Section 1-4 - Dynamics

Newton's Laws of Motion

The root of the word dynamics comes from the greek word for force.  For the moment, we shall employ our grade school definition for force, a push or a pull.  We shall study the effects of these forces on the motions of objects.

The history of the study of classical mechanics is interesting and involves a lot of very smart people.  For simplicity, certain views are often ascribed to one of three historical figures, whether those men actually held those beliefs or not:  Aristotle, Galileo, and Newton.  The notions students have when they start a course in classical mechanics are often Aristotelian; examples include that a force is necessary to keep an object in motion, or that a ball dropped from the mast of a moving ship will hit the deck astern of the base of the mast.  Part of the purpose of these classes is to disabuse students of these Aristotelian notions.  Oddly enough, studies show that even students who have performed well on course exams still hold the same Aristotelian notions at the end of the class that they started with.

Here is the simplified history of Mechanics:
Aristotle is often credited with being the first observational scientist.  He saw things in the world around him about which he was curious and attempted to explain them through logical arguments.  Generally, these explanations were wrong, mostly because he didn't or couldn't separate out processes which act concurrently.  For example, an object's natural motion is to fall toward the earth and come to rest.  Even if the object were subjected to a violent motion (such as being thrown), the natural motion will eventually take over.  This seems pretty logical, since one can see this behaviour daily.  Another example might be to slide a book across the table; very soon, the book will come to a stop and remain at rest.  Aristotle might have said something like this:
An object at rest will remain at rest unless acted on by a force.
An object in motion will come to rest unless acted on by a force.

Galileo is often called the first experimental physicist; he actually set up different scenarios in a lab to test his understanding of the laws of nature.  He was the first to try to separate out the different effects that could influence an object's motion so as to study them independently.  For example, we all probably know what it is that makes the sliding book come to rest: friction.  What if the table were made smoother?  How far would the book go before stopping?  And if it were smoother still?  And if it were perfectly smooth?  The book would slide forever.  Galileo's version of these laws might have been:
An object at rest remains at rest unless acted on by a force.
An object in motion will continue that motion unless acted on by a force.
The second part, in other words, means that  an object will maintain a constant velocity (acceleration will be zero) unless a force acts on it.
This last set of rules is now known as Newton's First Law of Motion.

Newton went a bit further.  He investigated how those forces altered the motion of an object and found that the acceleration is proportional to the force applied and inversely proportional to the mass.  We can write this (tentatively at least) as
a a F/m,  or more familiarly as F a ma.
If we choose the correct units, we can make the proportionality an equality.  Let the force necessary to accelerate one kilogram at one meter/second² be called one Newton.  So,
F = ma
Since acceleration is a vector quantity, so must be force; this should be obvious since one can discuss how hard to push and in what direction.  Additionally, we see that the mass of an object is not just a measure of how much material is present, but also of how difficult it is to change the motion of the object.

What if there is more than one force acting on an object?  We can add the forces as vectors to find the net force, F_Net  = S_i F_i.  Think of these situations:
Two twin footballers push equally hard against an opponent in opposite directions.  Will the opponent accelerate?  Are there forces applied to the opponent?  Is there a net force applied?

Our ultimate form of Newton's Second Law of Motion is
S_i F_i  = ma.
The second law can be verified using an apparatus similar to that used in lab: measure acceleration based on kinematic measurements and compare to the force applied.
N.B.: Include in this sum only the forces which act on the body under consideration.

Newton's Third Law of Motion seems to be the one students have the most trouble with, although it really is the easiest to understand:
If object A exerts a force on object B, then B exerts a force on A which is equal in magnitude but opposite in direction.
For the moment, we'll not try to justify this (we will later in an indirect way), but think about this scenario: A speeding car A hits a parked car B; the parked car B is accelerated forward because of the force exerted by A, while A slows down due to the force exerted backwards on it by B.
Two forces which fulfill this description are referred to as a third law pair.  To be a third law pair, the forces must fit the description given above, e.g., A pulls B and B pulls back on A.

