FHSST Physics/Forces/Newton's Laws of Motion

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Newton's Laws of MotionEdit

Our current laws of motion were discovered by Sir Isaac Newton. It is said that Sir Isaac Newton started to think about the nature of motion and gravitation after being struck on the head by a falling apple.

Newton discovered 3 laws describing motion:

First LawEdit

Newton's first law basically says that a force has to be applied to an object to make it move or to make it stop. The first part of that statement definitely makes sense. The only way I can make something move is to have something give it a push. The second part of that statement might not be quite as easy to just take as fact. We've all witnessed objects slow down when nobody is pushing them. How then can we say that the only way to stop an object's motion is with a force? The answer is that there are forces that we don't always see. Most of the time, the force that we don't see is the force of friction.

Friction is the force that resists motion when two things are sliding past one another. To understand what friction is, think about sandpaper. If you try to rub 2 pieces of sandpaper together, it will be hard to get them to slide. This same phenomenon happens between all objects to some degree. This frictional force is what slows objects down or stops their motion.

Second LawEdit

Definition: The time rate of change in momentum is proportional to the applied force and takes place in the direction of the force.

The law is represented in the following basic form (the system of measurement is chosen such that constant of proportionality is 1) :

\mathbf{F} = {d(m \mathbf{v}) \over dt}

The product of mass and velocity i.e. mv is called the momentum. The net force on a particle is, thus, equal to rate change of momentum of the particle with time. Generally mass of the object under consideration is constant and thus can be taken out of the derivative :

\mathbf{F} = m {d(\mathbf{v}) \over dt}
\mathbf{F} = m \mathbf{a}

For constant mass,


\mathbf{F} = m\mathbf{a}

Force is equal to mass times acceleration. This version of Newton's Second Law of Motion assumes that the mass of the body does not change with time, and as such, does not represent a general mathematical form of the Law. Consequently, this equation cannot, for example, be applied to the motion of a rocket, which loses its mass (the lost mass is ejected at the rear of the rocket) with the passage of time.

It makes sense that the direction of the acceleration is in the direction of the resultant force. If you push something away from you it doesn't move toward you unless of course there is another force acting on the object towards you!


Worked Example 16 Newton's Second LawEdit

Question: A block of mass 10 kg is accelerating at 2 m·s−2. What is the magnitude of the net force acting on the block?

Answer:

Step 1 :

We are given

  • the block's mass
  • the block's acceleration

all in the correct units.

Step 2 :

We are asked to find the magnitude of the force applied to the block. Newton's Second Law tells us the relationship between acceleration and force for an object. Since we are only asked for the magnitude we do not need to worry about the directions of the vectors:

\begin{matrix}F_{Net} &=& ma \\&=& 10\ \mbox{kg} \times 2 \mbox{ m} \cdot \mbox{s}^{-2} \\&=& 20\ \mbox{N}\end{matrix}

Thus, there must be a net force of 20 N acting on the box.


Worked Example 17 Newton's Second Law 2Edit

Question: A 12 N force is applied in the positive x-direction to a block of mass 100 mg resting on a frictionless flat surface. What is the resulting acceleration of the block?

Answer:

Step 1 :

We are given

  • the block's mass
  • the applied force

but the mass is not in the correct units.

Step 2 :

Let us begin by converting the mass:

\begin{matrix}100 \mbox{ mg} &=& 100 \times 10^{-3} \mbox{ g} = 0.1 \mbox{ g} \\1000 \mbox{ g} &=& 1 \ \mbox{kg} \\1 &=& 1kg\times \frac{1}{1000g} \\&=& \frac{1kg}{1000g} \\0.1g &=& 0.1g \times 1 \\&=& 0.1g \times \frac{1kg}{1000g} \\&=& 0.0001\ kg \\\end{matrix}

Step 3 :

We know that net force results in acceleration. Since there is no friction the applied force is the resultant or net force on the block (refer to the earlier example of the block pushed on the surface of the table). The block will then accelerate in the direction of this force according to Newton's Second Law.

Step 4 :

To determine the magnitude of the acceleration:

\begin{matrix}F_{Res} &=& ma \\12 \mbox{ N} &=& (0.0001 \mbox{ kg})a\\a &=& \frac{12 \mbox{ N}}{0.0001 \mbox{kg}}\\&=& 120000\frac{\mbox{N}}{\mbox{kg}} \\&=& 120000\frac{\mbox{kg} \cdot \mbox{m}}{\mbox{s}^2 \cdot \mbox{kg}} \\&=& 120000\frac{\mbox{m}}{\mbox{s}^2} \\&=& 1.2\times 10^{5}\ \mbox{ m} \cdot \mbox{s}^{-2}\end{matrix}

From Newton's Second Law the direction of the acceleration is the same as that of the resultant force. The final result is then that the block accelerates at 1.2\times10^{5}\ \mbox{ m} \cdot \mbox{s}^{-2} in the positive x-direction.


