’s more massive, the first mass (\(m_a\)) will be pulled up the plane; if \(m_b\) is less, then \(m_a\) will slide down the plane.

### Mass & Weight

What’s the difference between mass and weight? Mass is an intrinsic property of material; weight is the force due to gravity on a mass, and equals \(mg.\) In the metric (MKS) system, mass is measured in kilograms and weight (force) is measured in newtons. In the Imperial system — here’s the problem — weight and mass are used somewhat interchangably. The Imperial unit of weight is the pound (lb) is also thought of as mass. Some engineers talk about “lbf” — pound of force and “lbm” — pound of mass. I’ve also heard of a unit for mass called the

*slug,*which equals \(m = F_g/g\) or pounds divided by gravity. Just remember that pounds is a force, and is weight (\(mg\)) not mass. Confusing enough? This is one place where the metric (MKS) system really wins!

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