Question:  A book sits on a table.  There is a gravitational force exerted on the book by the earth (this is called the weight, see below) and a normal force of contact (see below) acting upward on the book from the table.  If the acceleration of the book is zero (it's not moving), then what can we say about the two forces just mentioned?

Weight

Why go through all this?  Well, suppose that I were to drop a ball and ask you, 'what is the acceleration due to gravity?'  You'd say 9.8 m/s² downward and be right.  Now, let me drop a smaller ball, and ask again, then a yet smaller ball, and so on.  So long as there actually is a ball to drop, the acceleration is 9.8 m/s² downward.  But what if there is no ball to be accelerated? What would be a_g then?  On the other hand, even if there were no ball the strength of the field would still be 9.8 N/kg.
Now instead, consider a ball resting on a table.  What is the acceleration?

Normal Force of Contact

Tension

Strings and ropes are often looped over wheels.  This does nothing more than change the direction of the tensions at the ends, so long as the wheel is frictionless and massless.  Situations where the wheel is not frictionless or massless will be treated later

Applications of Newton's Laws

Consider a man standing on a spring scale.  First of all, what does the scale actually measure?

Now, let's put the man and the scale in an elevator which is accelerating upward.  The diagram is similar to the one above.  Writing the second law results in:
N_M,S - W = ma
or
N_M,S - gm = ma
So,
N_M,S = ma + gm.
So, we see that if the elevator is accelerating upward, the scale reading will be higher than the man's weight, while it will be lower if the elevator's acceleration is downward.
See Problem 4-6.

Here is a problem we shall use as the model for presenting solutions.
Consider a block of mass M on a frictionless plane inclined at an angle q from the horizonal.  What will be the acceleration of the block?
First, a figure may help:

Next, a free body diagram

showing the forces acting on the block: first the weight which is straight downward, and then the normal force of contact perpendicular to the surface.  A common misconception is that the Normal force points upward, and is probably due to the fact that students start by trying problems like the first one addressed here.

Next, we'll pick a co-ordinate system.  There is a need for some experience here, but here is a hint: we certainly expect the block to accelerate along the plane, and not to either jump off the plane or burrow into it.  So, we'll try to choose a system such that the acceleration is along one of the axes.  It will be much easier to solve this problem if we orient the axes parallel and perpendicular to the plane.  It's not impossible to solve the problem otherwise, but it's a lot tougher mathematically.  If the problem is such that the acceleration is zero, then this aspect is not so important and other considerations need to be examined (see below).

Now, we can write the Second Law as
N + gm = ma,
which is not useful, since we then have one equation with two unknowns.  However, we saw in the last section that we can often consider the x and y motions separately:
N_x + (gm)_x = m a_x
N_y - (gm)_y = m a_y
In this case, N_y = N and N_x = 0.  Also, a_y = 0 (see comment above) and we can let a_x = a.
We must also decompose the weight:

So, (gm)_x = gm sinq and (gm)_y = gm cosq.
This all leaves us with:
gm sinq = m a
N - gm cosq = 0.
It turns out that the second equation is not much use, but figuring it out was probably good practice.  Looking at the first equation tells us that
a = g sinq.
Now, we can relate this information to other possible aspects of the problem using the kinematic relationships discussed in earlier sections.

In the previous example, we alluded to 'other considerations' in choosing a co-ordinate system.  As stated, it's generally best to align the axes so that the acceleration is along one of them; if the acceleration is zero, this is not so important, and a judicious choice of axes might reduce the amount of math we need to do.  Consider this problem:

Apply a force horizontally to the block of the previous problem so that the block remains stationary.  Find F and the normal force, N.

Let's start with a free body diagram:

Note that the Normal force is perpendicular to the surface between the block and the incline.  Now, if one were to follow the hint above about choosing a co-ordinate system, one would choose the (tilted) axes in blue in the next figure:

However, in this problem, the acceleration is zero ( the block is to be held stationary).  So, here is an example of those 'other considerations.'
 