Weight and MassEdit

You must have heard people saying ``My weight is 60 kg. This is actually correct, though physicists often have difficulty understanding it because it is mass that is measured in kilograms. This is a different meaning of the word weight from the one used in mechanics, where weight is the force of gravity exerted by the earth on an object with mass:

\begin{matrix}F_{weight} = mg\end{matrix}

As such, weight is measured in newtons.

If you compare this equation to Newton's Second Law you will see that it looks exactly the same with the a replaced by g. Thus, when weight is the only force acting on an object (i.e. when F_{weight}

Fweight is the resultant force acting on the object) the object has an acceleration g. Such an object is said to be in free fall. The value for g is the same for all objects (i.e. it is independent of the objects mass):

\begin{matrix}g=9.8 \mbox{m} \cdot \mbox{s}^{-2} \approx 10 \mbox{m} \cdot \mbox{s}^{-2}\end{matrix}

\begin{matrix}g=9.8 \mbox{m} \cdot \mbox{s}^{-2} \approx 10 \mbox{m} \cdot \mbox{s}^{-2}\end{matrix}

(4.3)

You will learn how to calculate this value from the mass and radius of the earth in Chapter ???. Actually the value of g varies slightly from place to place on the Earth's surface.

The reason that we often get confused between weight and mass, is that scales measure your weight (in newtons) and then display your mass using the equation above.


Worked Example 18 Calculating the resultant and then the accelerationEdit

Question: A block (mass 20 kg) on a frictionless flat surface has a 45 N force applied to it in the positive x-direction. In addition a 25 N force is applied in the negative x-direction. What is the resultant force acting on the block and the acceleration of the block?

Answer:

Step 1 :

We are given

  • the block's mass
  • Force F1 = 45 N in the positive x-direction
  • Force F2 = 25 N in the negative x-direction

all in the correct units.

Step 2 :

We are asked to determine what happens to the block. We know that net force results in an acceleration. We need to determine the net force acting on the block.

Step 3 :

Since F1 and F2 act along the same straight line, we can apply the algebraic technique of vector addition discussed in the Vectors chapter to determine the resultant force. Choosing the positive x-direction as our positive direction:

Positive x-direction is the positive direction:

\begin{matrix}F_{Res}&=& (+45 \ \mbox{N}) + (-25 \mbox{ N}) \\& = & +20 \ \mbox{N} \\&=& 20 \mbox{N}\ in\ the\ positive\ x-direction\end{matrix}

where we remembered in the last step to include the direction of the resultant force in words. By Newton's Second Law the block will accelerate in the direction of this resultant force.

Step 4 :

Next we determine the magnitude of the acceleration:

\begin{matrix}F_{Res} & =& ma \\20 \mbox{N} &=& (20 \mbox{kg})a\\a &=& \frac{20 \mbox{N}}{20 \mbox{kg}} \\&=& 1\frac{ \mbox{N}}{ \mbox{kg}} \\&=& 1\frac{ \mbox{kg} \cdot  \mbox{m}}{ \mbox{s}^2 \cdot  \mbox{kg}} \\&=& 1\ \mbox{ m} \cdot \mbox{s}^{-2} \\\end{matrix}

The final result is then that the block accelerates at 1\ \mbox{ m} \cdot \mbox{s}^{-2} in the positive x-direction (the same direction as the resultant force).


Worked Example 19 Block on inclineEdit

Question: A block (mass 10 kg) is released on an inclined plane. What is the resulting rate of acceleration of the block if the angle theta is 25 degrees and the coefficient of friction between the block and the plane is 0.25?

Fhsst forces11.png

Solution:

Step 1: The acceleration of the block will be parallel to the plane, so we break the problem into two parts - forces perpendicular to the plane (which must cancel out since there is no acceleration perpendicular to the plane) and parallel to the plane.