If one were to use the blue co-ordinate system, then both the Weight and Fapplied would need to be decomposed, giving a total of five terms to deal with in Newton's Second Law: so that, x:   F cosq - gm sinq = 0 y:   N - F sinq - gm cosq = 0 Re-arrange to get: F = gm sinq /cosq = gm tanq = 2*9.8*1.73 = 33.9 N N = F sinq + gm cosq = [gm tanq] sinq  + gm cosq = gm [tanq sinq  + cosq] =          =  2*9.8 [0.5 + 0.867*1.73 ] = 39.2 N

If one were to use the green co-ordinate system, then only the normal would need to be decomposed, giving a total of four terms to deal with in Newton's Second Law: so that,  x:   F - Nsinq = 0 y:   Ncosq  - gm = 0 Re-arrange to get N =  gm/cosq = 2*9.8/0.5 = 39.2 N F = Nsinq = [gm/cosq]sinq = gm tanq = 2*9.8*1.73 = 33.9 N

The difference in the amount of manipulation may be small, but this serves as an example of how a judicious choice of co-ordinate systems can help to minimize your effort.

Let's do one more to illustrate another point:
Consider two blocks as shown with the inclined surface being frictionless.

Since there are two bodies, we will have to have two free body diagrams, and two sets of Newton's second law equations.  There looks like a complication in choosing a co-ordinate system, though; no matter how the x and y axes are oriented, at least one acceleration will need to have two components, and there will have to be several equations relating the accelerations of each block to each other.  We can avoid this by using a 'fractured' co-ordinate system.  For example, if mass one rises one metre, mass two must slide down the incline by one metre (assuming the string can't stretch).  By using the system shown below,

we can minimize the tedium of relating all the necessary quantities and use simply Dx, v, and a to describe the motions of the masses along their respective x axes, while asserting that there is no motion in the y directions  (that is, Dx₁ = Dx₂ = Dx, v_1x = v_2x = v, a_1x = a_2x = a, v_1y = v_2y = 0, a_1y = a_2y = 0).

Let's do free body diagrams:

Note that the angle marked q in this diagram is the same as the original angle of inclination.

So, write Newton's second law for each mass:

x: T - gm₁ = m₁a_x = m₁a
y: no forces

x: -T + gm₂ sinq  = m₂a_x = m₂a
y: - gm₂ cosq + N = m₂a_y = 0

The y equation turns out to be of no interest for this problem.
The other two equations can be subtracted to eliminate the tension T:

T	-	gm1g	=	m1a
-T	+	gm2sinq	=	m2a
----------	-	----------	-	-------------
gm2sinq	-	gm1	=	(m1 + m2)a

a = [(m₂sinq - m₁)/(m₂+m₁)]g
Now, to find T, we substitute this answer back into one of the original equations.  It is good form to find T such that a does not appear explicitly:
T = gm₁ + m₁a = gm₁ + m₁[(m₂sinq - m₁)/(m₂+m₁)]g = gm₁[1 + [(m₂sinq - m₁)/(m₂+m₁)]
T = [m₁m₂g/(m₂+m₁)] [1 + sinq]

Now, you can consider more difficult problems by using these examples as models; while the mathematical manipulations may become more difficult, the approach is the same.

Example:
Consider the Atwood's Machine, comprising two masses connected by a massless string:

Find the tension in the string and the acceleration of the masses.  Give yourself no more than one minute to obtain the correct answer.

---

Friction

We started by considering an object at rest on the desk; clearly the sum of the forces acting on this object (weight and normal force from the desk) is zero, since there is no acceleration.  Then we applied a small force horizontally to the object, but we were mildly surprised that it did not accelerate; if Newton's second law is to remain correct, there must be yet another force acting oppositely to our applied force which causes the total horizontal force to be zero (second law, F_appl - F_mystery = ma = 0).

What's more, the magnitude of that force changes as we change our applied force; it's always just big enough to cancel our force.  That is, if we apply 2 newtons, it applies 2 newtons, if we apply 5 newtons, it applies 5 newtons.  We'll call this force friction.  A graph of this situation might look like this:

Furthermore, we saw that, if we continue to increase our applied force, there comes a point at which this frictional force reaches a maximum value; we know this because we can apply enough force to make the object move, and that requires a net force.  How big is this maximum frictional force and what quantites determine its value?

We did a demo which should have convinced you that the magnitude of the maximum frictional force is related to the nature of the surfaces that are pressed against one another, and also proportional to the normal force with which the surfaces are being forced together:

We measured the maximum value of the frictional force by attempting to move one metal cylinder (sitting on a mouse pad 'sled') and noticing what force was necessary to get the cylinder to start just moving.  We repeated the experiment with two cylinders and found that about double the force was required.  Does the maximum frictional force then depend on the mass of the object?  Not directly: we repeated the experiment with my hand pushing on the mousepad and found that the force necessary to budge the pad depended more directly on how hard I pushed down on the pad.  In other words, the maximum frictional force really depends on the normal force acting between the two surfaces.  By doubling the mass of the cylinders, we double the weight, double the normal force (second law), and thereby double the maximum possible frictional force (If this is not entirely convincing, we could repeat the experiment using a magnet on the mousepad, and vary the normal force by putting various strength magnets under the table to attract the riding magnet.).  So, we may tentatively write that
F_fMAX ~ N.
Second, we flipped the mousepad over so that the shiny side was in contact with the table and repeated the experiment.  We found again twice as much force was required to budge two cylinders on teh mousepad as to budge one, thus verifying the result above, but also suggesting that the maximum possible frictional force depends on the nature of the two surfaces in contact.  This effect is quantified with a number called the co-efficient of friction, m, such that
F_fMAX = mN.
This co-efficient is usually determined experimentally, and has different values for different combinations of surfaces; there should be a table of common combinations in your textbook.

To review so far:
There is a force of contact called friction which acts along the interface of two objects (as opposed to perpendicular to the interface, as for the normal force of contact).
This force is only as big as it needs to be to prevent the surfaces from sliding against one another, but only up to a maximum value that depends on the natures of the two surfaces and on how hard they are being pushed together:
F_f mN.

Now, what happens to the frictional force once the object starts to move (or more properly, once the surfaces start to slide against one another)?  It doesn't go to zero, since we know that friction will bring a book sliding across the table to rest.  We repeated the experiment outlined above, making our measurements after the cylinders were already moving.  First, we made an argument that, if we drag the mousepad and metal cylinders at constant speed across the table, the force applied by the string is equal in magnitude to the frictional force (second law, F_appl - F_f = ma = 0).  We found that putting another cylinder on the pad doubled the frictional force.  Is this because the mass is double?  No, since in the absence of friction, no applied force would be needed to keep the object moving, regardless of the mass on the pad.  It is instead because the (vertical) normal force N has increased to compensate the increased weight.  We tested this again by pushing down on the pad and sliding it across the table.  We also investigated the effect on the nature of the surfaces by turning the pad over so that its rough side was in contact with the table.  In the end, we developed this relationship, similar to that above:
F_f = mN.
We will find that, unlike the previous case, the frictional force due to the sliding surfaces is (approximately) constant; we test this by sliding the apparatus across the desk at various constant speeds.  We also find that this co-efficient is generally less than the one above; think of how much easier it is to keep a heavy object moving across the floor than to get it to move in the first place.  So we'll need to be able to distinguish them:  the symbol m_K represents the co-efficient of kinetic friction (surfaces are sliding), while m_S represents the co-efficient of static friction (surfaces are not sliding).
F_Sf m_SN.
F_Kf = m_KN.
Generally, m_S  > m_K.
Because of the inequality in the relationship for static friction (the static frictional force is only as big as it needs to be), we usually restrict ourselves to situations where the surfaces are 'about to slide,' or some similar condition so that we know that we are at the critical point when the equality holds true.