Step 2: We know that there is no acceleration in the direction perpendicular to the surface of the inclined plane, or in other words a_{\perp}=0. Therefore we know that:

\begin{matrix} \\ \sum{F_{\perp}} &=& 0 
\\  N - W_{\perp} &=& 0
\\  N - m \   g \  {\cos \theta} &=& 0
\\ N &=& m \  g \  {\cos \theta} 
\\ &=& (10 \  kg)(9.8 \  {m \over {s^2}}) \  {\cos 25^\circ} 
\\ &=& 88.82 \  N \end{matrix}

Step 3: Summing forces parallel to the plane yields:

\begin{matrix} \\ \sum{F_{\|}} &=& m \  g \  {\sin \theta} - \mu \  N 
\\  m a_{\|} &=& m \  g \  {\sin \theta} - \mu \  N 
\\  (10 kg) a_{\|} &=& (10 kg) (9.8 \  {m \over s^2}) \sin 25^\circ - (0.25) \  (88.82 \  N) 
\\ a_{\|} &=& (9.8 \  {m \over s^2}) \  \sin 25^\circ - (0.25) \  {{88.82 \ N} \over {10 \ kg}}
\\ &=& 5.5 \ {m \over {s^2}} \end{matrix}

The final result is that the block accelerates down the inclined plane with an acceleration of 5.5 \ m \cdot s^{-2}.

Third LawEdit

Definition: For every force one body applies to another (action) there is always an equal but opposite in direction force another body applies back to the first one (reaction.)

This law is a direct consequence of the Principle of Conservation of Linear Momentum.

Newton's Third Law is easy to understand but it can get quite difficult to apply it. An important thing to realise is that the action and reaction forces can never act on the same object and hence cannot contribute to the same resultant.


Worked Example 20 Identifying action-reaction PairsEdit

Question: Consider pushing a box on the surface of a rough table.

  1. Draw a force diagram indicating all of the forces acting on the box.
  2. Identify the reaction force for each of the forces acting on the box.

Answer:

  1. The following force diagram shows all of the forces acting on the box

Fhsst forces12.png


There is an important thing to realise which is related to Newton's Third Law. Think about dropping a stone off a cliff. It falls because the earth exerts a force on it (see Chapter ???) and it doesn't seem like there are any other forces acting. So is Newton's Third Law wrong? No, the reactionary force to the weight of the stone is the force exerted by the stone on the earth. This is illustrated in detail in the next worked example.


Worked Example 21 Newton's Third LawEdit

Question: A stone of mass 0.5 kg is accelerating at 10 m·s−2 towards the earth.

  1. What is the force exerted by the earth on the stone?
  2. What is the force exerted by the stone on the earth?
  3. What is the acceleration of the earth, given that its mass is 5.97 × 1027 kg?

Answer:

  1. Step 1 : We are given
    • the stone's mass
    • the stone's acceleration (g)
  1. By Newton's Third Law the stone must exert an equal but opposite force on the earth. Hence the stone exerts a force of 5 N towards the stone on the earth.
  2. We have
    • the force acting on the earth
    • the Earth's mass

File:Fhsst forces13.png

Fhsst magnify.png Newton first published these laws in Philosophiae Naturalis Principia Mathematica (1687) and used them to prove many results concerning the motion of physical objects. Only in 1916 were

Newton's Laws superseded by Einstein's theory of relativity.

File:Fhsst forces15.png

The next two worked examples are quite long and involved but it is very important that you understand the discussion as they illustrate the importance of Newton's Laws.


Worked Example 22 RocketsEdit

Question: How do rockets accelerate in space?

Answer:

  • Gas explodes inside the rocket.
  • This exploding gas exerts a force on each side of the rocket (as shown in the picture below of the explosion chamber inside the rocket).

File:Fhsst forces17.png

Note that the forces shown in this picture are representative. With an explosion there will be forces in all directions.

  • Due to the symmetry of the situation, all the forces exerted on the rocket are balanced by forces on the opposite side, except for the force opposite the open side. This force on the upper surface is unbalanced.
  • This is therefore the resultant force acting on the rocket and it makes the rocket accelerate forwards.

Systems and External ForcesEdit

The concepts of a system and an external forces are very important in physics. A system is any collection of objects. If one draws an imaginary box around such a system then an external force is one that is applied by an object or person outside the box. Imagine for example a car pulling a trailer.

Fhsst forces16.png

Concept of system is extremely important. Consider an example of a person holding a load (box) on his head and standing on a platform.

Now count the forces :

(i) Box presses the person down (ii) The person pushes the box up (reaction to box's weight) (iii) Box is pulled down by the earth (iv) Earth is pulled up by the box (v) Person presses the floor (vi) Floor pushes the person up (vii) Person is pulled down by earth and (viii) Earth is pulled up by the person.

Quite a mess. Which ones are internal forces and which ones are external forces? Identification of system comes handy here. We can select the box or the person as a system for analyzing forces or we may even consider box and person together as a single system. Only condition to watch while selecting a system is that all parts of the system should have same acceleration. If two parts are at different accelerations then each of them should be treated as a separate system.

Last modified on 14 July 2012, at 19:13