Here is a not very good figure which indicates approximately this behaviour.

Mastery Question

Mastery Question

More Example Problems

How did this happen?  Take a look at N and F_f:
N = - F sinq + W = 897 N.      F_f = m_KN = m_K [- F sinq + W] = (0.26)(-300*sin20^o + 1000) = 233 N.
Before, the normal acted to 'balance' the weight and the downward component of the applied force, but here the normal is 'assisted' by the now upward component of the applied force.  Lessening N lessens the frictional force.

Just for fun, let's take the second scenario and change it such that the box is moving to the left.  The second law analysis is the same as above, except for the direction of the frictional force (watch the red addition sign):
x:  F cosq +F_f = ma_x
y:  F sinq - W + N = ma_y = 0
F_f = m_KN
Re-arranging and substituting results in
a_x = [F cosq + F_f ]/m = [F cosq + m_KN]/m = [F cosq +m_K[ - F sinq + W]]/[W/g] = [300 cos20^o+ 0.26[ - 300 sin20^o + 1000]]/[1000/9.8] =  11.56 m/s².
We see that we could have obtained this result from the results of the last problem simply by reversing the sign of m_K.  This is a useful trick.

Friction problems can be very difficult to set up, depending on what information is given.  Let's look at a fairly standard problem and see what kinds of questions can be asked and how we might deal with them.  Consider once again two blocks connected by a light string over a wheel, with one block on an incline.

What kinds of questions could be asked?  One might wonder

• what is the largest value of m₁ (or the smallest value of m₂) for which m₁ will not slide down the plane?
• what is the smallest value of m₁ (or the largest value of m₂) for which m₁ will not slide up the plane?
• what is the smallest value of m_S  for which m₁ will not slide down (or up) the plane?
• what is the smallest (or largest) value for q for which m₁ will not slide up (or down) the plane?
• what is the direction and magnitude of the acceleration, assuming that the blocks are in motion?
• about many things not listed here.

Write Newton's second law, assuming that the blocks are moving, or about to move, with m₁ sliding down the incline.  Keep in mind, we may well be wrong about this.  Even worse, the blocks may be moving one way but accelerating the other!

For m₁,
x:   - T + gm₁sinq - F_f = m₁a_x
y:   N₁ - gm₁cosq = 0
F_f = mN₁

For m₂,
x:   T - gm₂ = m₂a_x
y: No forces

Note that we have left the co-efficient of friction unlabelled, since the relationships are true for both types, so long as in the static case, the blocks are 'just about to move.'
Now, the only force here for which the direction is not certain is the friction.  Here though is a trick: if it turns out that we were wrong about the direction of F_f, we can just replace the value of m with its negative; this is a math trick, not a physics trick.
Let's address the specific questions listed above in order.

If they aren't sliding just yet, a_x = 0.
 - T + gm₁sinq - F_f = 0
N₁ -gm₁cosq = 0
F_f = mN₁
T - gm₂ = 0
Through substitution and canceling out the g common to the remaining terms, we get
m₁sinq - mm₁cosq - m₂ = 0.
Now, let's try to answer some of the questions posed above.

• How big can m₁ be without sliding down the incline?  Alternately, how small can m₂ be without m₁ sliding down the incline?

• How small can m₁ be without sliding up the incline?   How big can m₂ be without m₁ sliding up the incline?  These equations weren't written for sliding up the plane; we need to reverse the direction of the frictional force, so replace m_S with (-m_S).

• What is the smallest value of m_S  for which m₁ will not slide down (or up) the plane?

• What is the largest (or smallest) value for q for which m₁ will not slide down (or up) the plane?

If the blocks are already in motion,

• What is the direction and magnitude of the acceleration, assuming that the blocks are in motion?

Here's another:
Consider a car traveling down a hill inclined at angle q,  The co-efficient of static friction between tires and road is is m_S.  What is the greatest deceleration the car can enjoy without skidding?
Let down the slope be the positive x-axis, so we're looking for a negative acceleration, here.
Write Newton's second law in each direction:
y:  N - gmcosq = ma_y = 0
x:  -F_f + gmsinq = ma_x = ma
F_f = m_SN (in the critical case)
With straightforward substiution, we see that
a = [sinq - m_Scosq]g.
First, we see that the mass of the car doesn't matter; the answer is the same for minis and for trucks.
Second, we see that, since we want a negative acceleration for the car to slow, if the hill's incline is steeper than a critical value, tanq = m_S, the car will accelerate down the hill regardless of whether the brakes are on or not.  Conversely, if the co-efficient of static friction is less than tanq, the same thing will happen.  What if the car did start to skid?  Since the co-efficient of kinetic friction is usually less than that of static friction, this would just make the situation worse.

Another example:
Three blocks (M₁ = 10 kg, M₂ = 5 kg, and M₃ = 3 kg) are connected by massless strings over massless, frictionless wheels as shown.  The acceleration of the 5 kg block is 2 m/s² to the left.  The two surfaces have the same co-efficient of kinetic friction, m_K.  Find the tensions (T₁ and T₂) in the strings and the co-efficient of kinetic friction.

There is something wrong with the way this problem is stated.  See if you can figure it out.

M₂:
x: T₂ + F_{f 2} - T₁ = M₂a
y: N₂ - gM₂ = 0
F_{f 2} = m_KN₂

M₃:
x: F_{f 3} - T₂ + gM₃sinq = M₃a
y: N₃ - gM₃cosq = 0
F_{f 3} = m_KN₃

For M₃, we'll solve for the Normal and substitute it in for Friction:
N₃ = gM₃cosq
F_{f 3} = m_KN₃ = m_KgM₃cosq
m_KgM₃cosq  - T₂ + gM₃sinq = M₃a

Let's do the same for M₂:
N₂ = gM₂
F_{f 2} = m_KN₂ = m_KgM₂
T₂ + m_KgM₂ - T₁ = M₂a

Now, let's add these two resulting equations to the one we have from M₁:
T₁ - gM₁ = M₁a
T₂ + m_KgM₂ - T₁ = M₂a
m_KgM₃cosq  - T₂ + gM₃sinq = M₃a
 

T1 - gM1	=	M1a
T2 + mKgM2 - T1	=	M2a
mKgM3cosq  - T2 + gM3sinq	=	M3a
-------------------------------------------------		-------------
mKgM3cosq  + gM3sinq + mKgM2 - gM1	=	[M1+M2+M3]a

This eliminated the tensions from consideration.  Re-arrange to get
m_K[M₃cosq  + M₂] + M₃sinq  - M₁ = [M₁+M₂+M₃](a/g)
m_K[M₃cosq  + M₂]  = [M₁+M₂+M₃]a/g - M₃sinq + M₁
m_K  = [[M₁+ M₂+ M₃](a/g) - M₃sinq + M₁]/[M₃cosq  + M₂]

m_K  = [[10 + 5 + 3](-2/9.8) - 3sin25^o+ 10]/[3cos25^o  + 5] = 0.66  (Note that, according to my co-ordinate system, the acceleration is -2m/s².)

If I'd gotten a negative number for m_K, then I'd know that I'd guessed the direction of motion incorrectly.

T₁ = M₁ (a + g) = 10*(-2+9.8) = 78 N

T₂ = - m_KM₂g + T₁ + M₂a = -0.66*5*9.8 + 78 + 5*(-2) = 35.7 N

Now, this really is a strange problem if we look more deeply than the authors intended.  Why didn't we get two answers, one corresponding to the blocks moving and accelerating to the left, the other corresponding to them moving to the right and slowing down, as we normally would?  The answer is in the specific numbers given in the problem; in the latter case, no friction at all gives the blocks an acceleration of -4.85m/s², greater than the given -2m/s², that is, there is no solution for the numbers given if the blocks are sliding to the right..